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Olenka [21]
3 years ago
14

On a nice summer day,Kim takes her niece Madison for a walk in her stroller.If they start from rest and accelerate at a rate of

0.5m/s2,what is their velocity after they have traveled 5.0
Physics
1 answer:
11111nata11111 [884]3 years ago
6 0

2.5m/s

Explanation:

Given parameters:

Initial velocity = 0m/s

Acceleration = 0.5m/s²

time of travel = 5s

Solution:

Final velocity = ?

Solution:

Acceleration can be defined as the change in velocity with time:

          Acceleration = \frac{Final velocity - Initial velocity}{time}

  From the equation above, the unknown is final velocity:

Final velocity - initial velocity = Acceleration x time

 since initial velocity = 0

   Final velocity = 0.5 x 5 = 2.5m/s

Learn more:

Acceleration brainly.com/question/3820012

#learnwithBrainly

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A small truck has a mass of 2145 kg. How much work is required to decrease the speed of the vehicle from 25.0 m/s to 12.0 m/s on
MAXImum [283]

Answer:

The work required is -515,872.5 J

Explanation:

Work is defined in physics as the force that is applied to a body to move it from one point to another.

The total work W done on an object to move from one position A to another B is equal to the change in the kinetic energy of the object. That is, work is also defined as the change in the kinetic energy of an object.

Kinetic energy (Ec) depends on the mass and speed of the body. This energy is calculated by the expression:

Ec=\frac{1}{2} *m*v^{2}

where kinetic energy is measured in Joules (J), mass in kilograms (kg), and velocity in meters per second (m/s).

The work (W) of this force is equal to the difference between the final value and the initial value of the kinetic energy of the particle:

W=\frac{1}{2} *m*v2^{2}-\frac{1}{2} *m*v1^{2}

W=\frac{1}{2} *m*(v2^{2}-v1^{2})

In this case:

  • W=?
  • m= 2,145 kg
  • v2= 12 \frac{m}{s}
  • v1= 25 \frac{m}{s}

Replacing:

W=\frac{1}{2} *2145 kg*((12\frac{m}{s} )^{2}-(25\frac{m}{s} )^{2})

W= -515,872.5 J

<u><em>The work required is -515,872.5 J</em></u>

3 0
3 years ago
A ball is thrown horizontally from the top of a building 14.9 m high. The ball strikes the ground at a point 107 m from the base
umka2103 [35]

Answer:

1) t=1.743 sec

2)Vo=61.388  m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the  initial velocity of the ball that is=61.388  m/sec

4)Vf=17.08 m/s

Explanation:

1)From second equation of motion we get

h=Vit+(1/2)gt^2

here in case(a): Vi=0 m/s,h=14.9m,,put these values in above equation to find the time the ball is in motion

14.9=(0)*t+(1/2)(9.8)t^2

t^2=14.9/4.9

t^2=3.040 sec

t=1.743 sec

2) s=Vo*t

Putting values we get

107=Vo*1.743

Vo=61.388  m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the  initial velocity of the ball that is=61.388  m/sec

4)From third equation of motion we know that

Vf^2-Vi^2=2gh

here Vi=0 m/s,h=14.9 m

Vf^2=Vi^2+2gh=0+2(9.8)(14.9)

Vf^2=292.04

Vf=17.08 m/s

8 0
3 years ago
The unit light-year is a measure of
Virty [35]
One light-year is the distance that light travels in vacuum
in one year. It's a unit of distance.
3 0
3 years ago
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Calculate the kinetic energy in joules of a 1,500 kg car that is moving at a speed of 42 km/h
Rzqust [24]
Data:
KE (Kinetic Energy) = ? (Joule)
m (mass) = 1500 Kg 
v (speed) = 42 Km/h
converting to m/s (42 / 3.6), we have: v (speed) = 11.6 m/s

Formula:
K_{E} =   \frac{1}{2} m*v^2

Solving:
K_{E} = \frac{1}{2} m*v^2
K_{E} =  \frac{1}{2} *1500*(11.6)^2
K_{E} = \frac{1}{2} *1500*134.56
K_{E} =  \frac{201840}{2}
\boxed{\boxed{K_{E} = 100920\:Joule}}\end{array}}\qquad\quad\checkmark





4 0
3 years ago
The muscular system and the skeletal system of the human body work together to
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 A. Allow movement.  
Muscles connect to your skeleton and they contract and move the skeleton along. <span>They help the process of movement happen in a smoother manner.</span>
4 0
3 years ago
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