A model rocket has a mass of 0.2 kg, with a motor that can provide a force of 200 N. A second model rocket is being built with the same motor, but it is being designed to accelerate one-fourth as much as the first rocket.What kind of change can be made in the design to achieve this objective?
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Answers
We can achieve this design by adding weight to the second rocket, such as by adding a heavier nose cone, fins, or other components.
What kind of change can be made in the design?The first model rocket has a mass of 0.2 kg and is powered by a motor that can provide a force of 200 N.
Using Newton's second law of motion, we can calculate the acceleration of the rocket:
F = ma
where;
F is the force applied to the rocket, m is the mass of the rocket, and a is the acceleration of the rocket.Rearranging the equation, we can solve for the acceleration:
a = F/m
Plugging in the values, we get:
a = 200 N / 0.2 kg
a = 1000 m/s^2
So the first model rocket accelerates at 1000 m/s².
The second model rocket is being designed to accelerate one-fourth as much as the first rocket. This means that its acceleration will be:
a' = 1/4 * a
a' = 1/4 * 1000 m/s²
a' = 250 m/s²
So we need to design the second rocket so that it accelerates at 250 m/s².
We can use the same motor, so the force applied to the rocket will still be 200 N. Therefore, we need to change the mass of the rocket to achieve the desired acceleration.
Using the equation for acceleration, we can solve for the mass needed to achieve an acceleration of 250 m/s²:
a' = F/m'
250 m/s² = 200 N / m'
m' = 200 N / 250 m/s^2
m' = 0.8 kg
So the mass of the second rocket needs to be 0.8 kg in order to achieve an acceleration of 250 m/s² with the same motor.
Thus, we can achieve this by adding weight to the second rocket, such as by adding a heavier nose cone, fins, or other components. Alternatively, we could also increase the size of the second rocket while keeping the same mass, which would decrease its acceleration but also increase its overall speed and altitude.
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Related Questions
A toy car with of mass m1 moves in a circular path of radius r on a flat, level table at constant linear speed v. The car is attached to a string that passes through a hole in the middle of the table. A brass ball of mass m2 is attached to the other end of the string and hangs vertically under the table. All frictional forces are considered to be negligible.Which equation correctly shows how the given variables are related? a. m2 = m1*v / r b. m1*g = m2*v*r^2c. m2*g = m1*v^2 / rd. m1*g = m2*v^2 / re. m2*g = m1*v*r^2f. m1 = m2*v / r
Answers
Given:
The mass of the car is,
\(m_1\)The mass of the brass ball is,
\(m_2\)The radius of the circular path is,
\(r\)and the constant linear speed of the car is,
\(v\)To find:
The correct equation of the variables
Explanation:
As the friction is negligible, the tension at the horizontal string is,
\(T=\frac{m_1v^2}{r}\)The tension in the vertical string is,
\(T=m_2g\)As there is no friction,
\(m_2g=\frac{m_1v^2}{r}\)Hence, the correct option is (c).
Answer:
C
Explanation:
Comment
What an interesting question! It is a good one to get to get to know how the answer was derived.
Equation
M2 g is the gravitational force . It's the one that is keeping the toy car going in a circle.
The circular force is F = m1*v^2/r. This force pulls on the string to hold m1 moving a circle.
The key to the question is that both forces have to be equal.
m2*g = m1*v^2/r
Solution
The only answer that works and looks like what I have written is C
A negative test charge experiences a force to the right as a result of an electric field. Which is the best conclusion to draw based on this description?
Answers
The most logical conclusion to make in light of the facts available is that the electric field is directed to the left. This conclusion may be drawn from the fact that a negative test charge, which is attracted to positive charges and repels other negative test charges, feels a force to the right.
The force applied to the negative test charge suggests the presence of positive charges in the electric field, indicating that the field lines begin with positive charges and end on negative charges since opposing charges attract.
The assumption that electric field lines originate from positive charges and end on negative charges is supported by this deduction. As a result, in this situation, the electric field is directed leftward.
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use equivalent ratios in higher terms to eliminate decimals from the following ratio 3.04 to 6 and ratio is ?
Answers
The equivalent ratio in higher terms to eliminate decimals from the ratio 3.04 to 6 is 76 to 150.
To eliminate decimals from the ratio 3.04 to 6, we can multiply both terms of the ratio by a common factor that will result in whole numbers.
First, let's convert 3.04 to a fraction:
3.04 = 3 + 0.04 = 3 + 4/100 = 3 + 1/25 = 75/25 + 4/100 = 76/25
Now, the ratio 3.04 to 6 can be written as:
3.04/6 = 76/25 / 6
To eliminate the decimal, we can multiply both the numerator and denominator by 25:
(76/25) * 25 / (6 * 25) = 76 / 150
Therefore, the equivalent ratio in higher terms to eliminate decimals from the ratio 3.04 to 6 is 76 to 150.
By multiplying both terms by 25, we effectively scale up the ratio to eliminate the decimal and create whole numbers. This allows us to express the ratio in higher terms without decimals. The final ratio 76 to 150 represents the same relationship as the original ratio 3.04 to 6, but in whole number form.
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is power and voltage the same thing yes or no
Answers
A metallic ring of mass 1 kg has moment of inertia 1 kg m² when rotating about one of its diameters. It is molten and
remoulded into a thin uniform disc of the same radius. How much will its moment of inertia be, when rotated about its own
axis.
Answers
Answer:
The moment of inertia of disc about own axis is 1 kg-m².
Explanation:
Given that,
Mass of ring m= 1 kg
Moment of inertia of ring at diameter \((I_{r})_{d}=1\ kg\ m^{2}\)
The radius of metallic ring and uniform disc both are equal.
So, \(R_{r}=R_{d}\)
We need to calculate the value of radius of ring and disc
Using theorem of perpendicular axes
\((I_{r})_{c}=2\times (I_{r})_{d}\)
Put the value into the formula
\((I_{r})_{c}=2\times1\)
\((I_{r})_{c}=2\ kg\ m^2\)
Put the value of moment of inertia
\(MR_{r}^2=2\)
\(R_{r}^2=\dfrac{2}{M}\)
Put the value of M
\(R_{r}^2=\dfrac{2}{1}\)
So, \(R_{r}^2=R_{d}^2=2\ m\)
We need to calculate the moment of inertia of disc about own axis
Using formula of moment of inertia
\(I_{d}=\dfrac{1}{2}MR_{d}^2\)
Put the value into the formula
\(I_{d}=\dfrac{1}{2}\times1\times2\)
\(I_{d}=1\ kg\ m^2\)
Hence, The moment of inertia of disc about own axis is 1 kg-m².
Need help please physics question
Answers
2500 J
cause at that time pe (potential energy) becomes 0 .
And we all know energy can neither be created nor be destroyed it's just gets converted. :)
find the rms speed of a sample of oxygen at 30° C and having a molar mass of 16 g/mol.
Answers
At 30°C, the rms speed of a sample of oxygen with a molar mass of 16 g/mol is approximately 482.34 m/s.
The root mean square (rms) speed of a gas molecule is a measure of the average speed of the gas particles in a sample. It can be calculated using the formula:
vrms = √(3kT/m)
Where:
vrms is the rms speed
k is the Boltzmann constant (1.38 x 10^-23 J/K)
T is the temperature in Kelvin
m is the molar mass of the gas in kilograms
To calculate the rms speed of oxygen at 30°C (303 Kelvin) with a molar mass of 16 g/mol, we need to convert the molar mass to kilograms by dividing it by 1000:
m = 16 g/mol = 0.016 kg/mol
Substituting the values into the formula, we have:
vrms = √((3 * 1.38 x 10^-23 J/K * 303 K) / (0.016 kg/mol))
Calculating this expression yields the rms speed of the oxygen sample:
vrms ≈ 482.34 m/s
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A cell of inter resistance of 0.5 ohm is connected to coil of resistance 4 ohm and 8 ohm joined in parallel.If there is current of 2A in 8 ohm,find the emf of the cell.
Answers
A cell of inter resistance of 0.5 ohm is connected to coil of resistance 4 ohm and 8 ohm joined in parallel.If there is current of 2A in 8 ohm, the electromotive force (emf) of the cell is approximately 14.5 volts.
To find the emf of the cell, we can apply Ohm's Law and Kirchhoff's laws to analyze the circuit.
Given:
Resistance of the coil, R1 = 4 ohm
Resistance of the other resistor, R2 = 8 ohm
Current passing through the 8-ohm resistor, I = 2A
First, let's analyze the parallel combination of the 4-ohm and 8-ohm resistors.
The total resistance of two resistors in parallel can be calculated using the formula:
1/Rp = 1/R1 + 1/R2
Substituting the given values, we have:
1/Rp = 1/4 + 1/8
1/Rp = 2/8 + 1/8
1/Rp = 3/8
Rp = 8/3 ohm
Now, let's consider the total resistance in the circuit, which includes the internal resistance of the cell (0.5 ohm) and the parallel combination of the resistors (8/3 ohm).
R_total = R_internal + Rp
R_total = 0.5 + 8/3
R_total = 1.833 ohm
Now, we can find the emf of the cell using Ohm's Law:
emf = I * R_total
emf = 2 * 1.833
emf ≈ 3.667 volts
Therefore, the emf of the cell is approximately 3.667 volts.
However, it is worth noting that the given current of 2A passing through the 8-ohm resistor does not affect the emf calculation since the emf of the cell is independent of the current in the circuit.
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A soldier is tasked with measuring the muzzle velocity of a new rifle. Knowing the principles of
projectile motion, he decides to perform a simple experiment at the indoor firing range. The soldier
hangs a target a distance of d = 109 m from the end of the barrel. The rifle is mounted so that the bullet?
exits moving horizontally at the same height as the bullseye. After six trials, the soldier tabulates the
values he measured for the vertical distance h from the bullseye to the bullet strike.
d
h
What is the most accurate muzzle velocity that the soldier can report to his sergeant?
Answers
Answer:
it could be that there is a big percentage
Explanation:
CAN SOMEONE DO THIS FOR ME!!!!!!!!!! PLEASEEEEEEEEEE
Answers
Describe the current at points X, Y and Z on the circuit.
Answers
At point X, the current is the same as the current entering the circuit, which is determined by the voltage of the power source and the resistance of the components in the circuit.
Describe the current at points X, Y and Z on the circuit?At point Y, the current is the same as the current leaving the circuit, which is determined by the voltage of the power source and the resistance of the components in the circuit. At point Z, the current is the sum of the currents at points X and Y, since the current at point Z is the sum of all the currents at the points entering and leaving the circuit.At point X, the current is the current that flows into the circuit at that point. This current will be determined by the total resistance in the circuit. If the total resistance of the circuit is low, then the current at point X will be higher. On the other hand, if the total resistance of the circuit is high, then the current at point X will be lower.At point Y, the current is the current that flows through the circuit between points X and Y. This current will be determined by the resistance of the components between X and Y. If the resistances of the components between X and Y are high, then the current at point Y will be lower. On the other hand, if the resistances of the components between X and Y are low, then the current at point Y will be higher.At point Z, the current is the current that flows out of the circuit at that point. This current will be determined by the total resistance in the circuit. If the total resistance of the circuit is low, then the current at point Z will be lower. On the other hand, if the total resistance of the circuit is high, then the current at point Z will be higher.To learn more about Electric Circuits refer to:
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What is the momentum of a 9kg object that has a velocity of 3 m/s
Answers
Hope it helps:)
Some dragonflies splash down onto the surface of a lake to clean themselves. After this dunking, the dragonflies gain altitude, and then spin rapidly at about 1100 rpm to spray the water off their bodies. When the dragonflies do this "spin-dry," they tuck themselves into a "ball" with a moment of inertia of 2.0×10−7kg⋅m2 . How much energy must the dragonfly generate to spin itself at this rate?
Answers
The dragonfly must generate approximately 4.8 × 10^-4 Joules of energy to spin itself at a rate of 1100 rpm.
Start by converting the rotational speed from rpm (revolutions per minute) to rad/s (radians per second). Since 1 revolution is equal to 2π radians, we can use the conversion factor:
Angular speed (ω) = (1100 rpm) × (2π rad/1 min) × (1 min/60 s)
ω ≈ 115.28 rad/s
The moment of inertia (I) is given as 2.0 × 10^-7 kg⋅m².
Use the formula for rotational kinetic energy:
Rotational Kinetic Energy (KE_rot) = (1/2) I ω²
Substituting the given values:
KE_rot = (1/2) × (2.0 × 10^-7 kg⋅m²) × (115.28 rad/s)²
Calculate the value inside the parentheses:
KE_rot ≈ (1/2) × (2.0 × 10^-7 kg⋅m²) × (13274.28 rad²/s²)
KE_rot ≈ 1.331 × 10^-3 J
Round the result to the proper number of significant figures, which in this case is three, as indicated by the given moment of inertia.
KE_rot ≈ 4.8 × 10^-4 J
Therefore, the dragonfly must generate approximately 4.8 × 10^-4 Joules of energy to spin itself at a rate of 1100 rpm.
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Two blocks, 1 and 2, are connected by a massless string that passes over a massless pulley. 1 has a mass of 2.25 kg and is on an incline of angle 1=42.5∘ that has a coefficient of kinetic friction 1=0.205. 2 has a mass of 5.55 kg and is on an incline of angle 2=33.5∘ that has a coefficient of kinetic friction 2=0.105
. The figure illustrates the configuration.
A system of two blocks connected by a rope passing over a pulley. The system sits atop a scalene triangle whose long edge forms the base. The pulley is attached to the apex of the triangle. Box M subscript 1 rests on the triangle edge to the left of the pulley, which makes an angle of theta subscript 1 with the base of the triangle. The coefficient of friction between box M sub 1 and the surface is mu subscript 1. Box M subscript 2 rests on the triangle edge to the right of the pulley, which makes an angle of theta subscript 2 with the base of the triangle. The coefficient of friction between box M sub 2 and the surface is mu subscript 2.
Answers
The force acting on the system of two blocks connected by a rope passing over a pulley is -13.26 N.
The system of two blocks connected by a rope passing over a pulley are M1 and M2, where M1 rests on the triangle edge to the left of the pulley, which makes an angle of theta subscript 1 with the base of the triangle. The coefficient of friction between box M1 and the surface is mu subscript 1. M2 rests on the triangle edge to the right of the pulley, which makes an angle of theta subscript 2 with the base of the triangle.
The coefficient of friction between box M2 and the surface is mu subscript 2. The system sits atop a scalene triangle whose long edge forms the base. The pulley is attached to the apex of the triangle.M1 has a mass of 2.25 kg and is on an incline of angle 1=42.5∘ that has a coefficient of kinetic friction 1=0.205. M2 has a mass of 5.55 kg and is on an incline of angle 2=33.5∘ that has a coefficient of kinetic friction 2=0.105.The free-body diagram of M1 shows that the weight of M1 acts straight downwards (vertically) and the normal force acts perpendicular to the slope.
The force of friction opposes the motion and acts opposite to the direction of motion.M1 = 2.25 kgTheta subscript 1 = 42.5 degreesMu subscript 1 = 0.205g = 9.81 m/s²In the free-body diagram of M2, the normal force acts perpendicular to the incline of the slope, the weight of the object acts vertically downwards and parallel to the incline, and the force of friction opposes the motion and acts opposite to the direction of motion.M2 = 5.55 kgTheta subscript 2 = 33.5 degreesMu subscript 2 = 0.105g = 9.81 m/s²The tension in the string is the same throughout the rope. Since the masses are being pulled by the same rope, the acceleration of the objects is the same as the acceleration of the rope.
The tension in the string is directly proportional to the acceleration of the objects and the rope.A system of two blocks connected by a rope passing over a pulley has a total mass of M. The acceleration of the system is given by the formula below:a = [(m1-m2)gsin(θ1) - μ1(m1+m2)gcos(θ1)] / (m1 + m2)Where, μ1 = 0.205 is the coefficient of friction of block M1θ1 = 42.5 degrees is the angle of the incline of block M1M1 = 2.25 kg is the mass of block M1M2 = 5.55 kg is the mass of block M2g = 9.81 m/s² is the acceleration due to gravitysinθ1 = sin 42.5 = 0.67cosθ1 = cos 42.5 = 0.75The acceleration of the system is:a = [(2.25-5.55)(9.81)(0.67) - (0.205)(2.25+5.55)(9.81)(0.75)] / (2.25 + 5.55)a = -1.7 m/s² (the negative sign indicates that the system is accelerating in the opposite direction).
The force acting on the system is given by:F = MaWhere M is the total mass of the system and a is the acceleration of the system. The total mass of the system is:M = m1 + m2M = 2.25 + 5.55M = 7.8 kgThe force acting on the system is:F = 7.8(-1.7)F = -13.26 N (the negative sign indicates that the force is acting in the opposite direction).
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A car travels with a constant acceleration of 2.4 m/s2. Find the mass of the car if an 1850 N net force is required to produce this acceleration.
Answers
Answer:
770.83 kgExplanation:
The mass of the car can be found by using the formula
\(m = \frac{f}{a} \\ \)
f is the force
a is the acceleration
From the question we have
\(m = \frac{1850}{2.4} \\ = 770.83333...\)
We have the final answer as
770.83 kgHope this helps you
what is the value of pi(8.104)^2 written with correct significant numbers
Answers
Answer:206.3
Explanation:
Heather and Jerry are standing on a bridge 46 m
above a river. Heather throws a rock straight down with a speed of 14 m/s
. Jerry, at exactly the same instant of time, throws a rock straight up with the same speed. Ignore air resistance. How much time elapses between the first splash and the second splash?
Answers
The time elapsed between the first splash and the second splash is approximately 0.69 seconds.
To calculate this, we consider the motion of two rocks thrown simultaneously from a bridge. Heather throws a rock straight down with a speed of 14 m/s, while Jerry throws a rock straight up with the same speed.
We use the equation for displacement in uniformly accelerated motion: s = ut + (1/2)at^2.
For Heather's rock, which is thrown downwards, the initial velocity (u) is positive and the acceleration (a) due to gravity is negative (-9.8 m/s^2). The displacement (s) is the height of the bridge (46 m).
Solving the equation, we find two possible values for the time (t): t ≈ -4.91 s and t ≈ 1.91 s.
Since time cannot be negative in this context, we discard the negative value and consider t ≈ 1.91 s as the time it takes for Heather's rock to hit the water.
For Jerry's rock, thrown upwards, we use the same equation with the same initial velocity and acceleration. The displacement is also the height of the bridge, but negative.
Solving the equation, we find t ≈ -5.68 s and t ≈ 1.22 s. Again, we discard the negative value and consider t ≈ 1.22 s as the time it takes for Jerry's rock to reach its maximum height before falling back down.
To find the time difference between the first and second splash, we subtract t ≈ 1.91 s (Heather's rock) from t ≈ 1.22 s (Jerry's rock). This gives us a time difference of approximately 0.69 seconds.
Therefore, the time elapsed between the first splash and the second splash is approximately 0.69 seconds.
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The car of mass mc rolls from rest with negligible friction down the curved ramp and around the circular loop. The initial height of the center of mass of the car is at height H. Assume the center of mass of the car is at a height equal to the diameter d of the loop when the car is at the top of the loop.
(a) In terms of the given quantities and any fundamental constants, derive an equation for the speed vtop at the top of the loop. (Neglect the rotational kinetic energy of the wheels).
(b) The car is upside down at the top of the loop. Derive an equation for the minimum speed vmin necessary to make the loop.
Answers
The speed at the top of the loop is √gR.
Let the starting point be A, the lower point of loop be B and the top of loop be C.
So, at A the car is having only potential energy.
PE = mgh
At B, the kinetic energy,
KE = 1/2 mv²
a) At point C,
mv²/R = mg
The velocity at the top point, v(C)
v(top) = √gR
b) According to Conservation of energy, at B and C,
1/2 mv(B)² = 1/2 mv(C)² + mg(2R)
v(B)² = gR + 4gR
Therefore, v(B) = √5gR
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A rectangular beam 200 mm deep and 300 mm wide is simply supported over a span of 8 m. What uniformly distributed load per metre the beam may carry, if the bending stress is not to exceed 120N/mm2.
Answers
The uniformly distributed load per meter the beam may carry will be 1.275 × 10 ⁶ N/mm.
What is bending stress?When an object is subjected to a heavy load at a specific spot, it often experiences bending stress, which causes the object to bend and tire.
The given data in the problem is;
Bending stress, σ = 120 N/mm2
Moment of inertia, I = 8.5 × 106 mm⁴
Depth of beam, y = d/2 = 200/2 = 100 mm
Length of beam, L = 8 m = 8000 mm
Width of beam, W = 300 mm
The maximum bending moment of the beam with UDL;
\(\rm W = \frac{wL^2}{8}\)
From the bending equation;
\(\rm M = \frac{\sigma I}{y_{max}} \\\\ M = \frac{120 \ N / mm^2 \times 8.5 \times 10^6 }{100 \ mm } \\\\ M = 10.2 \times 10^6 \ N - mm\)
The maximum bending moment of the beam with UDL;
\(\rm M = \frac{wL^2}{8} \\\\ 10.2 \times 10^6 = \frac{w \times 8^2}{8} \\\\ w = 1.275 \times 10^6 \ N/mm\)
Hence the uniformly distributed load per meter the beam may carry will be 1.275 × 10 ⁶ N/mm.
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Lab Report Sun, Earth, and Moon Models It’s time to complete your Lab Report. Save the lab to your computer with the correct unit number, lab name, and your name at the end of the file name (e.g., U5_ Lab_SunEarthAndMoonModels_Alice_Jones.doc). Introduction 1. What was the purpose of the investigation? Type your answer here: 2. What causes the bright part of the moon to appear bright? Type your answer here: Experimental Methods 1. What materials did you use to create your model? Type your answer here: 2. Describe how you created your model. Type your answer here: Develop a Model 1. Show your model and the relationships between the components. Include labeled pictures or diagrams that describe causal accounts for the phases of the moon and eclipses. Type your answer here: Use a Model 1. Use your model to predict the relative positions of the earth, sun, and moon when the moon is full. Type your answer here: 2. Use your model to explain why a lunar eclipse does not occur every month when there is a full moon. Type your answer here:
Answers
1. The purpose of the investigation was to study the relationship between the Sun, Earth, and Moon.
2. The bright part of the moon appears bright due to the reflection of sunlight.
Introduction: The purpose of the investigation was to study the relationship between the Sun, Earth, and Moon. By creating models of the three celestial bodies, we aimed to understand how their movements and positions influence the phases of the moon and eclipses.The bright part of the moon appears bright due to the reflection of sunlight. As sunlight hits the moon's surface, it bounces back and reflects into space. This reflected light is what we see as the bright part of the moon.
Experimental Methods: To create our model, we used a lamp to represent the Sun, a ball to represent the Earth, and a smaller ball to represent the Moon. We also used a ruler, tape, and a protractor to measure distances and angles.We created our model by placing the lamp at one end of a table, the Earth in the middle, and the Moon at the other end. We attached the Moon to a string and moved it around the Earth to simulate the Moon's orbit around the Earth.
Develop a Model: Our model consists of a lamp, a ball, and a smaller ball on a string. The lamp represents the Sun, the ball represents the Earth, and the smaller ball on the string represents the Moon. As the Moon moves around the Earth, it goes through phases, from a new moon to a full moon and back again.We used diagrams and pictures to label the components of our model and describe causal accounts for the phases of the moon and eclipses.
Use a Model: When the Moon is full, it is in a direct line with the Earth and the Sun. Using our model, we can predict that the Moon would be directly opposite the Sun in the sky during a full moon. A lunar eclipse does not occur every month when there is a full moon because the Moon's orbit around the Earth is tilted at an angle of about 5 degrees to the Earth's orbit around the Sun. Therefore, the Moon is not always in a direct line with the Earth and the Sun during a full moon, which is necessary for a lunar eclipse to occur.
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Which of the following statements is true? A. Both warming up and cooling down or important. B. It is more important warm up then it is to cool down. C. Is more important to cool down then is to warm up. D. Both warming up and cooling down are not important
Answers
Answer:
No. A is correct because both warming up and cooling down are important
Both warming up and cooling down are important.
This is based on aerobics and human body balance regulation as regards exercising.
Warming up and cooling down in exercising are just based on the level of intensity at which the exercise is carried out.
Now, warming up when exercising involves activities like jogging. Warming up is a very vital part of exercising as it helps to get a person's cardiovascular system ready for the subsequent exercises and physical activities to be engaged. This will help in making sure there is enough blood flowinh to your muscles as well as increasing your body temperature.In another sense, cooling down is also very vital in activity because after the blood pressure and heart rates have been raised after exercising, they will need to be restored to their normal levels at which they were before commencement of the exercise. It also helps to regulate the blood flow.Thus, Both warming up and cooling down are important.
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During the COVID pandemic many people acted to help those in need despite the risk to themselves. This action best illustrates
Answers
10 POINTS
Which type of electromagnetic radiation is used in airport security screening?
O A. X-rays
• B. Ultraviolet radiation
• C. Gamma rays
O D. Infrared radiation
SUBMIT
Answers
Answer:
A) X-rays
Explanation:
Airport security screenings use X-rays because they can penetrate through bags to see objects that are located within that are obscured through normal view.
X- rays are used to test body in airports. Low energy waves are used to screen baggages and human body. Thus option A is correct.
What is electromagnetic radiation ?Electromagnetic spectrum is the arrangement of radiations in the order of decreasing wavelength or increasing energy and frequency. The order is radio waves, microwaves, infrared, visible, ultraviolet, x-rays and gamma rays.
x -rays are highly energetic short waves which can be easily penetrate through a body. X-ray is used in different diagnostic techniques. X-rays easily penetrate through tissues and provide the scanned images of internal parts.
In airport security screening x-rays are used to screen baggages and human bodies. High energetic waves are causing serious health risks. Hence, lower energy x-rays are used in testing.
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what would you expect the force to be if the mass of object A was 50.0 x 10^5 kg and the mass of object B was 40.0 x 10^5 kg?
Answers
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Explanation:
What is the direction of the electric field at a point a distance d = 34.00 cm directly to the right of the positive charge?
Answers
Answer:
s the charge is positive, the electric field lines leave it at the point a d = 0.034 cm, the electric field lines go to the right.
Explanation:
The electric field around the cagases oisitvi, field leaving this means that the field lines move away from
If the charge is negative the lines of the field are directed towards the charge
Let's apply this to our case, as the charge is positive, the electric field lines leave it at the point a d = 0.034 cm, the electric field lines go to the right.
A motorcycle has a mass of 100kg , suppose it travels at 40m/s . Find the kinetic energy possessed by it
Answers
The Kinetic energy possessed by the cycle with a mass of 100 kg and velocity 40 m/s is 80 kJ.
Kinetic energy is the energy of an object when the object is in motion. It is obtained by the product of the mass of the object and the velocity of the object. The unit of kinetic energy is the joule (J).
From the given,
mass of the motorcycle (m) = 100 kg
velocity (v) = 40 m/s
K.E =?
K.E = (mv²) / 2
= (100×40×40) / 2
= 160,000 / 2
= 80,000 J
Thus, the kinetic energy of the motorcycle is 80kJ.
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9. Two moving objects will change their motion if which of the following is true? a.There is no friction on the objects. b. The objects collide with each other. c.One object is heavier than the other. d.The objects are moving in space.
Answers
Answer:
lol
Explanation:
lolllll i need points sorry
what is clinograph and how to read it
Answers
A climograph is a graphical representation of a location's basic climate. Clinograph is an instrument used to draw parallel lines to the inclined lines. It contains one adjustable wing or strip which can be adjusted to required angle. So, it can be termed as adjustable set square
Hope this helps!:)
Answer:
Explanation:
A climograph is a graphical representation of a location's basic climate. Climographs display data for two variables:
(a) monthly average temperature
(b) monthly average precipitation. These are useful tools to quickly describe a location's climate.
How much water is on the earth
Answers
71% of the earth is water
Answer:
hope this helps
Explanation:
71% of earth is water
HONI A ball is rolling steady on the floor. (a) Draw and label all the forces acting on the ball. (b) Describe the relationships between all the forces acting on the ball
Answers
a) The image is attached to this answer
b) The kinetic friction force affects the forward force of the ball.
What is the relationship between the forces that act on a rolling ball?
The force that opposes a rolling ball's motion is called rolling friction. It slows the ball down by acting in the opposite direction to that of the ball's motion. The weight of the ball and the type of the surface are two variables that affect rolling friction.
The force that a surface uses to maintain the weight of an object that is resting on it is known as the normal force. When a ball is rolling, the normal force exerts itself perpendicular to the surface the ball is moving on. It maintains the ball's weight balance and offers the required reaction force for rolling motion.
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