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Verdich [7]
2 years ago
8

At a given instant of time, a car and a truck are traveling side by side in adjacent lanes of a highway. The car has a greater v

elocity than the truck. Does the car necessarily have a greater acceleration? Explain.
Physics
1 answer:
miv72 [106K]2 years ago
8 0

Answer:

NO.

Explanation:

No, It does not mean that the acceleration of the car will be higher.

The car and the truck are moving in the same direction but may have any value of acceleration. Higher velocity of car cannot imply higher acceleration. Acceleration is change in velocity to the change in time. Even if the velocity of car is higher it does not mean its change in velocity will be higher.

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A 6.5 l sample of nitrogen at 25◦c and 1.5 atm is allowed to expand to 13.0 l. the temperature remains constant. what is the fin
ollegr [7]
Since the temperature of the gas remains constant in the process, we can use Boyle's law, which states that for a gas transformation at constant temperature, the product between the gas pressure and its volume is constant:
pV=k
which can also be rewritten as
p_1 V_1 = p_2 V_2 (1)
where the labels 1 and 2 mark the initial and final conditions of the gas.

In our problem, p_1 = 1.5 atm, V_1 =6.5 L and V_2 =13.0 L, so the final pressure of the gas can be found by re-arranging eq.(1):
p_2 = p_1  \frac{V_1}{V_2}= (1.5 atm) \frac{6.5 L}{13.0 L}=0.75 atm

Therefore the correct answer is
<span>1. 0.75 atm</span> 
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Nanotechnology and microtechnology devices are both very tiny. However, one property of nanotechnology distinguishes it from mic
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Nanotechnology displays particle quality
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Joe wanted to experiment with different factors that affect the freezing rate of water. He put two cups of water into each of tw
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Three different planet-star systems, which are far apart from one another, are shown above. The masses of the planets are much l
alex41 [277]

a) 4F0

b) Speed of planet B is the same as speed of planet A

Speed of planet C is twice the speed of planet A

Explanation:

a)

The magnitude of the gravitational force between two objects is given by the formula

F=G\frac{m_1 m_2}{r^2}

where

G is the gravitational constant

m1, m2 are the masses of the 2 objects

r is the separation between the objects

For the system planet A - Star A, we have:

m_1=M_p\\m_2 = M_s\\r=R

So the force is

F_A=G\frac{M_p M_s}{R^2}=F_0

For the system planet B - Star B, we have:

m_1 = 4 M_p\\m_2 = M_s\\r=R

So the force is

F=G\frac{4M_p M_s}{R^2}=4F_0

So, the magnitude of the gravitational force exerted on planet B by star B is 4F0.

For the system planet C - Star C, we have:

m_1 = M_p\\m_2 = 4M_s\\r=R

So the force is

F=G\frac{M_p (4M_s)}{R^2}=4F_0

So, the magnitude of the gravitational force exerted on planet C by star C is 4F0.

b)

The gravitational force on the planet orbiting around the star is equal to the centripetal force, therefore we can write:

G\frac{mM}{r^2}=m\frac{v^2}{r}

where

m is the mass of the planet

M is the mass of the star

v is the tangential speed

We can re-arrange the equation solving for v, and we find an expression for the speed:

v=\sqrt{\frac{GM}{r}}

For System A,

M=M_s\\r=R

So the tangential speed is

v_A=\sqrt{\frac{GM_s}{R}}

For system B,

M=M_s\\r=R

So the tangential speed is

v_B=\sqrt{\frac{GM_s}{R}}=v_A

So, the speed of planet B is the same as planet A.

For system C,

M=4M_s\\r=R

So the tangential speed is

v_C=\sqrt{\frac{G(4M_s)}{R}}=2(\sqrt{\frac{GM_s}{R}})=2v_A

So, the speed of planet C is twice the speed of planet A.

3 0
2 years ago
If a 40,000g cannon ball is 35m above the Earth's surface, how much potential energy does the cannon
mihalych1998 [28]

Answer: 21

Explanation:

because 9 + 10 = 21

8 0
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