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4 years ago

8

Two football players collide head-on in midair while trying to catch a thrown football. The first player is 98.5 kg and has an i

nitial velocity of 6.05 m/s, while the second player is 119 kg and has an initial velocity of −3.50 m/s. What is their velocity (in m/s) just after impact if they cling together? (Indicate the direction with the sign of your answer.)

1 answer:

romanna [79]4 years ago

7 0

Answer:

$v_f=0.825m/s$

Explanation:

We must use conservation of linear momentum before and after the collision, $p_i=p_f$

Before the collision we have:

$p_i=p_1+p_2=m_1v_1+m_2v_2$

where these are the masses are initial velocities of both players.

After the collision we have:

$p_f=(m_1+m_2)v_f$

since they clong together, acting as one body.

This means we have:

$m_1v_1+m_2v_2=(m_1+m_2)v_f$

Or:

$v_f=\frac{m_1v_1+m_2v_2}{m_1+m_2}$

Which for our values is:

$v_f=\frac{(98.5kg)(6.05m/s)+(119kg)(-3.5m/s)}{(98.5kg)+(119kg)}=0.825m/s$

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If the maximum acceleration that is tolerable for passengers in a subway train is 1.39 m/s2 and subway stations are located 780

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Answer:

The maximum speed a subway train can attain between stations is 32.93 m/s.

Explanation:

Given;

maximum tolerable acceleration = 1.39 m/s²

distance between subway train, d = 780 m

The distance available to accelerate between stations = ¹/₂ x 780 m = 390 m

Apply the following kinematic equation to determine the maximum speed;

v² = u² + 2ad

v² = 0 + 2ad

v² = 2(1.39)(390)

v² = 1084.2

v = √1084.2

v = 32.93 m/s

Therefore, the maximum speed a subway train can attain between stations is 32.93 m/s.

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4 years ago

A rotating turbine generates electricity to power a blow dryer what is the energy transformation

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Electric energy is transformed into mechanical energy of the dryer's engine and heat of the dryer's heater unit.

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4 years ago

A person walking in high heals can damages the floor by making small dimples in the floor since all their weight is concentrated

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Given :

Mass of person, m = 81 kg.

Area of high heal, A = 3.5 cm² = 0.00035 m².

To Find :

The pressure applied by the heal on the ground.

Solution :

We know, pressure is given by :

$P = \dfrac{mass}{Area}\\\\P = \dfrac{81}{0.00035} \ N/m^2\\\\P = 231428.57 \ N/m^2$

Hence, this is the required solution.

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3 years ago

A luggage handler pulls a suitcase of mass 19.6 kg up a ramp inclined at an angle 24.0 ∘ above the horizontal by a force F⃗ of m

Dvinal [7]

(a) 638.4 J

The work done by a force is given by

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the object

$\theta$ is the angle between the direction of the force and the displacement

Here we want to calculate the work done by the force F, of magnitude

F = 152 N

The displacement of the suitcase is

d = 4.20 m along the ramp

And the force is parallel to the displacement, so $\theta=0^{\circ}$ . Therefore, the work done by this force is

$W_F=(152)(4.2)(cos 0)=638.4 J$

b) -328.2 J

The magnitude of the gravitational force is

W = mg

where

m = 19.6 kg is the mass of the suitcase

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting,

$W=(19.6)(9.8)=192.1 N$

Again, the displacement is

d = 4.20 m

The gravitational force acts vertically downward, so the angle between the displacement and the force is

$\theta= 90^{\circ} - \alpha = 90+24=114^{\circ}$

Where $\alpha = 24^{\circ}$ is the angle between the incline and the horizontal.

Therefore, the work done by gravity is

$W_g=(192.1)(4.20)(cos 114^{\circ})=-328.2 J$

c) 0

The magnitude of the normal force is equal to the component of the weight perpendicular to the ramp, therefore:

$R=mg cos \alpha$

And substituting

m = 19.6 kg

g = 9.8 m/s^2

$\alpha=24^{\circ}$

We find

$R=(19.6)(9.8)(cos 24)=175.5 N$

Now: the angle between the direction of the normal force and the displacement of the suitcase is 90 degrees:

$\theta=90^{\circ}$

Therefore, the work done by the normal force is

$W_R=R d cos \theta =(175.4)(4.20)(cos 90)=0$

d) -194.5 J

The magnitude of the force of friction is

$F_f = \mu R$

where

$\mu = 0.264$ is the coefficient of kinetic friction

R = 175.5 N is the normal force

Substituting,

$F_f = (0.264)(175.5)=46.3 N$

The displacement is still

d = 4.20 m

And the friction force points down along the slope, so the angle between the friction and the displacement is

$\theta=180^{\circ}$

Therefore, the work done by friction is

$W_f = F_f d cos \theta =(46.3)(4.20)(cos 180)=-194.5 J$

e) 115.7 J

The total work done on the suitcase is simply equal to the sum of the work done by each force,therefore:

$W=W_F + W_g + W_R +W_f = 638.4 +(-328.2)+0+(-194.5)=115.7 J$

f) 3.3 m/s

First of all, we have to find the work done by each force on the suitcase while it has travelled a distance of

d = 3.80 m

Using the same procedure as in part a-d, we find:

$W_F=(152)(3.80)(cos 0)=577.6 J$

$W_g=(192.1)(3.80)(cos 114^{\circ})=-296.9 J$

$W_R=(175.4)(3.80)(cos 90)=0$

$W_f =(46.3)(3.80)(cos 180)=-175.9 J$

So the total work done is

$W=577.6+(-296.9)+0+(-175.9)=104.8 J$

Now we can use the work-energy theorem to find the final speed of the suitcase: in fact, the total work done is equal to the gain in kinetic energy of the suitcase, therefore

$W=\Delta K = K_f - K_i\\W=\frac{1}{2}mv^2\\v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(104.8)}{19.6}}=3.3 m/s$

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3 years ago

If heat Q is required to increase the temperature of a metal object from 4 ∘C to 7∘C, the amount of heat necessary to increase i

alisha [4.7K]

Answer:

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Explanation:

Case 1 :

$m$ = mass of the metal

$c$ = specific heat of the metal

$\Delta T$ = Change in temperature = 7 - 4 = 3 C

Amount of heat required for the above change of temperature is given as

$Q = m c \Delta T\\Q = m c (7 - 4)\\Q = 3 m c$

Case 2 :

$m$ = mass of the metal

$c$ = specific heat of the metal

$\Delta T$ = Change in temperature = 19 - 7 = 12 C

Amount of heat required for the above change of temperature is given as

$Q' = m c \Delta T\\Q' = m c (12)\\Q' = (4)(3 m c)\\Q' = 4Q$

Hence the correct choice is

4Q

6 0

3 years ago

Read 2 more answers