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ehidna [41]
3 years ago
11

6.) What force would evetually cause the ball to come to a stop after leaving the ramp?​

Physics
1 answer:
irina1246 [14]3 years ago
4 0

Answer:

The force in question is friction. There is also friction between the ground and part of the ball that touches the ground as it rolls. The friction force acts in the opposite direction to the motion of the ball, slowing it and eventually stopping it.

Explanation:

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the answer is A *but I am not sure*

6 0
3 years ago
The weight of a block on the inclined plane is 500 N and the angle of incline is 30 degrees. What is the magnitude of the force
yanalaym [24]

Answer:

250 N

433 N

Explanation:

N = Normal force by the surface of the inclined plane

W = Weight of the block = 500 N

f = static frictional force acting on the block

Parallel to incline, force equation is given as

f = W Sin30

f = (500) Sin30

f = 250 N

Perpendicular to incline force equation is given  

N = W Cos30

N = (500) Cos30

N = 433 N

3 0
3 years ago
On a cold winters day if you left a cup of water sitting outside it could freeze heat is transferred out of the water describe t
valkas [14]
The water molecules would slow down, and as they slow down, the heat created from their movement would cease.
4 0
4 years ago
The open-circuit voltage of a car battery is measured to be 12 V. During engine startup, the battery delivers 10 A to the starte
dmitriy555 [2]

Answer:

- the Thevenin equivalent circuit for the battery is uploaded below

- the battery delivered 117 watts of power to the starter motor

Explanation:

Given the data in the equation

diagram of the Thevenin equivalent circuit for the battery is uploaded below.

Current I = 10 A

Voltage 1 = 12 V

voltage 2 = 11.7 v

R = (V1 - V2) / I

R = (12-11.7)/10

R = 0.3 / 10

R = 0.03Ω

Thevenin equivalent circuit

R_{L} = V2 / I = 11.7 / 10

R_{L} = 1.17Ω

so, power delivered to the starter motor will be;

p = (V2)² / R_{L}

P = ( 11.7 V )² / 1.17Ω

p =  136.89 / 1.17

p =  117 watts

Therefore, the battery delivered 117 watts of power to the starter motor

8 0
3 years ago
How much heat is needed to raise the temperature of 2 kg of copper from 20º to 30ºC. The specific heat of copper is 390 J/kgºC.
ryzh [129]

Answer:

7800 J

Explanation:

Heat needed = mass of copper x specific heat of copper x change in temperature  

Change in temperature = 30ºC - 20ºC = 10ºC

Specific heat of copper = 390 J/kgºC

Mass of copper = 2 Kg

Substituting the given values in above equation, we get –  

Heat needed = 2 Kg x 390 J/kgºC x 10ºC

= 7800 J

3 0
4 years ago
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