The impulse experienced by a body is equivalent to the body’s change in?

1 answer:

5 0

<span>impulse =force*time=mass*acceleration*time=mass*... in momentum , I hope this helps you out!! Also have an amazing day and good luck on any further work !!!

#brainlyzkool</span>

You might be interested in

Show that the speed with which a projectile leaves the ground is equal to its speed just before it strikes the ground at the end

Rina8888 [55]

Answer:

Thus, the velocity at the time of strike is same as the velocity at the time of projection.

Explanation:

Let a projectile is projected vertically upwards with a speed of u and reaches to the maximum height H.

At maximum height , the speed is zero and then the projective comes back on the ground.

Use the third equation of motion

$v^2 = u^2 + 2 g h \\\\0 = u^2 - 2 g H\\\\\u =\sqrt{2gH}$

Now let the velocity at the time of strike is v'.

Use third equation of motion, here initial velocity is zero.

$v'^2 = 0 + 2 g H \\\\v = \sqrt{2gH}$

Thus, the velocity at the time of strike is same as the velocity at the time of projection.

8 0

3 years ago

Devise an exponential decay function that fits the given data, then answer the accompanying questions. Be sure to identify the

7nadin3 [17]

Answer:

22145.27733 ft

124984.76055 ft

Explanation:

The equation of pressure is

$P=P_0e^{-kh}$

where,

$P_0$ =Atmospheric pressure = 800 mbar

k = Constant

h = Altitude = 35000 ft

$P=\dfrac{1}{3}P_0$

$\dfrac{1}{3}P_0=P_0e^{-k35000}\\\Rightarrow \dfrac{1}{3}=e^{-k35000}\\\Rightarrow 3=e^{k35000}\\\Rightarrow ln3=k35000\\\Rightarrow k=\dfrac{ln3}{35000}\\\Rightarrow k=3.13\times 10^{-5}$

Now

$P=\dfrac{1}{2}P_0$

$ln2=kh\\\Rightarrow h=\dfrac{ln2}{k}\\\Rightarrow h=\dfrac{ln2}{3.13\times 10^{-5}}\\\Rightarrow h=22145.27733\ ft$

The altitude will be 22145.27733 ft

$P=0.02P_0$

$0.02P_0=P_0e^{-kh}\\\Rightarrow 0.02=e^{-3.13\times 10^{-5}h}\\\Rightarrow ln0.02=-3.13\times 10^{-5}h\\\Rightarrow h=\dfrac{ln0.02}{-3.13\times 10^{-5}}\\\Rightarrow h=124984.76055\ ft$