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3 years ago

What is the kinetic energy of a 144 gram baseball moving at a velocity of 38.7 meters per second?

Physics

1 answer:

gulaghasi [49]3 years ago

8 0

Answer:

The kinetic energy of the baseball is 107.83 J

Explanation:

Given;

mass of the baseball, m = 144 g = 0.144 kg

velocity of the baseball, v = 38.7 m/s

The kinetic energy of the baseball is given by;

K.E = ¹/₂mv²

where;

m is mass of the object

v is speed of the object

K.E = ¹/₂(0.144)(38.7)²

K.E = 107.83 J

Therefore, the kinetic energy of the baseball is 107.83 J

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A hiker, caught in a rainstorm might absorb 1 liter of water in her clothing. if it is windy so that this amount of water is eva

NeTakaya

Water evaporates at 100⁰C

So change in temperature = 100-20 = 80⁰C

Amount of water to be evaporated = 1 liter = 1L*1kg/liter = 1 kg

Specific heat of water is 1 calorie/gram ⁰C = 4.186 joule/gram =4186 J/kg

So heat required E = mcΔT = 1 * 4186 *80= 334880 J =334.88 kJ

So amount of heat require to evaporate water = 334.88 kJ

4 0

3 years ago

An apple falls from a tree and 2.5 seconds later hits the ground. How fast is the apple falling when it hits the ground? Neglect

denis-greek [22]

Answer:

Explanation:

<h2>Since the ball rises for 2.5 s, the time to fall is 2.5 s. The acceleration is 9.8 m/s2 everywhere, even when the velocity is zero at the top of the path. Although the velocity is zero at the top, it is changing at the rate of 9.8 m/s² downward.</h2>

6 0

2 years ago

PLEASE HELP I NEED THIS TO PASS THE EIGHTH GRADE AND I ONLY HAVE A COUPLE HOURS LEFT!!!!!!

zavuch27 [327]

Answer:

the pe at the top of the building: 784 J

the pe halfway through the fall: 392 J

the pe just before hitting the ground: 784 J

Explanation:

Pls brainliest me

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7 0

3 years ago

You wish to produce an emf of 41.0 mV using an inductor whose inductance is 13.0 H. You start with a current of 1.50 mA through

Molodets [167]

Answer:

The current through the inductor at the end of 2.60s is 9.7 mA.

Explanation:

Given;

emf of the inductor, V = 41.0 mV

inductance of the inductor, L = 13 H

initial current in the inductor, I₀ = 1.5 mA

change in time, Δt = 2.6 s

The emf of the inductor is given by;

$V = L\frac{di}{dt} \\\\V = \frac{L(I_1-I_o)}{dt} \\\\L(I_1-I_o) = V*dt\\\\I_1-I_o = \frac{V*dt}{L}\\\\I_1 = \frac{V*dt}{L} + I_o\\\\I_1 = \frac{41*10^{-3}*2.6}{13} +1.5*10^{-3}\\\\I_1 = 8.2*10^{-3} + 1.5*10^{-3}\\\\I_1 = 9.7 *10^{-3} \ A\\\\ I_1 = 9.7 \ mA$

Therefore, the current through the inductor at the end of 2.60 s is 9.7 mA.

6 0

3 years ago

An infinite sheet of charge, oriented perpendicular to the x-axis, passes through x = 0. It has a surface charge density σ1 = -2

pishuonlain [190]

Answer:

$E_{total}=4.82*10^6N/C$

vector with direction equal to the axis X.

Explanation:

We use the Gauss Law and the superposition law in order to solve this problem.

<u>Superposition Law:</u> the Total Electric field is the sum of the electric field of the first infinite sheet and the Electric field of the second infinite sheet:

$E_{total}=E_1+E_2$

<u>Thanks Gauss Law</u> we know that the electric field of a infinite sheet with density of charge σ is:

$E=\sigma/(2\epsilon_o)$

Then:

$E_{total}=(\sigma_1+\sigma_2)/(2\epsilon_o)=(-2.7*10^{-6}+88*10^{-6})/(2*8.85*10^{-12})=4.82*10^6N/C$

This electric field has a direction in the axis perpendicular to the sheets, that means it has the same direction as the axis X.

7 0

3 years ago

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