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katovenus [111]
4 years ago
5

A reckless coyote engineer straps on a pair of ice skates and attaches a rocket capable of producing 5.0 × 10 3 N of thrust to h

is back. Together, the coyote and the rocket have a mass of 120 kg. If the coyote starts at rest on level, frictionless ice and bends over such that the rocket thrust is directed parallel to the ice, what is his final speed if the rocket burns for 5.0 s?
Physics
1 answer:
Marianna [84]4 years ago
3 0

Answer:

208.33 m/s

Explanation:

t = Time taken

u = Initial velocity = 0

v = Final velocity

a = Acceleration

m = Mass of coyote and the rocket = 120 kg

F = Force of thrust produced by the rockets = 5\times 10^3\ N

F=ma\\\Rightarrow a=\frac{F}{m}\\\Rightarrow a=\frac{5\times 10^3}{120}\\\Rightarrow a=\frac{125}{3}\ m/s^2

v=u+at\\\Rightarrow v=0+\frac{125}{3}\times 5\\\Rightarrow v=208.33\ m/s

Velocity of the rocket at 5 seconds is 208.33 m/s

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A person absentmindedly walks off the edge of a tall cliff. They will fall 50 m into either
kifflom [539]

The man can survive, and he lands 12.1 m from the base of the cliff (into the lake)

Explanation:

The motion of the person is equivalent to the motion of a projectile, which consists of two separate motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

We start by analyzing the vertical motion, to find the time it take for the person to reach the ground level. We can use the following suvat equation:

s=ut+\frac{1}{2}gt^2

where

s = 50 m is the vertical distance covered (the height of the cliff)

u = 0 is the initial vertical velocity

g=9.8 m/s^2 is the acceleration of gravity

t is the time of flight

Solving for t,

t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(50)}{9.8}}=3.19 s

The person moved horizontally at a constant speed of

v_x = 3.8 m/s

So, the horizontal distance covered by the man during his flight will be

d_x = v_x t = (3.8)(3.19)=12.1 m

So, the man will land on the lake (which starts from 12 m), so he can survive.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

8 0
3 years ago
a loaded sack of total mass is 1000 gramme falls down from the floor of a lorry 200cm high, calculate the workdone by the gravit
Helen [10]

Answer:

W = 20 J

Explanation:

Given that,

The mass of a loaded sack, m = 1000 g = 1 kg

It falls down from the floor of a lorry 200 cm high, h = 2 m

We need to find the work done by the gravity. The work done by an object under the action of gravity is given by :

W = mgh

Substitute all the values,

W = 1 × 10 × 2

= 20 J

Hence, the required work done by gravity is equal to 20 J.

8 0
3 years ago
The purpose of many scientific investigations is to test a {n}
Gnoma [55]
Hypothesis , I believe the answer is hypothesis
3 0
3 years ago
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A fixed 11.2-cm-diameter wire coil is perpendicular to a magnetic field 0.53 T pointing up. In 0.10 s , the field is changed to
Karolina [17]

Answer:

The average induced emf in the coil is 0.0286 V

Explanation:

Given;

diameter of the wire, d = 11.2 cm = 0.112 m

initial magnetic field, B₁ = 0.53 T

final magnetic field, B₂ = 0.24 T

time of change in magnetic field, t = 0.1 s

The induced emf in the coil is calculated as;

E = A(dB)/dt

where;

A is area of the coil = πr²

r is the radius of the wire coil = 0.112m / 2 = 0.056 m

A = π(0.056)²

A = 0.00985 m²

E = -0.00985(B₂-B₁)/t

E = 0.00985(B₁-B₂)/t

E = 0.00985(0.53 - 0.24)/0.1

E = 0.00985 (0.29)/ 0.1

E = 0.0286 V

Therefore, the average induced emf in the coil is 0.0286 V

3 0
3 years ago
What is the uses of echoes​
Anestetic [448]

Answer:

Echoes are the reflection of sound from relatively flat object that is far enough away that you can discern the time difference. Echoes are used to measure distance, velocity, and the shape of objects. Echoes off gratings result in an unusual pinging sound

6 0
3 years ago
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