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Hoochie [10]
3 years ago
12

Which balloon would fall faster, water filled or murcury filled?

Physics
1 answer:
bazaltina [42]3 years ago
7 0
A mercury filled balloon would fall faster then water. Mercury is heavier.
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A fast-moving vw beetle traveling at 54 mph hit a mosquito hovering at rest above the road. which bug experienced the largest fo
allsm [11]
1. the insect
because of how fast the car is going and how small they are
3 0
3 years ago
Read 2 more answers
Acceleration problem <br> Show work plz
Dennis_Churaev [7]

Answer:

The answer to your question is: vo = 25 m/s

Explanation:

data

a = -7.5 m/s²

d = 42 m

vf = 0 m/s

vo = ?

Formula

vf² = vo² - 2ad

Substitution

0² = vo² - 2(7.5)(42)

We clear vo from the equation

vo² = 2(7.5)(42)  

vo² = 630               simplifying

vo = 25 m/s            result

3 0
3 years ago
A force of 10 N acts on an object with a mass of 10 kg. What is its acceleration? A. 1 m/s2 B. 10 m/s2 C. 2 m/s2 D. 5 m/s2
Akimi4 [234]

The acceleration exerted by the object of mass 10 kg is \mathbf{1} m / \boldsymbol{s}^{2}

Answer: Option A

<u>Explanation:</u>

According to Newton’s second law of motion, any external force acting on a body will be directly proportional to the mass of the body as well as acceleration exerted by the body. So, the net external force acting on any object will be equal to the product of mass of the object with acceleration exerted by the object. Thus,

                  Force = Mass \times Acceleration

So,

                 Acceleration=\frac{\text {Force}}{\text {Mass}}

As the force acting on the object is stated as 10 N and the mass of the object is given as 10 kg, then the acceleration will be

                  Acceleration =\frac{10 \mathrm{N}}{10 \mathrm{kg}}=1 \mathrm{m} / \mathrm{s}^{2}

So, the acceleration exerted by the object of mass 10 kg is \mathbf{1} m / \boldsymbol{s}^{2}

4 0
3 years ago
Please help me answer this question
Pavel [41]
It can either be all of them or just 1 and 3
4 0
2 years ago
What is the electric potential v due to the nucleus of hydrogen at a distance of 5.292×10−11 m ?
viktelen [127]

Answer: 27.21 V

Explanation:

The <u>electric potential</u> V_{E} due to a point charge is expressed as:

V_{E}=k\frac{q}{r}

Where:

k=9(10)^{9}\frac{Nm^{2}}{C^{2}} is the <u>electric constant</u>

q=1.6(10)^{-19}C is the <u>electric charge of the hydrogen nucleus</u>, which is positive

r=5.292(10)^{-11}m is the <u>distance</u>

Rewritting the equation with the known values:

V_{E}=9(10)^{9}\frac{Nm^{2}}{C^{2}}\frac{1.6(10)^{-19}C}{5.292(10)^{-11}m}

Finally:

V_{E}=27.21 V

5 0
3 years ago
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