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alexandr402 [8]
3 years ago
13

Are you doing work holding a heavy piece of furniture in place

Physics
1 answer:
Crazy boy [7]3 years ago
8 0

Answer:

No

Explanation:

Work Done is Force x Distance.

You are exerting a force to hold the piece of furniture in place, however work is not being done as you are not moving it. (as you are holding it in place)

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A 0.50-kg mass attached to the end of a string swings in a vertical circle (radius 2.0 m). When the mass is at the highest point
il63 [147K]

Answer:

31.1 N

Explanation:

m = mass attached to string = 0.50 kg

r = radius of the vertical circle = 2.0 m

v = speed of the mass at the highest point = 12 m/s

T = force of the string on the mass attached.

At the highest point, force equation is given as

T + mg =\frac{mv^{2}}{r}

Inserting the values

T + (0.50)(9.8) =\frac{(0.50)(12)^{2}}{2}

T = 31.1 N

7 0
3 years ago
Read 2 more answers
If 10. joules of work must be done to move 2.0 coulombs of charge from point A to point B in an electric field, the potential di
masya89 [10]
The equation for work (W) done by an electric field is:

W = qΔV

where q is the magnitude of the charge and ΔV is the potential difference. The question gives you W and q, so plug n' play to find ΔV:

10 = 2ΔV
ΔV = 5
8 0
3 years ago
A 10 kg blue cart moving to the right at 25 m/s collides with a 17 kg red cart moving in the opposite direction at 16 m/s. If, a
scoray [572]

Answer:

24.8m/s

Explanation:

Given data

m1= 10kg

u1=25m/s

m2=17kg

u2=16m/s

v1=10m/s

v2=??

Applying the conservation of linear momentum

m1u1+m2u2=m1v1+m2v2

substitute

10*25+17*16=10*10+17*v2

250+272=100+17v2

522=100+17v2

522-100=17v2

422=17v2

Divide both sides by 17

v2= 422/17

v2= 24.8 m/s

Hence the velocity of the red cart is 24.8m/s in the opposite direction of the blue cart

3 0
2 years ago
20 POINTS!!!! WILL MARK BRAINLIEST!!!!!
oee [108]
1. Because of gravity....

2. No you either feel still ( gravity) or is actually in movement....

please vote my answer branliest! Thanks
8 0
3 years ago
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What is the force of gravity (from the Earth) on the 700kg satellite if it’s 10km above the Earths surface?
joja [24]

Answer:

The force of gravity on a 700 kg satellite if its 10 km above Earth's surface is given by

    = {\frac{(6.674\times 10^{-11}N. m^2/kg^2)(5.97\times 10^{24}kg) }{(10\times10^3)^2} = 3984378 m / s^{2}

Explanation:

The force of gravity on a 700 kg satellite if its 10 km above Earth's surface is given by

    = {\frac{(6.674\times 10^{-11}N. m^2/kg^2)(5.97\times 10^{24}kg) }{(10\times10^3)^2} = 3984378 m / s^{2}

3 0
3 years ago
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