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Westkost [7]
3 years ago
13

a driver travels 135 km, east in 1.5 h, stops for 45 minutes for lunch, and then resumes driving for the next 2.0 h through a di

splacement of 215 km, east. what is the driver’s average velocity?
Physics
1 answer:
daser333 [38]3 years ago
7 0

Answer:

v = 98.75 km/h

Explanation:

Given,

The distance driver travels towards the east, d₁ = 135 km

The time period of the travel, t₁ = 1.5 h

The halting time, tₓ = 46 minutes

The distance driver travels towards the east, d₂ = 215 km

The time period of the travel, t₁ = 2 h

The average speed of the vehicle before stopping

                                    v₁ = d₁/t₁

                                        = 135/1.5

                                       = 90 km/h

The average speed of vehicle after stopping

                                    v₂ = d₂/t₂

                                         = 215/2

                                        = 107.5 km/h

The total average velocity of the driver

                                      v = (v₁ +v₂) /2

                                         = (90 + 107.5)/2

                                         = 98.75 km/h

Hence, the average velocity of the driver, v = 98.75 km/h

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Answer:

31677.2 lb

Explanation:

mass of hammer (m) = 3.7 lb

initial velocity (u) = 5.8 ft/s

final velocity (v) = 0

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acceleration due to gravity (g) 32 ft/s^{2}

force = m x ( a + g )

where

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  • g is the acceleration due to gravity = 32 ft/s^{2}
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       from v = u + at

       a = (v-u)/ t

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we can substitute all required values into force= m x (a+g)

force = 3.7 x (8529.4 + 32) = 31677.2 lb

       

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3 years ago
A force of 25N acts on a mass of 5.0kg, initially at rest. Calculate the distance travelled before achieving a velocity of 20m/s
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Answer:

40m

Explanation:

let's calculate the acceleration first

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rearranging to find acceleration:

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we can now use the formula v^2 = u^2 + 2as where v = final velocity, u = initial velocity, a = acceleration and s = distance

rearranging v^2 = u^2 + 2as the distance is

s = (v^2 - u^2) ÷ 2a

v = 20, u = 0, a = 5

s = (20^2 - 0^2) ÷ (2 × 5) = 40m

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Explanation:

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