These two magnitudes are related in the following way. Suppose an object is moving at a certain speed $v_1$ and changes it to $v_2$ . The impulse is numerically equivalent to the change of linear momentum. Let's recall the momentum is given by

$p=mv$

The initial and final momentums are, respectively

$p_1=mv_1,\ p_2=mv_2$

The change of momentum is

$\Delta p=p_2-p_1=m(v_2-v_1)$

It is numerically equal to the Impulse J

$J=\Delta p$

$J=m(v_2-v_1)$

We are given

$m=2498\ kg,\ v_1=17.1\ m/s,\ v_2=2.6\ m/s$

The impulse the car experiences during that time is

$J=2498(2.6-17.1)=2498(-14.5)$

J=-36221 Kg.m/s

The magnitude of J is

J=36221 Kg.m/s

8 0

3 years ago

A 0.600 kg block is attached to a spring with spring constant 15 N/m. While the block is sitting at rest, a student hits it with

gulaghasi [49]

Answer:

8.8 cm

31.422 cm/s

Explanation:

m = Mass of block = 0.6 kg

k = Spring constant = 15 N/m

x = Compression of spring

v = Velocity of block

A = Amplitude

As the energy of the system is conserved we have

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{0.6\times 0.44^2}{15}}\\\Rightarrow A=0.088\ m\\\Rightarrow A=8.8\ cm$

Amplitude of the oscillations is 8.8 cm

At x = 0.7 A

Again, as the energy of the system is conserved we have

$\dfrac{1}{2}kA^2=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2\\\Rightarrow v=\sqrt{\dfrac{k(A^2-x^2)}{m}}\\\Rightarrow v=\sqrt{\dfrac{15(0.088^2-(0.7\times 0.088)^2)}{0.6}}\\\Rightarrow v=0.31422\ m/s$

The block's speed is 31.422 cm/s

4 0

3 years ago

Convert 172 to a scientific notation

gtnhenbr [62]

Answer:

1.72×10^2

Explanation:

Is the answer correct ?

6 0

2 years ago

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