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sleet_krkn [62]
3 years ago
6

When you lift a bowling ball with a force of 61.1 N, the ball accelerates upward with an acceleration a. If you lift with a forc

e of 70.9 N, the ball's acceleration is 2.07a. Calculate the weight of the bowling ball.
Physics
1 answer:
gtnhenbr [62]3 years ago
7 0

Answer:

the weight of the ball is w = 51.94 N ( mass = 5.3 kg)

Explanation:

Following Newton's second law:

net force = mass * acceleration = weight/gravity * acceleration

then denoting 1 and 2 as the first and second lift

F₁ - w= w/g *a₁

F₂ -w = w/g *a₂ = w/g * 2.07a

dividing both equations

(F₂- w)/(F₁ -w)= 2.07

(F₂- w) = 2.07 * (F₁ -w)

1.07*w = 2.07*F₁ - F₂

w = (2.07*F₁ - F₂ )/ 1.07

replacing values

w = (2.07*61.1 N - 70.9 N )/ 1.07  = 51.94 N

then the weight of the ball is w = 51.94 N ( mass = 5.3 kg)

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Answer:

(i) false

(ii) true

(iii) true

(iv) false

Explanation:

(i) The ratio of Cp and Cv is not constant for all the gases. It is because the value of cp and Cv is different for monoatomic, diatomic and polyatomic gases.

So, this is false.

(ii) For monoatomic gas

Cp = 5R/2, Cv = 3R/2

So, thier ratio

Cp / Cv = 5 / 3 = 1.67

This statement is true.

(iii) for diatomic gases

Cp = 7R/2, Cv = 5R/2

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This statement is true.

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6 0
3 years ago
Electrons and protons are basic particles (and together make atoms) that have charge. What is the electric charge of an electron
trapecia [35]

Answer:

hi, here's the answer

Explanation:

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8 0
3 years ago
To convert centimeters to kilometers, which conversion factors would you need?
Lelechka [254]

Answer:

The last option is the only correct one if you like to multiply

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Explanation:

Each fraction in the last two options has a value of 1

example

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and

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only one of the two fractions in each of the top two options has a value of 1.

3 0
2 years ago
The greater the pull of gravity on an object, the greater the what of that object
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The compressor on an air conditioner draws 66.1 a when it starts up. the start-up time is about 0.388 s. how much charge passes
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I= \frac{Q}{\Delta t}
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Q=I \Delta t

for the compressor in our problem, the intensity of current is I=66.1 A, while the time is \Delta t= 0.388 s, so the amount of charge that crosses a certain point of the circuit during this time is
Q=I \Delta t=(66.1 A)(0.388 s)=25.65 C
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