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Setler79 [48]
3 years ago
14

The young modulus of an elastic material is 2.5×10 raised to power 3 newton per meter square. What magnitude of load will cause

an extension of 200 cm in 20cm length of the material given that it has a diameter of 15m
the young modulus of an elastic material is  {2.5 \times 10}^{3}   nm ^{ - 2}  .  What magnitude of load will cause an extension of 200 cm in 20cm length of the material given that it has a diameter of 15m
​
Physics
1 answer:
katrin2010 [14]3 years ago
5 0

Answer:.

200

Explanation:

.

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Newtons 1st law of motion states that the object will continue to move at its present speed and direction until an outside force acts upon it.
 
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A student conducts an experiment in which a cart is pulled by a variable applied force during a 2 s time interval. In trial 1, t
fredd [130]

Answer:

The answer is "Including all three studies of 0s to 2s, that shift in momentum is equal".

Explanation:

Its shift in momentum doesn't really depend on the magnitude of its cars since the forces or time are similar throughout all vehicles.

Let's look at the speed of the car

F = m a\\\\a =\frac{F}{m}

We use movies and find lips

\to v = v_0 + a t\\\\\to v = v_0 + (\frac{F}{m}) t

The moment is defined by

\to p = m v

The moment change

\Delta p = m v - m v_0

Let's replace the speeds in this equation

\Delta p = m (v_0 + \frac{F}{m t}) - m v_0\\\\\Delta p = m v_0  + F t - m v_0\\\\\Delta p = F t

They see that shift is not directly proportional to the mass of cars since the force and time were the same across all cars.

5 0
3 years ago
A quarterback passes a football from height h = 2.1 m above the field, with initial velocity v0 = 13.5 m/s at an angle θ = 32° a
alisha [4.7K]

Answer:

a)    x = v₀² sin 2θ / g

b)    t_total = 2 v₀ sin θ / g

c)    x = 16.7 m

Explanation:

This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity

        sin θ = v_{oy} / vo

        cos θ = v₀ₓ / vo

         v_{oy} = v_{o} sin θ

         v₀ₓ = v₀ cos θ

         v_{oy} = 13.5 sin 32 = 7.15 m / s

         v₀ₓ = 13.5 cos 32 = 11.45 m / s

a) In the x axis there is no acceleration so the velocity is constant

         v₀ₓ = x / t

          x = v₀ₓ t

the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero

          v_{y} = v_{oy} - gt

          0 = v₀ sin θ - gt

          t = v_{o} sin θ / g

         

we substitute

       x = v₀ cos θ (2 v_{o} sin θ / g)

       x = v₀² /g      2 cos θ sin θ

       x = v₀² sin 2θ / g

at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,

b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time

at the highest point the vertical speed is zero

          v_{y} = v_{oy} - gt

          v_{y} = 0

           t = v_{oy} / g

           t = v₀ sin θ / g

as the time to get on and off is the same the total time or flight time is

           t_total = 2 t

           t_total = 2 v₀ sin θ / g

c) we calculate

          x = 13.5 2 sin (2 32) / 9.8

          x = 16.7 m

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