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enot [183]
3 years ago
14

James and Juan were at the highest point in the football stadium, dropping water balloons on people below. Many of the balloons

did not break, so the boys added more water to the balloons. What impact did adding water have on the balloons' potential energy? A) It had no impact; it just made the balloons bigger. B) The gravitational potential energy increased because the mass increased. C) The gravitational potential energy decreased because the diameter increased. D) The gravitational potential energy remained constant because they did not increase the height.
Physics
2 answers:
aleksley [76]3 years ago
6 0
Correct the answer is b
Fofino [41]3 years ago
4 0
When you add more water to the balloon, it makes it heavier. Therefore it would weigh the balloon down ( increasing mass) and increasing the energy to plummet down. So the answer is B.
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A particular car can go from rest to 90 km/h in 10 s. What is its acceleration? (Report your answer in km/h*s)
Romashka-Z-Leto [24]

Answer: 9 km/h

Explanation:i’m pretty sure

5 0
3 years ago
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A 55-kg block, starting from rest, is pushed a distance of 5.0 m across a floor by a horizontal force Fp whose magnitude is 140
max2010maxim [7]

Answer:

The answer is "151.25 J and -547.64 J".

Explanation:

u = 0\\\\v = 2.35\  \frac{m}{sec}\\\\d = 5.0 \ m\\\\

Using formula:

v^2 = u^2 + 2 \times a \times d\\\\2.35^2 = 0^2 + 2 \times a \times 5\\\\a = \frac{2.35^2}{10} \\\\

   = 0.55 \ \frac{m}{sec^2}\\\\

F_{net} = m \times a\\\\F_{net} = 55 \times 0.55 = 30.25\ N\\\\

Calculating the Work by net force

W = F_{net}\times d\\\\W = 30.25 \times 5 = 151.25 \ J\\\\

The above work is converted into thermal energy.

Now,

F_{net} = F_p - F_f\\\\F_p = 140 \ N\\\\F_f = u_k\times m \times g = u_k \times 55 \times 9.81\\\\F_f = 539.55 \times u_k\\\\30.25 = 140 - u_k \times 55 \times 9.81\\\\u_k = \frac{(140 - 30.25)}{(55\times 9.81)}\\\\uk = 0.203 = \text{Coefficient of friction}\\\\W_f = -F_f \times d\\\\W_f = -0.203 \times 55 \times 9.81 \times 5\\\\Work\ done\  by\ friction = -547.64 \ J

6 0
3 years ago
A light bulb is connected to a 2V supply and experiences a current of 6.4A. What is the power rating of the bulb?
Olenka [21]

Answer:

12.8 Watts

Explanation:

P = VI

P = (2 V) (6.4 A)

P = 12.8 W

8 0
3 years ago
The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, what should be the focal length an
yuradex [85]

Answer:

The focal length of the appropriate corrective lens is 35.71 cm.

The power of the appropriate corrective lens is 0.028 D.

Explanation:

The expression for the lens formula is as follows;

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}

Here, f is the focal length, u is the object distance and v is the image distance.

It is given in the problem that the given lens is corrective lens. Then, it will form an upright and virtual image at the near point of person's eye. The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, the corrective lens is used.

Put v= -71.4 cm and u= 24.0 cm in the above expression.

\frac{1}{f}=\frac{1}{24}+\frac{1}{-71.4}

\frac{1}{f}=0.028

f= 35.71 cm

Therefore, the focal length of the corrective lens is 35.71 cm.

The expression for the power of the lens is as follows;

p=\frac{1}{f}

Here, p is the power of the lens.

Put f= 35.71 cm.

p=\frac{1}{35.71}

p=0.028 D

Therefore, the power of the corrective lens is 0.028 D.

3 0
3 years ago
BRAINLIEST IF CORRECT
Sidana [21]

Hello There!

Sokka is here to help!!

The answer is...

<h2>D. Counter-arguments lead to circular logic in your argument.</h2>

Because, I am right. :)

Hopefully, this helps you!!

Sokka

3 0
2 years ago
Read 2 more answers
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