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DedPeter [7]
2 years ago
14

The go kart driver does 1234J of work when hitting a barrier. Assuming the barrier moved 100 meters from its original location,

what was the force at which the driver hit the barrier?
Physics
2 answers:
leva [86]2 years ago
6 0
Force=1234J*100m
So the answer is 123400

bezimeni [28]2 years ago
4 0
The guy in front of me is correct
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List several examples of applied force, normal force, and friction that you’ve observed in your life.
grandymaker [24]

applied forces would be push for example.

normal forces would seem to be a force such as gravity.

friction for example when you try to slide on carpet but the fabric or whatever its made of stops you.

3 0
3 years ago
Properties of light that define light as a wave?
MAVERICK [17]
The first choices are correct, because the second choices could happen by things other than light.
6 0
2 years ago
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Can someone help me ​
Julli [10]
The answer is the first one
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3 years ago
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At a rock concert, the sound intensity 1.0 m in front of the bank of loudspeakers is 0.10 W/m². A fan is 30 m from the loudspeak
Klio2033 [76]

To solve this problem we will apply the concepts related to the Area, the power and the proportionality relationships between intensity and distance.

The expression for sound power is,

P = AI

Here,

A = Area

I = Intensity

P = Power

At the same time the area can be written as,

A = \frac{\pi d^2}{4}

Now the intensity is inversely proportional to the square of the distance from the source, then

I \propto \frac{1}{r^2}

The expression for the intensity at different distance is

\frac{I_1}{I_2}= \frac{r^2_2}{r_1^2}

Here,

I_1 = Intensity at distance 1

I_2 = Intensity at distance 2

r_1 = Distance 1 from light source

r_2 = Distance 2 from the light source

If we rearrange the expression to find the intensity at second position we have,

I_2 = I_1 (\frac{r_1^2}{r_2^2})

If we replace with our values at this equation we have,

I_2 = (0.10W/m^2)(\frac{1.0m^2}{30.0m^2})

I_2 = 1.11*10^{-4} W/m^2

Now using the equation to find the area we have that

A = \frac{\pi (8.4*10^{-3}m)^2}{4}

A = 5.5*10^{-5}m^2

Finally with the intensity and the area we can find the sound power, which is

P = AI

P = (5.5*10^{-5}m^2)(1.11*10^{-4}W/m^2)

P = 6.1*10^{-9}J/s

Power is defined as the quantity of Energy per second, then

E = 6.1*10^{-9}J

8 0
3 years ago
Find the final temperature of 375 grams of tea (c = 4.184 J/g°C) if its initial temperature is 95°C just before it is placed in
Minchanka [31]

Answer:

the final temperature of the tea is 7.39⁰C.

Explanation:

Given;

mass of the tea, m = 375 g

specific heat capacity of the tea, C = 4.184 JJ/g°C

initial temperature of the tea, t₁ = 95°C

the final temperature of the tea, t₂ = ?

Energy lost by the refrigerator, Q = 137,460 J

The energy lost by the refrigerator is given by the following formula;

-Q = mc(t₂ - t₁)

-137,460 =375 x 4.184(t₂ - 95°C)

-137,460 = 1569(t₂ - 95°C)

t_2-95 = \frac{-137,460}{1569} \\\\t_2-95 = -87.61\\\\t_2 = -87.61 + 95\\\\t_2 = 7.39 \ ^0 C

Therefore, the final temperature of the tea is 7.39⁰C.

4 0
3 years ago
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