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den301095 [7]
3 years ago
12

What is the mass of a 735-N weight on Earth?

Physics
1 answer:
Allisa [31]3 years ago
3 0
W = mg,      Assuming g ≈ 9.8 m/s² on the earth surface.

735 N =  m* 9.8

735/9.8 =  m

75 = m

Mass , m = 75 kg.  B.
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. If measurements of a gas are 75L and 300 kilopascals and then the gas is measured a second time and found to be 50L, describe
julia-pushkina [17]
By Boyle's law:

P₁V₁ = P₂V₂

300*75 = P<span>₂*50

</span>P<span>₂*50= 300*75
</span>
P<span>₂ = 300*75/50 = 450
</span>
P<span>₂ = 450 kiloPascals.

The pressure has increased as a result of compression of gas.

Boyle's Law supports this observation.</span>
3 0
3 years ago
Read 2 more answers
Given: Q1 = +10 uc = 1.0 x 10-5C
ser-zykov [4K]

Answer:

-0.038 N

Explanation:

F=K Q1 Q2/r^2 by COULOMB'S LAW

F= 9×10^9×1×10^-5×-1.5×10^-5/(6)^2

F= -0.038 N

5 0
3 years ago
Can someone please give me the (Answers) to this? ... please ...<br><br> I need help….
IceJOKER [234]

#1.

<em>Car </em>1<em> weighs </em>300 kilograms<em> and is moving right at </em>3 meters per second (m/s)

  • v1 (before) = 3 m/s

  • v2 (before) = -1 m/s

  • v1 (after) = 0.5 m/s

#2.

Law of conservation of momentum

momentum before collorion = momentim after collosion

MV + mv = MV' + mv'

1500x25+ 1000x5

37500 + 15000

6 0
2 years ago
Anyone know 1-5? Will give brainliest
baherus [9]

Answer:

1. 200 metres West

2. Dividing distance by time

3. Speed was unchanged

4. 1800 metres

5. 15 seconds

Explanation:

Running 500 metres West puts you 500 Metres west from the start. Then running 300 metres east puts you 200metres from where you started.

Since displacement is the distance you are from your original position 200 Metres West is the answer

2. Distance = Speed x Time

Rearrange that to get Speed = Distance / Time

3. Acceleration is when the rate of increase or decrease of speed or the direction is changing. When the speed or sirection dont change acceleration is 0

4. Distance = Speed x Time

60 x 30 = 1800 metres

5. Time = Distance / Speed

300 / 20 = 15 seconds

7 0
3 years ago
A plastic rod has been bent into a circle of radius R = 8.00 cm. It has a charge Q1 = +2.70 pC uniformly distributed along one-q
vovikov84 [41]

Answer:

Part a)

V = -1.52 V

Part b)

V = -1.16 V

Explanation:

Part a)

Electric potential is a scalar quantity

so here we can say that total potential due to a ring on its center is given as

V = \frac{kQ}{R}

here we know that

Q = Q_1 - 6Q_1

Q = - 5 Q_1

Q_1 = 2.70 pC

now we have

V = \frac{(9\times 10^9)(-5\times 2.70 \times 10^{-12})}{0.08}

V = -1.52 V

Part b)

Potential on the axis of the ring is given as

V = \frac{kQ}{\sqrt{r^2 + R^2}}

V = \frac{(9\times 10^9)(-5\times 2.70\times 10^{-12})}{\sqrt{0.08^2 + 0.0671^2}}

V = -1.16 V

8 0
3 years ago
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