Ask question

Ask question

All categories

3 years ago

5

A 125 kg linebacker runs to the right at 2.5 m/s and an 80 kg quarterback runs to the left at 3 m/s. They collide in midair and

the linebacker holds on to the quarterback so they move off together. What is their velocity (magnitude and direction) immediately after this collision

1 answer:

Alexxx [7]3 years ago

6 0

Answer: 0.35 m/s to the right.

Explanation:

Let's define M1 and V1 as the mass and velocity of the linebacker, and define the right as the positive side.

M1 = 125kg

V1 = 2.5m/s

M2 and V2 as the velocity and mass of the quarterback.

M2 = 80kg

V2 = -3m/s

where V2 has a minus sign because the quarterback is running to the left.

We know that after the collision, both players keep stick togheter, so we can think this as an perfectly inleastic collision, where the velocity of both players can be obtained by:

v = (M1*V1 + M2*V2)/(M1 + M2) = (125kg*2.5m/s - 80kg*3m/s)/(125kg + 80kg)

v = (72.5/205) m/s = 0.35m/s

Because this velocity is positive, this means that both players are moving to the right.

You might be interested in

What is the best insulator?

Aleonysh [2.5K]

Answer:

E

Explanation:

All others conduct electricity and heat.

6 0

3 years ago

Read 2 more answers

A 8.8 cm diameter circular loop of wire is in a 1.04 T magnetic field. The loop is removed from the field in 0.30 s . Assume tha

denis23 [38]

Answer:

0.021 V

Explanation:

The average induced emf (E) can be calculated usgin the Faraday's Law:

$E = - \frac{N*\Delta \phi}{\Delta t}$

<u>Where:</u>

<em>N = is the number of turns = 1 </em>

<em>ΔΦ = ΔB*A </em>

<em>Δt = is the time = 0.3 s </em>

<em>A = is the loop of wire area = πr² = πd²/4 </em>

<em>ΔB: is the magnetic field = (0 - 1.04) T </em>

Hence the average induced emf is:

$E = - \frac{\Delta B*A}{\Delta t} = - \frac{(0- 1.04 T) \pi (0.088 m)^{2}}{4*0.3 s} = 0.021 V$

Therefore, the average induced emf is 0.021 V.

I hope it helps you!

8 0

3 years ago

7. A mother pushes her 9.5 kg baby in her 5kg baby carriage over the grass with a force of 110N @ an angle

jasenka [17]

Weight of the carriage $=(m+M)g =142.1\ N$

Normal force $=Fsin(\theta) + W = 197.1\ N$

Frictional force $=\mu N=27.59\ N$

Acceleration $=4.66\ m\ s^{-2}$

Explanation:

We have to look into the FBD of the carriage.

Horizontal forces and Vertical forces separately.

To calculate Weight we know that both the mass of the baby and the carriage will be added.

So Weight(W) $=(m+M)\times g =(9.5+5)\ kg \times 9.8 =142.1\ Newton\ (N)$

To calculate normal force we have to look upon the vertical component of forces, as Normal force is acting vertically.We have weight which is a downward force along with $F_x$ , force of $110\ N$ acting vertically downward.Both are downward and Normal is upward so Normal force $=Summation\ of\ both\ forces$

Normal force (N) $= Fsin(\theta)+W=110sin(30) + 142.1 =197.1\ N$

Frictional force (f) $=\mu N=0.14\times 197.1 =27.59\ N$

To calculate acceleration we will use Newtons second law.

That is Force is product of mass and acceleration.

We can see in the diagram that $F_y=Horizontal$ and $F_x=Vertical$ component of forces.

So Fnet = Fy(Horizontal) - f(friction) $= m\times a$

Acceleration (a) = $\frac{Fcos(\theta)-\mu N}{mass(m)} =\frac{(95.26-27.59)}{14.5}= 4.66\ m\ s{^2 }$

So we have the weight of the carriage, normal force,frictional force and acceleration.

3 0

2 years ago

A conservative force does the same work regardless of the path taken.

ella [17]

A conservative force is a force that when work is done against this force the work done does not depend on the path taken only the initial and final position.

5 0

3 years ago

Read 2 more answers

The propeller of a World War II fighter plane is 2.30 m in diameter. (a) What is its angular velocity in radians per second if i

Sergio [31]

Answer with Explanation:

We are given that

Diameter of fighter plane=2.3 m

Radius= $r=\frac{d}{2}=\frac{2.3}{2}=1.15 m$

a.We have to find the angular velocity in radians per second if it spins=1200 rev/min

Frequency= $\frac{1200}{60}=20 Hz$

1 minute=60 seconds

Angular velocity= $\omega=2\pi f$

Angular velocity= $2\times \frac{22}{7}\times 20=125.7 rad/s$

b.We have to find the linear speed of its tip at this angular velocity if the plane is stationary on the tarmac.

$v=r\omega=1.15\times 125.7=144.56 m/s$

c.Centripetal acceleration= $\omega^2 r=(125.7)^2(1.15)=18170.56 m/s^2$

Centripetal acceleration== $\frac{18170.56\times g}{9.81}=1852.25 g m/s^2$

7 0

3 years ago