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anygoal [31]
3 years ago
5

A girl launches a toy rocket from the ground. The engine experiences an average thrust of 5.26 N. The mass of the engine plus fu

el before liftoff is 25.0 g, which includes fuel mass of 12.3 g. The engine fires for a total of 1.90 s.A. What is the average exhaust speed of the engine?B. This engine is placed in a rocket body of mass 60.5 g. What is the magnitude of the final velocity of the rocket if it were to be fired from rest in outer space by an astronaut on a spacewalk?
Physics
1 answer:
Triss [41]3 years ago
7 0

Answer: The answer for A is - v = 786.93 m/s

The answer for B is - v = 122.40 m/s

Explanation:

a) To find the average exhaust speed (v) of the engine we can use the following equation:

F = vΔm

Where:

F: is the thrust by the engine = 5.26 N

Δm: is the mass of the fuel = 12.7 g

Δt: is the time of the burning of fuel = 1.90 s

v = F×ΔT/ΔT

b) To calculate the final velocity of the rocket we need to find the acceleration.

The acceleration (a) can be calculated as follows:

a = F/M

In the above equation, m is an average between the mass of the engine plus the rocket case mass and the mass of the engine plus the rocket case minus the fuel mass:

m = (m_{engine} + m_{rocket}) + (m_{engine} + m_{rocket} - m_{fuel})}{2} = {2*m_{engine} + 2*m_{rocket} - m_{fuel}}{2} = 2*25.0 g + 2*63.0 g - 12.7 g}{2} = 81.65 g

Now, the acceleration is:

a = 5.26 N/81.65-t 10^³kg} = 64.42 m*s^²

Finally, the final velocity of the rocket can be calculated using the following kinematic equation:

v= v_{0} + at = 0 + 64.42 m*s^{-2}*1.90 s = 122.40 m/s

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Nuetrik [128]

Answer: A student walks 50 meters east, 40 meters north, 35 meters east, and then 20 m south. Then the magnitude and direction of the student's total displacement will be 87.32 m along the direction of AD or in east-south direction.

Explanation: To find the correct answer, we need to know about the Displacement of a body in motion.

<h3>What is displacement of a body in motion?</h3>
  • The displacement is the shortest distance between initial and final positions of a body.
  • It's a vector quantity, and can positive, negative, or zero.
  • The magnitude of displacement is less than or equal to the distance travelled.
<h3>How to solve the problem?</h3>
  • At first, we can draw a diagram showing the motion of the body.
  • From the diagram, the displacement of the body will be equal to the distance between point A and D.
  • To solve this, we can use Pythagoras theorem.

AD=AC+CD\\AC^{2} =50^{2} +11^2\\AC=51.19 m\\Similarly,\\CD^2=35^2+9^2\\CD=36.13 m\\thus, \\AD=51.19+36.13=87.32 m

Thus, from the above calculations, we can conclude that, the displacement of the body will be equal to 87.32 m along the direction of AD or in east-south direction.

Learn more about the Displacement here:

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3 0
2 years ago
a wire with mass per unit length 75 g/m runs horizontally at right angles to a uniform horizontal 0.12 T magnetic field. what am
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Answer:

The amount of current that must flow through the wire for it to be suspended against gravity by magnetic force = 6.125 A

Explanation:

Force on a wire carrying current in an electric field is given by

F = (B)(I)(L) sin θ

For this question,

The magnetic force must match the weight of the wire.

F = mg

mg = (B)(I)(L) sin θ

(m/L)g = (B)(I) sin θ

Mass per unit length = 75 g/m = 0.075 kg/m

B = magnetic field = 0.12 T

I = ?

g = acceleration due to gravity = 9.8 m/s

θ = angle between wire's current direction and magnetic field = 90°

0.075 × 9.8 = 0.12 × I sin 90°

I = 0.075 × 9.8/0.12 = 6.125 A

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3 years ago
A block of mass 12.2 kg is sliding at an initial velocity of 3.9 m/s in the positive x-direction. The surface has a coefficient
Studentka2010 [4]

Answer:

Explanation:

a) Force of friction = μ R where μ is coefficient of kinetic friction and R is reaction force

R = mg where m is mass of the block

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deceleration = force / mass = 20.68 / 12.2 = 1.7 m /s²

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c )

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v = 0 , u = 3.9 m /s

a = 1.7 m /s

0 = 3.9² - 2 x 1.7 x s

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Anvisha [2.4K]

Answer:

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