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inna [77]
3 years ago
7

An OSU linebacker of mass 110.0 kg sacks a UM quarterback of mass 85.0 kg. Just after they collide, they are momentarily stuck t

ogether, and both are moving at a speed of 2.60 m/s. If the quarterback was at rest just before he was sacked, how fast was the linebacker moving just before the collision
Physics
1 answer:
expeople1 [14]3 years ago
3 0

Answer:

The initial speed of the linebacker just before the collision was 4.6 m/s.

Explanation:

Mass of OSU linebacker, m = 110 kg

Mass of UM quarterback, m' = 85 kg

Just after they collide, they are momentarily stuck together, the common speed is, V = 2.6 m/s

Initial speed of quarterback is 0 as it was at rest initially, u' = 0

Let u was the initial speed of the linebacker just before the collision. As they struck together, the momentum remains conserved. So,

mu+m'u'=(m+m')V\\\\mu=(m+m')V\\\\u=\dfrac{(m+m')V}{m}\\\\u=\dfrac{(110+85)\times 2.6}{110}\\\\u=4.6\ m/s

So, the initial speed of the linebacker just before the collision was 4.6 m/s.

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A 3.91 kg cart is moving at 5.7 m/s when it collides with a 4 kg cart which was at rest. They collide and stick together.
Nesterboy [21]

Answer:

<em>The velocity after the collision is 2.82 m/s</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

It states the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is  

P=mv.  

If we have a system of two bodies, then the total momentum is the sum of the individual momentums:

P=m_1v_1+m_2v_2

If a collision occurs and the velocities change to v', the final momentum is:

P'=m_1v'_1+m_2v'_2

Since the total momentum is conserved, then:

P = P'

Or, equivalently:

m_1v_1+m_2v_2=m_1v'_1+m_2v'_2

If both masses stick together after the collision at a common speed v', then:

m_1v_1+m_2v_2=(m_1+m_2)v'

The common velocity after this situation is:

\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}

There is an m1=3.91 kg car moving at v1=5.7 m/s that collides with an m2=4 kg cart that was at rest v2=0.

After the collision, both cars stick together. Let's compute the common speed after that:

\displaystyle v'=\frac{3.91*5.7+4*0}{3.91+4}

\displaystyle v'=\frac{22.287}{7.91}

\boxed{v' = 2.82\ m/s}

The velocity after the collision is 2.82 m/s

6 0
2 years ago
A ship, carrying freshwater to a desert island in the caribbean, has a horizontal cross-sectional area of 2800 m2 at the waterli
Alex17521 [72]
<span>Due that we already know the horizontal cross-sectional area of the ship, which is 2800 m2 and we are going to understand that value keeps constant for the whole 9.5 of height of the ship from the waterline till the new waterline after unloading, then we just need to calculate the volume as follows: V = A * H , where V is volume, A is area and H is height V= 2,800 * 9.5 = 26,600 m3 So this volum of 26,600 cubic meters is the volum of freshwater delivered in the island.</span>
3 0
3 years ago
If a volume of air is warmed, it expands and tends to
Butoxors [25]

if a volume of air is warmed it expands due to increased translational kinetic energy as it expands it will start to cool.

<h3>When does temperature increase volume?</h3>

We can then conclude that at constant pressure, temperature and volume are directly proportional: temperature increases, volume increases; decrease temperature, decrease volume.

In this case, the higher the temperature, the greater the kinetic energy that acts on the molecules of this gas, so when the gas expands, these molecules find more space and collide less, which will cause the gas to cool.

See more about volume at brainly.com/question/1578538

#SPJ12

7 0
2 years ago
A light horizontal spring has a spring constant of 138 N/m. A 3.35 kg block is pressed against one end of the spring, compressin
BARSIC [14]

Answer:

U_k = 0.113

Explanation:

using the law of the conservation of energy:

E_i -E_f=W_f

\frac{1}{2}Kx^2=NU_kd

where K is the spring constant, x is the spring compression, N is the normal force of the block, U_k is the coefficiet of kinetic friction and d is the distance.

Also, by laws of newton, N is calculated by:

N = mg

N = 3.35 kg * 9.81 m/s

N = 32.8635

So, Replacing values on the first equation, we get:

\frac{1}{2}(138)(0.123)^2= (32.8635)U_k(0.281m)

solving for U_k:

U_k = 0.113

8 0
3 years ago
the other end of the pipe. With the aide of the pipe, does the applied force produce a smaller torque, a greater torque, or the
pav-90 [236]

Answer:

With the addition of the pipe we have a greater torque.

Explanation:

We need to complete the description of the problem, searchin in internet we have:

"Sometimes, even with a wrench, one cannot loosen a nut that is frozen tightly to a bolt. It is often possible to loosen the nut by slipping one end of a long pipe over the wrench handle and pushing at the other end of the pipe. With the aid of the pipe, does the applied force produce a smaller torque, a greater torque, or the same torque on the nut?"

With the addition of the pipe we have a greater torque, as it increases the distance or radius of torque.

We know that torque is defined, as the product of force by distance, in this way we have:

T = F * d

where:

T = torque [N*m]

F = force [N]

d = distance [m]

We can see in the above equation, that increasing the distance increases torque proportionally.

3 0
3 years ago
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