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3 years ago

The work done on a box is 532 joules. The force applied to the box was 48 N. What was the displacement of the box? *

Physics

1 answer:

Semenov [28]3 years ago

8 0

Explanation:

Work = force × displacement

532 J = 48 N × d

d ≈ 11 m

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Two gliders on an air track collide in a perfectly elastic collision. Glider A has a mass of 1.1 kg and is initially travelling

Eva8 [605]

m1= mass 1 = 1.1 kg

Vi1 = initial velocity 1 = 2.7 m/s

m2= 2.4 kg

V2i = -1.9 m/s

We assume east as positive and west as negative.

Apply the formulas:

Vf1 = ?

$vf1=(\frac{m1-m2}{m1+m2})Vi1+(\frac{2m2}{m1+m2})Vi2$

Replacing:

$Vf1=\frac{(1.1-2.4)}{(1.1+2.4)}2.7+\frac{(2\times2.4)}{(1.1+2.4)}-1.9$ $Vf1=(\frac{-1.3}{3.5})2.7+(\frac{4.8}{3.5})-1.9$ $Vf1=-1-2.6=-3.6\text{ m/s}$

Answer: 3.6 m/s west

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1 year ago

A heater has a resistance of 10.0 Ω. It operates on a 12.0 V. What is the current through the resistor?

irakobra [83]

Answer:

1.2 amps :)

Explanation:

A heater has a resistance of 10.0 Ω. It operates on a 12.0 V. What is the current through the resistor?

Known:

R = 10.0 Ω

V = 12.0 V

Unknown:

I = ???

I = V/R

= 12.0 V / 10.0 Ω

= 1.2 amps

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3 years ago

Velocity of B rays ?<br>

stich3 [128]

Answer:

is from 9 x 107 m/sec to 27 x 107 m/s

6 0

3 years ago

Read 2 more answers

A woman with a mass of 60 kg climbs a set of stairs that are 3m high How much gravitational potential energy does she gain a res

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540 j

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2 years ago

A Foucault pendulum consists of a brass sphere with a diameter of 31.0 cm suspended from a steel cable 11.0 m long (both measure

kozerog [31]

Answer:

43.7 °C

Explanation:

$\alpha_b$ = Coefficient of linear expansion of brass = $18\times 10^{-6}\ ^{\circ}C$

$\alpha_s$ = Coefficient of linear expansion of steel = $11\times 10^{-6}\ ^{\circ}C$

$L_{0b}$ = Initial length of brass = 31 cm

$L_{0s}$ = Initial length of steel = 11 m

$\Delta L$ = Total change in length = 3 mm

Total change in length would be

$\Delta L=\Delta L_b+\Delta L_s\\\Rightarrow \Delta L=L_{0b}\alpha_b\Delta T+L_{0s}\alpha_b\Delta T\\\Rightarrow \Delta T=\frac{\Delta L}{L_{0b}\alpha_b+L_{0s}\alpha_b}\\\Rightarrow \Delta T=\frac{0.003}{0.31\times 18\times 10^{-6}+11\times 10^{-6}\times 11}\\\Rightarrow \Delta T=23.7\ ^{\circ}C$

$\Delta T=23.7\\\Rightarrow T_f-T_i=23.7\\\Rightarrow T_f=23.7+T_i\\\Rightarrow T_f=23.7+20\\\Rightarrow T_f=43.7\ ^{\circ}C$

The final temperature is 43.7 °C

6 0

3 years ago