Physics homework help.

During a football game, a receiver has just caught a pass and is standing still. Before he can move, a tackler, running at a vel

amm1812

Answer:

Explanation:

Using the law of conservation of momentum which states that 'the sum of momentum of bodies before collision is equal to their sum after collision. The bodies will move together with a common velocity after collision.

Momentum = Mass * Velocity

<u>Before collision;</u>

Momentum of receiver m1u1= 0 kgm/s (since the receiver is standing still)

Momentum of the tackler

m2u2 = 2.60*122 = 317.2 kgm/s

where m2 and u2 are the mass and velocity of the tacker respectively.

Sum of momentum before collision = 0+317.2 = 317.2 kgm/s

<u>After collision</u>

Momentum of the bodies = (m1+m2)v

v = their common velocity

m1 = mass of the receiver

Momentum of the bodies = (122+m1)(1.30)

Momentum of the bodies = 158.6+1.30m1

According to the law above;

317.2 = 158.6+1.30m1

317.2-158.6 = 1.30m1

158.6 = 1.30m1

m1 = 158.6/1.30

m1 = 122kg

The mas of the receiver is 122kg

5 0

3 years ago

A physical science teacher wants to demonstrate Newton's laws. She has 2 skateboards and takes them out to a flat parking lot. T

insens350 [35]

Answer:

C

Explanation:

if the mass is smaller the acceleration is larger

3 0

3 years ago

In a circus act, a uniform board (length 3.00 m, mass 25.0 kg ) is suspended from a bungie-type rope at one end, and the other e

tester [92]

Answer:

Force of Rope = 122.5 N

Force of Rope = 480.2N

Explanation:

given data

length = 3.00 m

mass = 25.0 kg

clown mass = 79.0 kg

angle = 30°

solution

we get here Force of Rope on with and without Clown that is

case (1) Without Clown

pivot would be on the concrete pillar so Force of Rope will be

Force of Rope × 3m = (25kg)×(9.8ms²)×(1.5m)

solve it and we get

Force of Rope = 122.5 N

and

case (2) With Clown

so here pivot is still on concrete pillar and clown is standing on the board middle and above the centre of mass so Force of Rope will be

Force of Rope × 3m = (25kg+73kg)×(9.8ms²)×(1.5m)

solve it and we get

Force of Rope = 480.2N

4 0

4 years ago

10. A change in

pogonyaev

Answer:

d velocity will be the one according to me

6 0

3 years ago

A bowling ball with a circumference of 27 in. weighs 14.8 lb and has a radius of gyration of 3.43 in. If the ball is released wi

Marrrta [24]

Answer:

Distance covered is 37.63 ft

Solution:

As per the question:

Circumference of the bowling ball, C = 27 in

Weight of the bowling ball, w = 14.8 lb

Radius of gyration, k = 3.43 in

Velocity of release of the ball, v' = 23 ft/s = 276 in/s

Coefficient of friction, $\mu = 0.19$

Now,

We know that the circumference of the circle is given by:

C = $2\pi R$

27 = $2\pi R$

$R = \frac{27}{2\pi } = 4.29\ in$

Also, in case of rolling, we know that:

Angular velocity, $\omega = \frac{v}{R}$

$v = \omega R$

Now, applying the conservation of angular momentum along the floor:

Initial angular momentum = Final angular momentum

$m\omega' = m\omega + I\omega$

Moment of Inertia, I = $\frac{2}{5}mR^{2} = mk^{2}$

$mv'R = mvR + mk^{2}\times \frac{v}{R}$

$v'R = vR + k^{2}\times \frac{v}{R}$

$276\times 4.29 = 4.29v + 3.43^{2}\times \frac{276}{4.29}$

$1184.04 - 756.903 = 4.29v$

v = 99.56 in/s

Now,

Friction force, f = $\mu mg$

Also, acceleration of the ball can be computed as:

$\mu mg = - ma$

$a = - \mu g = - 0.19\times 386.09 = - 73.357\ in/s^{2}$

Now, the distance, d covered by the ball before rolling without slipping:

$v^{2} = v'^{2} + 2ad$

$99.56^{2} = 276^{2} + 2\times (- 73.357)d$

$99.56^{2} - 276^{2} = 2\times (- 73.357)d$

d = 451.65 in = 37.53 ft

7 0

3 years ago