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3 years ago

a track coach with a meter stick and a stopwatch is trying to determine if a student is walking with constant speed.he should...

Physics

1 answer:

Aleksandr [31]3 years ago

7 0

Give you a device that tracks your speed?

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Help asap!

patriot [66]

Answer:

220 V

Explanation:

$r_{1} = 20 ohm\\r_{2} = 30ohm\\r_{3} = 60ohm$

As the resistors are connected in series,

$R_{eq} = r_{1} +r_{2} +r_{3}$

= 20 + 30 + 60

= 110 ohms

V = IR

= 2 * 110

= 220 V

Hope it helps

7 0

2 years ago

You take the same 3 kg metal block and slide it along the floor, where the coefficient of friction is only 0.4. You release the

spin [16.1K]

Answer:

1.52905 seconds

4.58715 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

$\mu$ = Coefficient of friction = 0.4

g = Acceleration due to gravity = 9.81 m/s²

a = Acceleration = $\mu g$

Equation of motion

$v=u+at\\\Rightarrow t=\frac{v-u}{\mu g}\\\Rightarrow t=\frac{0-6}{-0.4\times 9.81}\\\Rightarrow t=1.52905\ s$

It will take 1.52905 seconds for the block to slow down

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2\mu g}\\\Rightarrow s=\frac{0^2-6^2}{2\times 0.4\times -9.81}\\\Rightarrow s=4.58715\ m$

The block will travel 4.58715 m before it stops

7 0

3 years ago

What's the real power of a purely resistive circuit in which the RMS voltage measures 80 volts and the RMS current measures 5 am

Nat2105 [25]

Answer:

D.400 watt

Explanation:

B/c Pave=IrmsVrms
Pave=(5)(80)
Pave=400 watt

where Pave = average power

Irms=RMS current &

Vrms=RMS voltage

7 0

1 year ago

Which of the following would cause the greatest decrease in gravitational force between the earth and the moon? an increase in t

Sphinxa [80]

a decrease in the distance between the earth and the moon

8 0

2 years ago

Two equal charges are 2m2m apart. If the charges and the distance are divided by two, how is the force between the charges affec

lara31 [8.8K]

Answer:

The force between the charges are not affected.

Explanation:

Given;

distance between two equal charges, R = 2m

The force between the charges is given by;

$F = \frac{kq^2}{R^2}\\\\F_1 = \frac{kq_1^2}{R_1^2}\\\\When\ the \ charges \ and \ the \ distance \ are \ divided \ by \ two \ (q_2 = \frac{q_1}{2}, \ R_2 = \frac{R_1}{2} )\\\\ F_2 = \frac{kq_2^2}{R_2^2}\\\\F_2 = \frac{k(q_1/2)^2}{(R_1/2)^2}\\\\F_2= \frac{4k*q_1^2}{4*R_1^2}\\\\F_2 = \frac{k*q_1^2}{R_1^2}\\\\F_2 = F_1$

Therefore, the force between the charges are not affected.

7 0

2 years ago