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OleMash [197]
3 years ago
10

When a net force of 17.0 newtons is applied to a dictionary placed on a frictionless table, it accelerates by 3.75 meters/second

Æ. What is the mass of the book?
Physics
1 answer:
4vir4ik [10]3 years ago
8 0
Force = (mass) x (acceleration)    (Newton's second law of motion)

Divide both sides of the equation by 'acceleration', and you have

Mass = (force) / (acceleration)

Mass = 17 newtons / 3.75 meters per second-sqrd = 4.533 kilograms (rounded)
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Repeat the experiment to make sure it gives the same results.

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A jet taxiing down the runway receives word that it must return to the gate. The jet is traveling 37.6 m/s when the pilot receiv
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Answer:

7 m/s^2

Explanation:

Given that the jet is traveling 37.6 m/s when the pilot receives the message. 

And it takes the pilot 5.37 s to bring the plane to a halt.

Acceleration of the plane can be calculated by using first equation of motion

V = U - at

Since the plane is going to stop, the final velocity V = zero.

And the acceleration will be negative

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0 = 37.6 - 5.37a

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a = 37.6 / 5.37

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Therefore, the acceleration of the plane to bring the plane to a halt is 7 m/s^2

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3 years ago
The parallax method of measuring star distances gives most accurate results when the gap between two observations of a star is a
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Which type of mine involves digging tunnels and shafts deep underground?
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The mine involves digging tunnels and shafts deep underground is the underground mining.

<h3>What is mining?</h3>

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7 0
2 years ago
Can someone solve this problem and explain to me how you got it​
evablogger [386]

Answer:

question5: F=74312.5N

question6: charge at the end of antenna=0.37N

Explanation:

Coulomb's law: the magnitude of the force of attraction or repulsion due to two charges is proportional to the product of the magnitude of the charges and inversely proportional to the square of distance between the charges.

⇒F\alpha\frac{q1*q2}{r^{2}}

∴F=k\frac{q1*q2}{r^{2}}

where F is the force of attraction or repulsion

k is Coulumb's constant=9*10^{9}Nm^{2}C^{-2}

q1 and q2 are the magnitude of the charges

r is the distance between two charges

The force between the two charges is attractive if they are of different polarity

The force between the two charges is repulsive if they are of same polarity

Question5:

Given: q1=0.041 C, q2=0.029 C, r=12 m

therefore by Coulumb's law,

F=9*10^{9}*\frac{0.041*0.029}{12^{2}}

F=74312.5N

Question6:

Given: q1=3*10^{-18}C, r=5 m, F=4*10^{-11}N

therefore by Coulumb's law,

4*10^{-11}=9*10^{9}*\frac{3*10^{-18}*q2}{5^{2}}

⇒q2=\frac{4*10^{-11}*25}{9*10^{9}*3*10^{-18}} \\=0.37C

4 0
3 years ago
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