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Sergeeva-Olga [200]
3 years ago
7

What happens to your image as you walk away from a plane mirror on a wall? It gets smaller.

Physics
1 answer:
IRISSAK [1]3 years ago
6 0

Explanation:

It stays the same size

........

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A glider of mass 0.170 kg moves on a horizontal frictionless air track. It is permanently attached to one end of a massless hori
vivado [14]

Answer:

Answer:

a.  1.594 m/s = v

b. 1.274 m/s = v

Explanation:

A) First calculate the potential energy stored in the spring when it is compressed by 0.180 m...

U = 1/2 kx²

Where U is potential energy (in joules), k is the spring constant (in newtons per meter) and x is the compression (in meters)

U = 1/2(13.0 N/m)(0.180 m)² = 0.2106 J

So when the spring passes through the rest position, all of its potential energy will have been converted into kinetic energy.  K = 1/2 mv².

 0.2106 J  = 1/2(0.170 kg kg)v²

0.2106 J  = (0.0850 kg)v²

2.808m²/s² = v²

1.594 m/s = v

(B)  When the spring is 0.250 m from its starting point, it is 0.250 m - 0.180 m = 0.070 m past the equilibrium point.  The spring has begun to remove kinetic energy from the glider and convert it back into potential.  The potential energy stored in the spring is:

U = 1/2 kx² = 1/2(13.0 N/m)(0.070 m)² = 0.031J

Which means the glider now has only 0.2106 J  - 0.031J = 0.1796 J of kinetic energy remaining.

K = 1/2 mv²

0.1796 J = 1/2(0.170 kg)v²

0.138 J = (0.0850 kg)v²

1.623 m²/s² = v²

1.274 m/s = v

5 0
3 years ago
In physics, a is a group of related objects that interact with each other and form a complex whole.
just olya [345]

Answer: a system

Explanation: just did the test

4 0
3 years ago
Read 2 more answers
Consider a grill with the lid closed to be a closed system. The propane provides chemical energy. The propane is ignited to prod
katrin [286]

'A' and 'C' talk about energy being created and destroyed.  That can't happen.

'D' trailed off in the middle, and we don't know WHAT it was talking about.

'B' is the only correct statement.


6 0
3 years ago
An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the
garik1379 [7]

(a) 6.04 rev/s

The speed of the ball is given by:

v=\omega r

where

\omega is the angular speed

r is the distance of the ball from the centre of the circle

In situation 1), we have

\omega=8.13 rev/s \cdot 2\pi = 51.0 rad/s

r = 0.600 m

So the speed of the ball is

v=(51.0 rad/s)(0.600 m)=30.6 m/s

In situation 2), we have

\omega=6.04 rev/s \cdot 2\pi = 37.9 rad/s

r = 0.900 m

So the speed of the ball is

v=(37.9 rad/s)(0.900 m)=34.1 m/s

So, the ball has greater speed when rotating at 6.04 rev/s.

(b) 1561 m/s^2

The centripetal acceleration of the ball is given by

a=\frac{v^2}{r}

where

v is the speed

r is the distance of the ball from the centre of the trajectory

For situation 1),

v = 30.6 m/s

r = 0.600 m

So the centripetal acceleration is

a=\frac{(30.6 m/s)^2}{0.600 m}=1561 m/s^2

(c) 1292 m/s^2

For situation 2 we have

v = 34.1 m/s

r = 0.900 m

So the centripetal acceleration is

a=\frac{v^2}{r}=\frac{(34.1 m/s)^2}{0.900 m}=1292 m/s^2

5 0
3 years ago
A runner whose mass is 54 kg accelerates from a stop to a speed of 7 m/s in 3 seconds. (A good sprinter can run 100 meters in ab
Darya [45]

Answer:

a. F=126N

b. E_K=1323J

Explanation:

Given:

m=54kg

v=7 m/s

t= 3s

The runner force average to find given the equations

a.

F=m*a

a=\frac{v}{t}

F=m*\frac{v}{t}=54kg*\frac{7m/s}{3s}

F=126N

b.

Work done by the system by this force so

W=F*d

W=E_K

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

E_K=\frac{1}{2}*54kg*(7m/s)^2

E_K=1323J

6 0
2 years ago
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