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fiasKO [112]
3 years ago
6

Kevin draws a figure that has 4 sides all sides have the same length his figure has no right angels what figure does he draw

Physics
1 answer:
Svetllana [295]3 years ago
4 0
Diamond/ rhombus/ parallelogram
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If the toy car weighs 250g and has a kinetic energy of 2.0 J, what is its velocity
sergejj [24]

Explanation:

Kinetic energy (J) = 2j

mass= 250g

velocity=?

1kg=1000g

mass= 250/1000

mass=0.25kg

Kinetic energy (J) = ½ x mass (kg) x [velocity]² (m/s)

2=1/2 × 0.25× [velocity]²

2=0.125× [velocity]²

[velocity]² = 2/0.125

[velocity]²=16

velocity= (16)^1/2

velocity= 4 m/s

3 0
3 years ago
he block is released, and it slides 2.0 m (from the point at which it is released) across a horizontal surface before friction s
alex41 [277]

Answer:

0.245

Explanation:

When the block is released, the initial elastic potential energy stored in the spring is entirely converted into kinetic energy of the block.

Therefore, we can calculate the initial speed of the block:

\frac{1}{2}kx^2 = \frac{1}{2}mv^2

where the term on the left is the potential energy and where the term on the right is the kinetic energy, and where

k = 4500 N/m is the spring constant

x = 8.0 cm = 0.08 m is the compression of the spring

m = 3.0 kg is the mass of the block

v is the initial velocity

Solving for v,

v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(4500)(0.08)^2}{3.0}}=3.1 m/s

Then, after the block is released, all its kinetic energy is converted into thermal energy as the block slows down, due to friction. Therefore, the work done by friction is equal to the initial kinetic energy of the block.

The force of friction is

F=\mu mg

where

\mu is the coefficient of friction

g=9.8 m/s^2 is the acceleration of gravity

So the work done by it is (in magnitude)

W=Fd=\mu mg d

where

d = 2.0 m is the distance covered

Therefore,

\frac{1}{2}mv^2 = \mu mg d

And solving for \mu,

\mu = \frac{v^2}{2gd}=\frac{3.1^2}{2(9.8)(2.0)}=0.245

8 0
3 years ago
A force of 140 140 newtons is required to hold a spring that has been stretched from its natural length of 40 cm to a length of
icang [17]

Answer:

The work done in stretching the spring is 0.875 J.

Explanation:

Given that,

Force = 140 N

Natural length = 60-40 = 20 cm

Stretch length of the spring = 65-60 = 5 cm

We need to calculate the spring constant

Using formula of Hooke's law

F= kx

140=k\times20\times10^{-2}

k=\dfrac{140}{20\times10^{-2}}

k=700

We need to calculate the work done

W=\int_{a}^{b}{kx}dx

=\int_{0}^{0.05}{700x}dx

On integration

W=700\times(\dfrac{x^2}{2})_{0}^{0.05}

W=700\times(\dfrac{(0.05)^2}{2}-0)

W=0.875\ J

Hence, The work done in stretching the spring is 0.875 J.

3 0
3 years ago
A ____ is a region where jets of gas from young stars impact and heat the gas surrounding the young star.
podryga [215]

Answer:

Hi, There! my name is Jay And I'm here to help!

<h2>Question</h2>

A ____ is a region where jets of gas from young stars impact and heat the gas surrounding the young star.

<h2>Answer</h2>

bipolar outflow

Explanation:

A bipolar outflow is a stream of matter in two opposing directions from a object, Most likely a star.

-------------------------------------------------------------------------------------------------------------

Hope this Helps!

Take Care!

Have a great day!

-Jay-

3 0
2 years ago
Read 2 more answers
What kind of eclipse occurred at the beginning of the week, bringing a colorful sight to stargazers in the americas, africa, eur
Archy [21]

Answer:

none

Explanation:

4 0
1 year ago
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