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andrezito [222]
2 years ago
5

Question 23 of 32

Physics
1 answer:
-BARSIC- [3]2 years ago
4 0

Answer:

2m/s

     

Explanation:

Given parameters:

Mass of truck  = 9000kg

Kinetic energy = 18000J

Unknown:

Speed  = ?

Solution:

Kinetic energy is the energy due to the motion of a body;

  K.E  = \frac{1}{2} mv²  

 So;

 m is the mass

  v is the speed

   2 KE = mv²

  2 x 18000 = 9000 x v²

     v²  = 4

     v = 2m/s

     

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Which example identifies a change in motion that produces acceleration?
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A spy satellite uses a telescope with a 1.7-m-diameter mirror. It orbits the earth at a height of 180 km.
WINSTONCH [101]

Answer: the minimum spacing that must be there between two objects on the earth's surface if they are to be resolved as distinct objects by this telescope 6.45 cm

Explanation:

Given that;

diameter of the mirror d = 1.7 m

height h = 180 km = 180 × 10³ m

wavelength λ = 500 nm = 5 × 10⁻⁹ m

Now Angular separation from the peak of the central maximum is expressed as;

sin∅= 1.22 λ / d

sin∅ = (1.22 × 5 × 10⁻⁹) / 1.7

 sin∅ = 3.588 × 10⁻⁷

we know that;

 sin∅  = object separation / distance from telescope

object separation =   sin∅ × distance from telescope

object separation = 3.588 × 10⁻⁷  × 180 × 10³

object separation =6.45 × 10⁻² m

then we convert to centimeter

object separation = 6.45 cm

Therefore the minimum spacing that must be there between two objects on the earth's surface if they are to be resolved as distinct objects by this telescope 6.45 cm

5 0
3 years ago
slader Question: A Model Rocket Is Launched Straight Upward With An Initial Speed Of 50m/s. Iit Accelerates With A Constant Upwa
xenn [34]

Answer:

Maximum height reached by the rocket is

y_{max} = 308 m

total time of the motion of rocket is given as

T = 16.44 s

Explanation:

Initial speed of the rocket is given as

v_i = 50 m/s

acceleration of the rocket is given as

a = 2 m/s^2

engine stops at height h = 150 m

so the final speed of the rocket at this height is given as

v_f^2 - v_i^2 = 2 a d

v_f^2 - 50^2 = 2(2)(150)

v_f = 55.68 m/s

so maximum height reached by the rocket is given as the height where its final speed becomes zero

so we will have

v_f^2 - v_i^2 = 2 a d

0 - 55.68^2 = 2(-9.81)(y - 150)

y_{max} = 308 m

Now the total time of the motion of rocket is given as

1) time to reach the height of 150 m

v_f - v_i = at

55.68 - 50 = 2 t

t_1 = 2.84 s

2) time to reach ground from this height

\Delta y = v_y t + \frac{1}{2}gt^2

-150 = 55.68 t - \frac{1}{2}(9.81) t^2

t_2 = 13.6 s

so total time of the motion of rocket is given as

T = 13.6 + 2.84 = 16.44 s

3 0
3 years ago
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