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MaRussiya [10]
3 years ago
12

A 1000 kg car pushes a 2000 kg car that has a dead battery. When the driver steps on the accelerator, the drive wheels of the ca

r push backward against the ground with a force of 4500 N.
Physics
1 answer:
butalik [34]3 years ago
4 0
So what is the question
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In a game of tug of war, team one pulls to the right with a force of 500 newtons and team two pulls to the left with a force of
seraphim [82]

Answer:

Explanation:

There is no set way to do this. All you have to do is define left and right. Left will be minus and right will be the opposite --- plus.

That is completely arbitrary. It could be the other way around. It does not matter.

Left is minus so: - 600 N   is the force going left.

Right plus so: + 500 N

Now just add.

Net Force = +500 - 600

Net Force = - 100 N

So the Net Force is - 100 N going to the left.

8 0
3 years ago
What charges are needed in the objects to attract both objects?
kvasek [131]

Answer:

both

Explanation:

8 0
3 years ago
Read 2 more answers
Bro what does this mean ? I don’t get this
zhannawk [14.2K]

CO2 is carbon dioxide a colorless gas

3 0
3 years ago
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A 16 g piece of Styrofoam carries a net charge of -8.6 µC and floats above the center of a large horizontal sheet of plastic tha
PilotLPTM [1.2K]

Answer:

the charge per unit area on the plastic sheet is - 3.23 x 10⁻⁷ C/m²

Explanation:

given information:

styrofoam mass, m = 16 g = 0.016 kg

net charge, q = - 8.6 μC

to calculate the charge per unit area on the plastic sheet, we can use the following equation:

F_{e} = mg

where

F_{e} = the force between the electric field

m = mass

g = gravitational force

F_{e} =qE

where

q = charge

E = electric field

and

E = σ/2ε₀

where

ε₀ = permitivity

thus

F_{e} =qE

mg = qσ/2ε₀

σ = (2mg ε₀)/q

  = 2 (0.016) (9.8)  (8.85 x 10⁻¹²)/( - 8.6 x 10⁻⁶)

  = - 3.23 x 10⁻⁷ C/m²

8 0
3 years ago
61. A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A
Mademuasel [1]

Answer:

Part a)

percentage = 21.3%

Part b)

percentage = 2.13 \times 10^{-5}%

Explanation:

As we know that total power used in the room is given as

P = P_1 + P_2 + P_3 + P_4

here we have

P_1 = (110)(3) = 330 W

P_2 = 100 W

P_3 = 60 W

P_4 = 3 W

P = 330 + 100 + 60 + 3

P = 493 W

Part a)

Since power supply is at 110 Volt so the current obtained from this supply is given as

110\times i = 493

i = 4.48 A

now resistance of transmission line

R = \frac{\rho L}{A}

R = \frac{(2.8 \times 10^{-8})(10\times 10^3)}{\pi(4.126\times 10^{-3})^2}

R = 5.23 \ohm

now power loss in line is given as

P = i^2 R

P = (4.48)^2(5.23)

P = 105 W

Now percentage loss is given as

percentage = \frac{loss}{supply} \times 100

percentage = \frac{105}{493} \times 100

percentage = 21.3%

Part b)

now same power must have been supplied from the supply station at 110 kV, so we have

110 \times 10^3 (i ) = 493

i = 4.48\times 10^{-3} A

now power loss in line is given as

P = i^2 R

P = (4.48 \times 10^{-3})^2(5.23)

P = 1.05 \times 10^{-4} W

Now percentage loss is given as

percentage = \frac{loss}{supply} \times 100

percentage = \frac{1.05 \times 10^{-4}}{493} \times 100

percentage = 2.13 \times 10^{-5}%

6 0
3 years ago
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