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JulijaS [17]
3 years ago
12

A car accelerates uniformly in a straight line

Physics
1 answer:
erik [133]3 years ago
3 0

Answer:

21.59 m/s

Explanation:

recall that one of the equations of motions can be expressed as

v² = u² + 2as

where,

v = final velocity (we are asked to find this)

u = initial velocity = 0m/s (because it says that it starts from rest)

a = acceleration = 3.7m/s²

s = distance travelled = 63 m

simply substitute the known values above into the equation:

v² = u² + 2as

v² = 0² + 2(3.7)(63)

v² = 466.2

v = √466.2

v = 21.59 m/s

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A skateboarder initially has 5 kJ of kinetic energy. As she freewheels along a flat section of path, she does 400 J of work agai
sweet [91]

The final kinetic energy of the skateboarder after she freewheels and did work against friction on the flat section of the path is 4,600 J.

<h3>Conservation of energy</h3>

The final kinetic energy of the stakeboarder is determined by applying the principle of conservation of energy as shown below;

ΔK.E = -W

K.Ef - K.Ei = -W

where;

  • K.Ef is the final kinetic energy
  • K.Ei is the initial kinetic energy
  • W is work done

K.Ef = K.Ei - W

K.Ef = 5,000 J - 400 J

K.Ef = 4,600 J

Thus, the final kinetic energy of the skateboarder is 4,600 J.

Learn more about kinetic energy here: brainly.com/question/25959744

4 0
2 years ago
How does the direction of current flow in the coil affect the orientation of the magnetic field produced by the electromagnet
Lynna [10]

Answer:

The magnetic field produced by an electric current is always oriented perpendicular to the direction of the current flow. And.Direction of magnetic field is governed by the 'right hand thumb rule, The right hand rule states that: to determine the direction of the magnetic force on a positive moving charge, ƒ, point the thumb of the right hand in the direction of v, the fingers in the direction of B, and a perpendicular to the palm points in the direction of Force . Similar to the situation with electric field lines, the greater the number of lines (or the closer they are together) in an area the stronger the magnetic field.

8 0
3 years ago
What is the gauge pressure at point A when spigot C is closed?
Nesterboy [21]

Given,

\begin{gathered} l_1=0.2m \\ l_2=0.07m \\ h=0.47m \\ \rho=\frac{1g}{cm^3} \end{gathered}

Atm

3 0
2 years ago
A particle undergoes two displacements. The first has a
Novay_Z [31]
First you need to draw the picture of the problem to better understand it. Like the one bellow. 

In this task you have 2 sides of triangle and we can calculate angle between them. Angle between them is 120 - 35 = 85 degrees.

Once you have those 3 variables you can calculate third side of triangle using cosine law.
a - second displacement
b - first displacement
c- resultant displacement.

a^{2} =  b^{2}+ c^{2}-2*b*c*cos85
now we just need to calculate this.
a^{2} = 38439.458
a = 196

now, we use cosine law again to find the angle between second and first displacement.
\alpha = 45.36 degrees

The angle marked with "?" in the graph is our direction angle. We will call it \beta

\beta =90-30-45.36 = 14.64

Second displacement has magnitude of 196 and a direction of -14.64 with positive x axis
3 0
3 years ago
A small block of mass m = 0.032 kg can slide along the frictionless loop-the-loop, with loop radius R = 12 cm. The block is rele
Lunna [17]

Answer:

Part a)

W = 0.15 J

Part b)

W = 0.11 J

Part c)

U = 0.19 J

Part d)

U = 0.038 J

Part e)

U = 0.075 J

Part f)

It is independent of the speed of the object so all part answers will remain the same

Explanation:

Part a)

As we know that Point P is at height 5R while point Q is at height R

so the work done by gravity from P to Q is given as

W = mg(5R - R)

W = 0.032(9.8)(4)(0.12)

W = 0.15 J

Part b)

When it reaches to the top of the loop then its final height from ground is

h = 2R

so work done from P to Q is given as

W = mg(5R - 2R)

W = 3mgR

W = 0.11 J

Part c)

Potential energy at P point is given as

U = mgH

U = 0.032(9.8)(5)(0.12)

U = 0.19 J

Part d)

Potential energy at Q point is given as

U = mgH

U = 0.032(9.8)(0.12)

U = 0.038 J

Part e)

Potential energy at top point is given as

U = mgH

U = 0.032(9.8)(2)(0.12)

U = 0.075 J

Part f)

Since all the answer from part a) to part e) depends only upon the position of the object.

So here we can say that it is independent of the speed of the object so all part answers will remain the same

8 0
4 years ago
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