All categories

2 years ago

A 40 Watt lightbulb is on for 3 minutes. How many joules of work does it do?

Physics

1 answer:

ella [17]2 years ago

6 0

Answer:

Explanation:

Power (p)= 40 watt

time (t)= 3 minutes = 180 seconds

Energy (E)= ?

we know

P = E/t

40 = E / 180

E = 7200 Joules

Therefore it does 7200 joules of work.

hope it helps :)

You might be interested in

URGENT!!!! WILL GIVE BRAINLIEST!!!!!! 15 POINTS

Artyom0805 [142]

It's the angle made by the incident ray when it's perpendicular to the surface. (Perpendicular lines are the lines that form a graph or like a 90-degree angle)

3 0

3 years ago

Read 2 more answers

6. Which of these contain muscles that are not under the mind's control? (Select all that apply.)

Arada [10]

Answer:

a c

Explanation:

edu 2021

3 0

2 years ago

The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

Also;

$\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}$

Thus;

$k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}$

where;

$\delta$ = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

$k = 104.82\ \ N/m$

Thus; the spring constant = 104.82 N/m

The angular velocity can be calculated by also using the conservation of energy;

$T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0$

$\frac{m(a+b)^2}{3} \omega_3^2 + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0$

$\frac{1.53(0.6+0.6)^2}{3} \omega_3^2 + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0$

$0.7344 \omega_3^2 = 2.128$

$\omega _3 = \sqrt{\frac{2.128}{0.7344} }$

$\omega _3 =1.70 \ rad/s$

Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s

7 0

2 years ago

12) Which is MOST LIKELY a solid at room temperature?

ella [17]

Gold
Silver
Iron
Platinum
Goes on and on

4 0

3 years ago

Two states of matter are described below. State A: Cannot be compressed and retains its shape State B: Highly compressible Which

telo118 [61]

Answer:

State A = piece of metal; State B = air

Explanation:

For the three main states of matter here's how it breaks down.

Solid - Cannot be compressed and retains its shape

Liquid - Cannot be compressed and does not retain its shape

Gas - Compressible and does not retain its shape.

Knowing this State A has to be solid. Only one of the options has A as a solid, so that's the answer. Worth knowing state B is a gas though, only one compressible, just like solid is the only one that retains its shape.

7 0

3 years ago

Read 2 more answers