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mrs_skeptik [129]
3 years ago
10

Give me two examples of a pushing force and two examples of a pulling force:

Physics
1 answer:
77julia77 [94]3 years ago
3 0

Answer:

A pushing force example could be a button, like on a keyboard.

Another example would be a piston.

An example of a pulling force could be a lever.

Another example could be rope.

Hope this helps! c:

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A 20-kg block slides down a fixed rough curved track The block has a speed of 5 0 m/s after its height above a horizontal surfac
Crank

Answer:

U = 102.8 J (100 J to two significant digits)

Explanation:

potential energy converted = 20(9.8)(1.8) = 352.8 J

kinetic energy at base of track = ½(20)5.0² = 250 J

energy (work) of friction 352.8 - 250 = 102.8 J

8 0
3 years ago
A particle moving along the x-axis has its position described by the function x=(3.00t3−1.00t 2.00)m where t is in s. at t = 4.0
prohojiy [21]
X =(3.00x4.00 x3-1.00t x 2.00) x m

x= (12.00x3- 1.00 x2.00) x m
x= 36.00 -1.00 x 2.00) x m
x = (36.00 -2.00) x m
x =( 34.00) x m
x =34.00   times m
7 0
3 years ago
The amount of steering wheel movement needed to turn will ____________ the faster you go.
Naddika [18.5K]

Answer:

The answer to your question is Decrease

4 0
3 years ago
Read 2 more answers
An electric light is plugged into a 120-V outlet. If the current in the bulb is 0.50 A, how much electrical energy does the bulb
ioda

Answer:

= 54,000 Joules or 54 kJ

Explanation:

Electrical energy is given by the formula;

E = VIt; where V is the potential difference in volts, I is the current and t is the time in seconds.

Therefore;

Electrical energy = 120 V × 0.50 A × 15 ×60 seconds

                            = 54,000 Joules

Thus; the electrical energy is 54,000 joules or 54 kJ

7 0
3 years ago
You are standing a distance of 17.0 meters from the center of a merry-go-round. The merry-go-round takes 9.50 seconds to go comp
GenaCL600 [577]

Answer:

(a)11.24 m/s

(b)7.44 m/s

(c)409 N

(d)539.55\mu

(e) 0

Explanation:

The period for 1 circle 2\pi of the merry go around is 9.5s. It means the angular speed is:

\omega = \theta / t = 2\pi / 9.5 \approx 0.661 rad/s

(a)The speed is

v = \omega * R = 0.661 * 17 = 11.24 m/s

(b) Centripetal acceleration:

a = \frac{v^2}{R} = \frac{11.24^2}{17} = 7.44 m/s^2

(c) Magnitude of the force that keeps you go around at this acceleration

F = ma = 55 * 7.44 = 409 N

(d) let the coefficient of friction by \mu. The frictional force shall be this coefficient multiplied by normal force reverting gravity of the man

F_f = mg\mu = 55*9.81\mu = 539.55\mu

3 0
3 years ago
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