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3 years ago

5

Which object would have more potential energy, a bowling ball stored on a high shelf, or a base ball on the same shelf? Why? Whi

ch one would you rather have fall on your head?

1 answer:

Anettt [7]3 years ago

7 0

Answer:

A bowling ball stored on a high shelf has more potential energy

A base ball on the same shelf should rather fall on my head

Explanation:

Potential energy =mgh where m= mass, g= gravitational force and h is height

The bowling ball is bigger and heavier than the base ball

which means more mass . The remaining parameters are constant.

This means the bowling ball has more potential energy.

Since the base ball is lighter then it’s preferable to land on someone’s head due to the lesser force it will expert and lesser pain too.

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A rectangular block floats in pure water with 0.400 in. above the surface and 1.60 in. below the surface. When placed in an aque

labwork [276]

Answer:

specific gravity = 0.8

specific gravity of solution = 2

Explanation:

given data

rectangular block above water = 0.400 in

rectangular block below water = 1.60 in

material floats below water = 0.800 in

solution

first we get here specific gravity of block that is

specific gravity = block vol below ÷ total block vol × specific gravity water ..............1

put here value we get

specific gravity = $\frac{1.60}{1.60+0.400}$ × 1

specific gravity = 0.8

and now we get here specific gravity of solution that is express as

specific gravity of solution = total block vol ÷ block vol below × specific gravity block ........................2

put here value we get

specific gravity of solution = $\frac{1.60+0.400}{0.800}$ × 0.8

specific gravity of solution = 2

6 0

3 years ago

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

The propeller of an airplane is at rest when the pilot starts the engine; and its angular acceleration is a constant value. Two

Maurinko [17]

Answer:0.318 revolutions

Explanation:

Given

Initially Propeller is at rest i.e. $\omega _0=0 rad/s$

after $t=10 s$

$\omega =10 rad/s$

using $\omega =\omega _0+\alpha t$

$10=0+\alpha \cdot 10$

$\alpha =1 rad/s^2$

Revolutions turned in 2 s

$\theta =\omega _0t+\frac{\alpha t^2}{2}$

$\theta =0+\frac{1\times 2^2}{2}$

$\theta =2 rad$

To get revolution $\frac{\theta }{2\pi }$

= $\frac{2}{2\pi}=0.318\ revolutions$

3 0

3 years ago

Read 2 more answers

A sample of a gas has a volume of 639 cm3 when the pressure is 75.9 kPa. What is the volume of the gas when the pressure is incr

const2013 [10]

Answer:

$388 cm^3$

Explanation:

For this problem, we can use Boyle's law, which states that for a gas at constant temperature, the product between pressure and volume remains constant:

$pV=const.$

which can also be rewritten as

$p_1 V_1 = p_2 V_2$

In our case, we have:

$p_1 = 75.9 kPa$ is the initial pressure

$V_1 = 639 cm^3$ is the initial volume

$p_2 = 125 kPa$ is the final pressure

Solving for V2, we find the final volume:

$v_2 = \frac{p_1 V_1}{p_2}=\frac{(75.9)(639)}{125}=388 cm^3$

7 0

3 years ago

If an asteroid were going to impact Earth, would you prefer to know in advance or let the NEO remain a mystery until impact? Why

ivanzaharov [21]

I would like to know in advance

6 0

2 years ago