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Nana76 [90]
3 years ago
10

A) How many joules of energy does a 100-watt light bulb use per hour? Express your answer using two significant figures.

Physics
1 answer:
kenny6666 [7]3 years ago
5 0

A) 3.6\cdot 10^5 J

The power used by an object is defined as

P=\frac{E}{t}

where

E is the energy used

t is the time elapsed

In this problem, we have

P = 100 W is the power of the light bulb

t = 1 h = 3600 s is the time elapsed

Solving for E, we find the amount of energy used by the light bulb:

E=Pt = (100 W)(3600 s)=3.6\cdot 10^5 J

B) 1.1 \cdot 10^2 m/s

The kinetic energy of an object is given by

K=\frac{1}{2}mv^2

where

m is the mass

v is the speed of the object

In this problem, we have a man of mass

m = 65 kg

we want its kinetic energy to be

E=3.6\cdot 10^5 J

Therefore, we can calculate its speed from the previous formula:

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.6\cdot 10^5 J)}{65 kg}}=105.2 m/s = 1.1 \cdot 10^2 m/s

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4 0
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

5 0
2 years ago
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Several types of radiation may be emitted during radioactive decay. The equation represents
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I think the answer is B
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A 50.1 kg diver steps off a diving board and drops straight down into the water. The water provides an average net force of resi
Varvara68 [4.7K]

Answer:

The total distance is 16.9 m

Explanation:

We understand work in physics as certain force exerted through certain distance. To reach that point below the water, the work done by the diver must be equal to the work done by the water's force of resistance. Therefore, we determine both work expressions and we solve the equation for the diver distance, which is the total distance between the diving board and the stopping point underwater.

W_{d}: work done by diver\\W_{R}: work due the force of resistance\\W_{d} = W_{R}\\F_{d}\times d_{d}=F_R \times d_{R}\\d_{d}= \frac{F_R \times d_R}{F_d}= \frac{F_R \times d_R}{m_d \times g}= \frac {1598 N \times 5.2 m}{50.1 kg \times 9.81 m/s^2}\\d_{d}= 16. 91 m

7 0
2 years ago
How much heat is given off when 210.0g of water at 0 degrees freezes into ice?
Sergio [31]

In the freezing physical change, when 210.0 g of water a 0 degrees freezes into ice, it gives off 71.0 kJ of heat.

<h3>What is freezing?</h3>

It is a physical change in which liquids give off heat to form solids.

We have 210.0 g of water at 0°C. We can calculate the amount of heat given off when it freezes into ice using the following expression.

Q = ΔH°fus × m

Q = 0.334 kJ/g × 210.0 g = 70.1 kJ

where,

  • Q is the heat released.
  • ΔH°fus is the latent heat of fusion.
  • m is the mass.

In the freezing physical change, when 210.0 g of water a 0 degrees freezes into ice, it gives off 71.0 kJ of heat.

Learn more about freezing here: brainly.com/question/40140

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5 0
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