Physics, asked by Thanu7479, 1 day ago

a truck moving with velocity of 90 km/h.the driver applies the brakes and the truck takes 5 second to stop .if the mass of the truck along with luggage is 1500kg then calculate the amount of fdorce exerted by the brakes on the truck

Answers

Answered by manohargund4126
0

Answer:

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Explanation:

108 km/h and it takes 4 s to stop after the brakes are applied. Calculate the force exerted by the brakes on the motorcar if its mass along with the passengers is 1000 kg.

108 km/h and it takes 4 s to stop after the brakes are applied. Calculate the force exerted by the brakes on the motorcar if its mass along with the passengers is 1000 kg.Medium

Solution

verified

verifiedVerified by Toppr

verifiedVerified by TopprForce=mass*acceleration

verifiedVerified by TopprForce=mass*accelerationf=m×

verifiedVerified by TopprForce=mass*accelerationf=m× t

verifiedVerified by TopprForce=mass*accelerationf=m× tv−u

verifiedVerified by TopprForce=mass*accelerationf=m× tv−u

verifiedVerified by TopprForce=mass*accelerationf=m× tv−u

verifiedVerified by TopprForce=mass*accelerationf=m× tv−u f=1000×

verifiedVerified by TopprForce=mass*accelerationf=m× tv−u f=1000× 4

verifiedVerified by TopprForce=mass*accelerationf=m× tv−u f=1000× 40−108×

verifiedVerified by TopprForce=mass*accelerationf=m× tv−u f=1000× 40−108× 18

verifiedVerified by TopprForce=mass*accelerationf=m× tv−u f=1000× 40−108× 185

verifiedVerified by TopprForce=mass*accelerationf=m× tv−u f=1000× 40−108× 185

verifiedVerified by TopprForce=mass*accelerationf=m× tv−u f=1000× 40−108× 185

verifiedVerified by TopprForce=mass*accelerationf=m× tv−u f=1000× 40−108× 185

verifiedVerified by TopprForce=mass*accelerationf=m× tv−u f=1000× 40−108× 185

verifiedVerified by TopprForce=mass*accelerationf=m× tv−u f=1000× 40−108× 185 =1000×(−7.5)

verifiedVerified by TopprForce=mass*accelerationf=m× tv−u f=1000× 40−108× 185 =1000×(−7.5)∴f=−7500N

verifiedVerified by TopprForce=mass*accelerationf=m× tv−u f=1000× 40−108× 185 =1000×(−7.5)∴f=−7500Nopposing force is denoted by negative sign.

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