A 23.8 kg space rock is pushed with a force and accelerates at 8.97 m/s/s. How much force was applied?

Suppose the equation of the velocity of a car at any given time is Vx= 30m/s+(2.5m/s^3)t^2 Find the change in the velocity of th

mario62 [17]

Answer:

time=4, answer choice A

time=5.5, answer choice B

Explanation:

This is taken from the kinematic equation, v=vo+(1/2)at^2

Velocity at time 4 = 30+(2.5)(4)^2

Velocity at time 5.5 = 30+(2.5)(5.5)^2

Velocity at time 4 seconds = 70 m/s

Velocity at time 5.5 seconds = 105.63 m/s

3 0

3 years ago

_____ is amount of “ground” an object moves from its starting point and ______ is how much “ground” and object covers in total.

laiz [17]

Displacement is the amount of "ground" and object moves from it starting point and distance is how much "ground" an object covers in total. Displacement by definition is moving of something from its place or position and distance by definition is the amount of space between two things (in this case the starting position and ending position).ΥΤ-ΤΟΝΙХWΙLSΟΝ

7 0

3 years ago

Read 2 more answers

How much force would I need to through my 10 pound dog against the wall to kill it?

Olin [163]

Answer:

Probably dont kill your dog

Explanation:

4 0

4 years ago

A car starts from rest at a stop sign. It accelerates at 2.0 m/s2 for 6.2 seconds, coasts for 2.5 s , and then slows down at a r

BigorU [14]

Answer:

the distance between the stop signs is 120.7 m.

Explanation:

The car moved in three stages;

(1) It accelerates from rest at 2.0 m/s² for 6.2 seconds

(2) it moved at a constant speed for 2.5 s

(3) it finally decelerate at the rate of 1.5m/s²

(1) The distance moved by the car during the first stage;

s₁ = ut + ¹/₂at²

s₁ = 0 + ¹/₂ (2)(6.2)²

s₁ = 38.44 m

(2) The distance moved by the car during the second stage;

calculate the constant speed of the car,

v = u + at

v = 0 + 2 x 6.2

v = 12.4 m/s

The distance moved by the car as it coasts for 2.5s: s₂ = vt

s₂ = 12.4 x 2.5

s₂ = 31 m

(3) The distance moved by the car during the third stage;

When the car stops, the final velocity is zero.

v² = u² + 2as₃

a = -1.5 m/s², since the car slowed down or decelerated.

0 = 12.4² + (2 x - 1.5)s₃

0 = 153.76 - 3s₃

3s₃ = 153.76

s₃ = 153.76 / 3

s₃ = 51.253 m

The total distance moved by the car from the start to stop = s₁ + s₂ + s₃

d = 38.44 m + 31 m + 51.253 m

d = 120.7 m

Therefore, the distance between the stop signs is 120.7 m.

8 0

3 years ago

A rigid container equipped with a stirring device contains 1.5 kg of motor oil. Determine the rate of specific energy increase w

jenyasd209 [6]

To solve this problem we will apply the first law of thermodynamics which details the relationship of energy conservation and the states that the system's energy has. Energy can be transformed but cannot be created or destroyed.

Accordingly, the rate of work done in one cycle and the heat transferred can be expressed under the function,

$\dot{U} = \dot{Q}-\dot{W}$