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3 years ago

If an object has zero acceleration, does it have to have zero velocity?

Physics

1 answer:

adelina 88 [10]3 years ago

3 0

Answer:

Yes, the velocity would also be zero.

Explanation:

Acceleration is the change in velocity over time, therefore, there has to be a change in velocity for something to accelerate. which means without acceleration, the object has no velocity.

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A 91.5 kg football player running east at 2.73 m/s tackles a 63.5 kg player running east at 3.09 m/s. what is their velocity aft

pentagon [3]

Answer:

2.877 m/s

Explanation:

According to the laws of conservation of linear momentum,

the momentum of the moving objects before impact is equal to the momentum of the objects after impact (Assuming no external forces were applied)

Let both players are tackled and moving in V velocity

M and m - masses of the players
U and u - velocities of them respectively (both velocities are towards east direction )

momentum before impact = momentum after impact

→MU + →mu = →(M+ m )v

91.5 * 2.73 + 63.5 * 3.09 = (91.5 + 63.5) * V

→V = 2.877 m/s (To East)

3 0

3 years ago

Read 2 more answers

Explain why liquid particles at a high pressure would need more energy to change to a gas than liquid particles at a low pressur

jenyasd209 [6]

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8 0

3 years ago

Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.

SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

(b) m1 = 915kg;

m2 = 605kg;

m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

$F_{pull}=m_{T}.a$

$m_{T}=\frac{F_{pull}}{a}$

$m_{T}=\frac{3615}{1.516}$

$m_{T}=2384.5$

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

For $m_{1}$ :

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

For $m_{2}$ :

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

$m_{2} = \frac{F_{23}-F_{12}}{a}$

$m_{2} = \frac{2304-1387}{1.516}$

$m_{2}$ = 605kg

For $m_{3}$ :

$m_{3} = m_{T} - (m_{1}+m_{2})$

$m_{3} = 2384.5-1520.0$

$m_{3}$ = 864.5kg

8 0

3 years ago

A street light is on top of a 9 foot pole. Joe, who is 3 feet tall, walks away from the pole at a rate of 4 feet per second. At

Gekata [30.6K]

Answer:2 ft/s

Explanation:

Given

Length of Pole is 9 ft

Length of Joe is 3 ft

Joe walks away from Pole at the rate 4 ft/s

Let Joe is x m away from Pole so its shadow length is x'

From Similar triangle concept

$\frac{x'}{x+x'}=\frac{3}{9}$

3x'=x+x'

x=2x'

and it is given $\frac{\mathrm{d} x}{\mathrm{d} t}=4 ft/s$

Differentiating

$\frac{\mathrm{d} x}{\mathrm{d} t}=2\frac{\mathrm{d} x'}{\mathrm{d} t}$

$4=2\times \frac{\mathrm{d} x'}{\mathrm{d} t}$

$\frac{\mathrm{d} x'}{\mathrm{d} t}=2 ft/s$

6 0

3 years ago

A ball is thrown upward with a speed of 28.2 m/s.A. What is its maximum height?B. How long is the ball in the air?C. When does t

Ede4ka [16]

Answer:

(A) The maximum height of the ball is 40.57 m

(B) Time spent by the ball on air is 5.76 s

Explanation:

Given;

initial velocity of the ball, u = 28.2 m/s

(A) The maximum height

At maximum height, the final velocity, v = 0

v² = u² -2gh

u² = 2gh

$h = \frac{u^2}{2g}\\\\h = \frac{(28.2)^2}{2*9.8}\\\\h = 40.57 \ m$

(B) Time spent by the ball on air

Time of flight = Time to reach maximum height + time to hit ground.

Time to reach maximum height = time to hit ground.

Time to reach maximum height is given by;

v = u - gt

u = gt

$t = \frac{u}{g}$

Time of flight, T = 2t

$T = \frac{2u}{g}\\\\ T = \frac{2*28.2}{9.8}\\\\ T = 5.76 \ s$

As the ball moves upwards, the speed drops, then the height of the ball when the speed drops to 12m/s will be calculated by applying the equation below.

v² = u² - 2gh

12² = 28.2² - 2(9.8)h

12² - 28.2² = - 2(9.8)h

-651.24 = -19.6h

h = 651.24 / 19.6

h = 33.23 m

Thus, at 33.23 m the speed will be 12 m/s

6 0

3 years ago