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3 years ago

A 2,100 kg car is lifted by a pulley. If the cable breaks at 4.50 m, what is the velocity of the car when it hits the ground

1 answer:

4 0

The problem involves the conversion of potential energy to kinetic energy as the object falls from rest. Energy is conserved, so the equation used is:

PEi + KEi = PEf + KEf

Since the object is falling from rest, the initial kinetic energy is zero. Also, since the object hits the ground at its final position, the final potential energy is zero. This leaves:

PEi = KEf
mgh = 1/2 mv^2

*cancel out mass on both sides of the equation

gh = 0.5v^2
v = sqrt(2gh) = sqrt(2*9.81*4.5) = 9.40 m/s --> final ans.

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Acceleration increases over time once a force is applied to the object. Given a force of 10.0 Newtons, which mass will have the

Zolol [24]

Mass is indirectly proportional to acceleration, so, lighter the object greater would be it's acceleration...

A) 0.10 kg is lightest among them, so it's your answer

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3 years ago

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Heat transfer between two substances is affected by specific heat and the

IgorLugansk [536]

The correct answer is D. Amount of time and area of physical contact between the substances.

Explanation:

Heat transfer refers to the flow of thermal energy or heat between two or more objects. This process involves multiple factors and implies heat from the hottest object goes to the coldest one until there is an equilibrium. To begin, heat transfer depends on the amount of thermal energy in the objects because objects must have a different amount of thermal energy for heat to flow.

Besides this, the amount of energy that flows depends on the time and the contact between the substances of objects. Indeed, objects need to be in contact or close to each other for heat to transfer, and the time needs to be enough for the process to occur. For example, if you place a pot over the fire just for a few seconds it is likely the heat transferred is minimal, which does not occur if you leave the pot more time. At the same time if the pot is in close contact with fire more heat will be transferred.-

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3 years ago

10. A boat traveling at 9.5 m/s reduces its acceleration at a rate of 2.5 m/s2. What is the final speed of the boat

Fiesta28 [93]

Answer:

6.75m/s

Explanation:

using the second equation of motion, the time is calculated.

and with the formula a= (v - u)/t

where a is acceleration but in this case it's deceleration (and should be negated as you solve it ) .

v is final velocity

u is initial velocity

t is time taken

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3 years ago

The electric eye that prevents an elevator door from closing when you are in the doorway relies upon which physics principle?

goblinko [34]

It's Photoelectric Effect, I just a test with this same question. I am not good for explaining exactly how, but I was right.

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3 years ago

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A spherical bowling ball with mass m = 3.6 kg and radius R = 0.101 m is thrown down the lane with an initial speed of v = 8.7 m/

luda_lava [24]

Answer:

1) The magnitude of the angular acceleration = 67.92 rad/ $s^{2}$

2) Magnitude of the linear acceleration = 2.744 m/ $s^{2}$

3) How long does it take the bowling ball to begin rolling without slipping = 0.906 s

4) How long does it take the bowling ball to begin rolling without slipping = 6.75 m

5) the final velocity is 6.21 m/s

Explanation:

the given information :

Bowling mass m = 3.6 kg

Radius = 0.101 m

Initial speed $v_{0}$ = 8.7 m/s

Coefficient of kinetic friction μ = 0.28

1) he magnitude of the angular acceleration of the bowling ball is

F = m a

$F_{g}$ = μ N , N = m g

$F_{g}$ = μ m g

1) The magnitude of the angular acceleration of the bowling ball as it slides down the lane:

momen inersia of Bowling ball I = (2/5) m $R^{2}$

torque τ = I α

τ = F R

I α = F R

(2/5) m $R^{2}$ α = μ m g R

α = (5 μ g / 2R) μ g R

= (5 x 0.28 x 9.8/ 2 x 0.101)

= 67.92 rad/ $s^{2}$

2) Magnitude of the linear acceleration of the bowling ball as it slides down the lane

F = - $F_{g}$ , $F_{k}$ is the force of kinetic friction

m a = - μ m g, remove m

the magnitude of linear accelaration is

a = μ g

= (0.28) (9.8)

= 2.744 m/ $s^{2}$

3) The bowling ball takes time to begin rolling without slipping:

The linear speed, $v_{t}$ = $v_{0}$ - a t

$v_{t}$ = $v_{0}$ - μ g t

the angular speed, ω = ω0 + α t

ω = ω0 + (5 μ g/2R ) t

$v_{t}$ = ω R

$v_{0}$ - μ g t = ω0 R + (5 μ g/2R ) t R

7 μ g t/2 = $v_{0}$ + ω0 R

hence,

t = (2 $v_{0}$ + ω0 R)/ 7 μ g

ω0 = 0 (no initial spin), therefore

t = 2 $v_{0}$ / 7 μ g

= 2 x 8.7 / 7 (0.28) (9.8)

= 0.906 s

4) How long it takes for the bowling ball to begin rolling without slipping, S

S = $v_{0}$ t - (1/2) a $t^{2}$

= (8.7) (0.906) - (1/2) (2.744) $0.906^{2}$

= 6.75 m

5) The final velocity

$v_{t}$ = $v_{0}$ - a t

$v_{t}$ = 8.7 - (2.744) (0.906)

$v_{t}$ = 6.21 m/s

4 0

3 years ago