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3 years ago

1.How can you conserve water in the kitchen?

Physics

1 answer:

PIT_PIT [208]3 years ago

4 0

Instead of washing vegetables in running water, fill a bowl of water and wash inside the bowl.

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A 720 g softball is traveling at 15.0 m/s when caught. If the force of the glove on the ball is 520 N, what is the time it takes

Sholpan [36]

Answer:

The time it takes the ball to stop is 0.021 s.

Explanation:

Given;

mass of the softball, m = 720 g = 0.72 kg

velocity of the ball, v = 15.0 m/s

applied force, F = 520 N

Apply Newton's second law of motion, to determine the time it takes the ball to stop;

$F = ma = \frac{mv}{t} \\\\t = \frac{mv}{F} \\\\t = \frac{0.72 \ \times \ 15}{520} \\\\t = 0.021 \ s \\$

Therefore, the time it takes the ball to stop is 0.021 s.

5 0

3 years ago

All galaxies follow the law of gravity. True or False

Juliette [100K]

The answer is true. All the galaxies in the universe follow the law of gravity.

Based from the book, It's about Time: the Illusion of Einstein’s Time Dilation Explained,

Einstein had explained that all the heavenly bodies in the universe follow the same scientific laws that are similar to our solar system. The stars and planets are held by the principles of inertia and gravity

8 0

2 years ago

high school physics, no need detail explain, just give the answer, but you have to make sure thank you

Goshia [24]

Answer:many questions add point

6 0

3 years ago

A 93 kg zebra is traveling 13 m/s east. What is the zebra’s momentum?

Harlamova29_29 [7]

Equation: Mass x Velocity = Momentum

Answer: 93 x 13 = 1,209

3 0

3 years ago

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

Distance of the center from each corners $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

Force by each of the charges at the remaining corners:

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

Now, net electric field in the vertical direction:

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

Now, net electric field in the horizontal direction:

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

8 0

3 years ago