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harkovskaia [24]
3 years ago
5

Jerome plays middle linebacker for souths varsity football team. In a game against cross-town rival north, he delivered a hit to

north's 82-kg running back, changing his eastward velocity of 5.6 m/s into a westward velocity of 2.5 m/s. Determine that momentum change of the running back
Physics
1 answer:
Makovka662 [10]3 years ago
7 0

The momentum change of the running back is - 664.2 kg m/s or 664 west.

<u>Explanation:</u>

Momentum is defined as the change in velocity of any object along with its mass. So mathematically, momentum can be derived using the product of mass with the change in velocity.

           \text { Momentum change }=m \times \Delta v

As here mass is given as 82 kg and the initial velocity was 5.6 m/s and final velocity is 2.5 m/s.

      Initial Momentum = m \times \text { initial velocity }=82 \times 5.6=459.2 \mathrm{kg} \mathrm{m} / \mathrm{s}

      Final Momentum = m \times \text { final velocity }=82 \times(-2.5)=-205 \mathrm{kg} \mathrm{m} / \mathrm{s}

      Momentum change = Final Momentum - Initial Momentum

      Momentum change = - 205 - 459.2 = - 664.2 kg m/s or 664 west

Thus, the changing momentum is -664.2 kg m/s. The negative sign indicates that the momentum is acting in the opposite direction on changing in the direction of velocity.

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Anastaziya [24]
M=F/A
Which means 30 divided by 5 m/s is 6kg(mass)
8 0
3 years ago
What is the frequency of a wave that has a period of vibration of 2 seconds?
34kurt

Answer:

The answer is 0.5 Hz

Explanation:

Its pretty easy to get the answer. One hertz (Hz) is equal to one cycle or period per second. So, just divide the period by the number of seconds.

1 period/2 secs = 1/2 Hz or 0.5 Hz

7 0
3 years ago
A ball is kicked at an angle of 35° with the ground.a) What should be the initial velocity of the ball so that it hits a target
stiks02 [169]

Answer:

a.18.5 m/s

b.1.98 s

Explanation:

We are given that

\theta=35^{\circ}

a.Let v_0 be the initial velocity of the ball.

Distance,x=30 m

Height,h=1.8 m

v_x=v_0cos\theta=v_0cos35

v_y=v_0sin\theta=v_0sin35

x=v_0cos\theta\times t=v_0cos35\times t

t=\frac{30}{v_0cos35}

h=v_yt-\frac{1}{2}gt^2

Substitute the values

1.8=v_0sin35\frac{30}{v_0cos35}-\frac{1}{2}(9.8)(\frac{30}{v_0cso35})^2

1.8=30tan35-\frac{6574.6}{v^2_0}

\frac{6574.6}{v^2_0}=21-1.8=19.2

v^2_0=\frac{6574.6}{19.2}

v_0=\sqrt{\frac{6574.6}{19.2}}=18.5 m/s

Initial velocity of the ball=18.5 m/s

b.Substitute the value then we get

t=\frac{30}{18.5cos35}

t=1.98 s

Hence, the time for the ball to reach the target=1.98 s

7 0
3 years ago
Let v1, , vk be vectors, and suppose that a point mass of m1, , mk is located at the tip of each vector. The center of mass for
g100num [7]

Answer:

Explanation:

Center of mass is give as

Xcm = (Σmi•xi) / M

Where i= 1,2,3,4.....

M = m1+m2+m3 +....

x is the position of the mass (x, y)

Now,

Given that,

u1 = (−1, 0, 2) (mass 3 kg),

m1 = 3kg and it position x1 = (-1,0,2)

u2 = (2, 1, −3) (mass 1 kg),

m2 = 1kg and it position x2 = (2,1,-3)

u3 = (0, 4, 3) (mass 2 kg),

m3 = 2kg and it position x3 = (0,4,3)

u4 = (5, 2, 0) (mass 5 kg)

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Now, applying center of mass formula

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Xcm = (m1•x1+m2•x2+m3•x3+m4•x4) / (m1+m2+m3+m4)

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Xcm = (2.2, 1.7, 0.8) m

This is the required center of mass

6 0
3 years ago
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tankabanditka [31]
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Letter B
3 0
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