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3 years ago

What is the rate of acceleration due to gravity on Earth?

Physics

1 answer:

Katena32 [7]3 years ago

3 0

Answer:

Negative 9.8 meters per second squared

Explanation:

The negative is for the direction (down, towards the center of the earth). Often this can be estimated as -10 m/s^2 to make calculations easier.

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Which equation is correct according to Ohm’s law? Which equation is correct according to Ohm’s law? A.) V = IR B.) I = R/V C.) R

vodomira [7]

Answer:

$V = IR$

Explanation:

Required

Which equation represents ohm's law?

Literally, ohm's law implies that current (I) is directly proportional to voltage (V) and inversely proportional to resistance (R).

Mathematically, this can be represented as:

$I\ \alpha\ \frac{V}{R}$

Convert the expression to an equation

$I\ =\ \frac{V}{R}$

Multiply both sides by R to make V the subject

$I\ * R\ =\ \frac{V}{R} * R$

$I\ * R\ =V$

Reorder

$V = I\ * R$

$V = IR$

<em>Option (a) is correct; Others are not</em>

6 0

3 years ago

Read 2 more answers

____ are solid while lines stretching across one or more lanes in the same direction, indicating the proper place to come to a s

Natalka [10]

Stop lines are solid white lines painted across the traffic lanes at intersections and pedestrian crosswalks, indicating the exact place to stop.

8 0

3 years ago

Your electric drill rotates initially at 5.21 rad/s. You slide the speed control and cause the drill to undergo constant angular

RUDIKE [14]

Answer:

The drill's angular displacement during that time interval is 24.17 rad.

Explanation:

Given;

initial angular velocity of the electric drill, $\omega _i$ = 5.21 rad/s

angular acceleration of the electric drill, α = 0.311 rad/s²

time of motion of the electric drill, t = 4.13 s

The angular displacement of the electric drill at the given time interval is calculated as;

$\theta = \omega _i t \ + \ \frac{1}{2}\alpha t^2\\\\\theta = (5.21 \ \times \ 4.13) \ + \ \frac{1}{2}(0.311)(4.13)^2\\\\\theta = (21.5173 ) \ + \ (2.6524)\\\\\theta =24.17 \ rad$

Therefore, the drill's angular displacement during that time interval is 24.17 rad.

6 0

3 years ago

A 265 g mass attached to a horizontal spring oscillates at a frequency of 3.40 Hz . At t =0s, the mass is at x= 6.20 cm and has

lara [203]

Answer:

The phase constant is 7.25 degree

Explanation:

given data

mass = 265 g

frequency = 3.40 Hz

time t = 0 s

x = 6.20 cm

vx = - 35.0 cm/s

solution

as phase constant is express as

y = A cosФ ..............1

here A is amplitude that is = $\sqrt{(\frac{v_x}{\omega })^2+y^2 }$ = $\sqrt{(\frac{35}{2\times \pi \times y})^2+6.2^2 }$ = 6.25 cm

put value in equation 1

6.20 = 6.25 cosФ

cosФ = 0.992

Ф = 7.25 degree

so the phase constant is 7.25 degree

5 0

3 years ago

A car is travelling at 70 mph. It travels for half an hour. How far has it gone?

rusak2 [61]

Answer:

35 miles

Explanation:

Since half of 70 is 35.....

6 0

3 years ago

Read 2 more answers