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3 years ago

A current of 0.92 A flows past a point in a circuit? How much charge, in units of Coulombs, passes that point in one minute?

Physics

1 answer:

Alisiya [41]3 years ago

7 0

<h3><u>Answer;</u></h3>

<u> = 55.2 Coulombs </u>

<h3><u>Explanation</u>;</h3>

We can determine Charge using the formula

Q =It, where Q is the amount of charge in Coulombs, I is the current in amperes and t is the time in seconds.

I = 0.92 amperes, t = 1 minute or 60 seconds

Charge = 0.92 × 60

<u> = 55.2 Coulombs </u>

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What is the velocity of an object that has been in free fall for 1.5s?

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Answer:

D. 15 m/s downward

Explanation:

v = at + v₀

v = (-9.8 m/s²) (1.5 s) + (0 m/s)

v = -14.7 m/s

Rounded to two significant figures, the answer is D, 15 m/s downward.

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3 years ago

An iron nail is driven into a block of ice by a single blow of a hammer. The hammerhead has a mass of 0.5 kg and an initial spee

larisa [96]

Answer:

The ice melts mass is:

$m_g=7.6x10^{-3} g$

Explanation:

Kinetic Energy

$K_E = 1/2*m*v^2$

$K_E = 1/2*0.5kg*(3.2m/s)^2$

$K_E=2.56 kg*m^2/s^2$

Heat gained by ice= mass(g) x 80 cal

( 1 cal = 4.184 *10^7er or g cm^2/ sec^2)

Assuming no loss in heat, in the motion so both continue with temperature 0~C

To find so the mass (gm) of ice melted

$m_g= 1/2 *(0.5*1000)*(3.2m/s)^2* 100*100 / (80*4.184*10^7)$

$m_g=7.6x10^{-3} g$

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3 years ago

What kind of frequency do radio waves have?<br><br>A. High frequency <br><br>B. Low frequency

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B low frequency it is the lowest frequency

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3 years ago

You pull straight up on the string of a yo-yo with a force 0.35 N, and while your hand is moving up a distance 0.16 m, the yo-yo

jarptica [38.1K]

Answer:

a) 0.138J

b) 3.58m/S

c) (1.52J)(I)

Explanation:

a) to find the increase in the translational kinetic energy you can use the relation

$\Delta E_k=W=W_g-W_p$

where Wp is the work done by the person and Wg is the work done by the gravitational force

By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

$Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J$

the change in the translational kinetic energy is 0.138J

b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

$\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2$

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

$v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}$

the new speed is 3.58m/s

c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

$\frac{E_{fr}}{E_{fr}}=\frac{1/2I\omega_f^2}{1/2I\omega_o^2}=\frac{\omega_f^2}{\omega_o^2}\\\\\omega_f=\frac{v_f}{r}\\\\\omega_o=\frac{v_o}{r}\\\\\frac{E_{fr}}{E_{fr}}=\frac{v_f^2}{v_o^2}=\frac{(3.58m/s)}{(2.9m/s)^2}=1.52J$

hence, the change in Er is about 1.52J times the initial rotational energy

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2 years ago

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dsp73

Answer:

Paramedic

Explanation:

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2 years ago

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