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allsm [11]
3 years ago
8

You push a hockey puck that is initially at rest on slick ice by applying a constant force until the puck reaches a final veloci

ty of 1 m/s. On the second attempt, you want the hockey puck to reach the same final velocity by applying a force that is twice as large.
1. Therefore, you must exert the force for a time interval that is
A. shorter than the time interval of your first attempt.
B. longer than the time interval of your first attempt.
C. the same as the time interval of your first attempt.
2. After the hockey puck has reached the final velocity, you suddenly stop pushing it. The hockey puck:
A. stops abruptly
B. reduces speed gradually
C. continues at constant velocity
D. increases speed gradually
E. reduces speed abruptly
Physics
1 answer:
stellarik [79]3 years ago
4 0

Answer:

1. A

2. B or C

Explanation:

1.

F=ma, meaning that if you use two times more force on a constant mass, the acceleration must double. Acceleration is change in velocity, which means that if you are aiming for the same final velocity the change must happen in half of the time. Therefore, the correct answer is choice A.

2.

By Newton's first law, an object in motion will stay in motion unless an external force acts on it. Since there is nothing pushing the puck in the other direction, the puck will either keep on going for at a constant velocity or will reduce its speed gradually, depending on whether or not this ice is considered to be frictionless. Hope this helps!

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Answer:

82780.42123 m/s

14.45 days

Explanation:

m = Mass of the planet

M = Mass of the star = 0.85\times 1.989\times 10^{30}\ kg=1.69065\times 10^{30}\ kg

r = Radius of orbit of planet = 0.11\times 149.6\times 10^{9}\ m=16.456\times 10^{9}\ m

v = Orbital speed

The kinetic and potential energy balance is given by

\frac{GMm}{r^2}=\frac{mv^2}{r}\\\Rightarrow v=\sqrt{\dfrac{Gm}{r}}\\\Rightarrow v=\sqrt{\dfrac{6.67\times 10^{-11}\times 1.69065\times 10^{30}}{16.456\times 10^{9}}}\\\Rightarrow v=82780.42123\ m/s

The orbital speed of the star is 82780.42123 m/s

The orbital period is given by

t=\frac{2\pi r}{v}\\\Rightarrow t=\dfrac{2\pi \times 16.456\times 10^{9}}{82780.42123}\\\Rightarrow t=1249040.48419\ seconds=\dfrac{1249040.48419}{24\times 60\times 60}=14.45\ days  

The orbital period is 14.45 days

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3 years ago
Group 17 is called the halogen family, and the group to its right is called the noble gases. How are these elements alike and ho
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-The group 7 elements are also known as the halogens. They include fluorine, chlorine, bromine and iodine, which all have seven electrons in their outer shell.

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3 characteristics of physical change
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Answer:

Explanation:

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Chegg ) Alice owns 20 grams of a radioactive isotope that has a half-life of ln(4) years. (a) Find an equation for the mass m(t)
stepladder [879]

Answer:

m(t)=20e^{-0.5t}

Explanation:

Given:

Initial mass of isotope (m₀) = 20 g

Half life of the isotope (t_{1/2}) = (ln 4) years

The general form for the radioactive decay of a radioactive isotope is given as:

m(t)=m_0e^{-kt}

Where,

m(t)\to mass\ after\ 't'\ years\\t\to years\ passed\\k\to rate\ of\ decay\ per\ year

So, the equation is: m(t)=20e^{-kt}

At half-life, the mass is reduced to half of the initial value.

So, at t=t_{1/2},m(t)=\frac{m_0}{2}. Plug in these values and solve for 'k'. This gives,

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Hence, the equation for the mass remaining is given as:

m(t)=20e^{-0.5t}

8 0
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