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3 years ago

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A 4-kg mass moving with speed 2 m/s, and a 2-kg mass moving with a speed of 4 m/s, are gliding over a horizontal frictionless su

rface. Both objects encounter the same horizontal force, which directly opposes their motion, and are brought to rest by it. Which statement best describes their respective stopping distances?
A) Both masses travel the same distance before stopping.
B) The 2-kg mass travels twice as far as the 4-kg mass before stopping
C) The 2 kg mass travels greater than twice as far
D) The 4-kg mass travels twice as far as the 2-kg mass before stopping

1 answer:

Akimi4 [234]3 years ago

3 0

Answer:

B) The 2-kg mass travels twice as far as the 4-kg mass before stopping

Explanation:

As we know that both mass have same horizontal force opposite to their motion

So we will have

$a = \frac{F}{m}$

$a_1 = \frac{F}{4}$

$a_2 = \frac{F}{2}$

now the stopping distance of an object moving with initial speed v is given as

$v_f^2 - v_i^2 = 2(-a) d$

$d = \frac{v^2}{2a}$

so here we have

$d_1 = \frac{2^2}{\frac{F}{4}}$

$d_1 = \frac{16}{F}$

for other object we have

$d_2 = \frac{4^2}{\frac{F}{2}}$

$d_2 = \frac{32}{F}$

So correct answer will be

B) The 2-kg mass travels twice as far as the 4-kg mass before stopping

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Answer:

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<em>This problem has been solved!See the answerA student weighs 1200N. They are standing in an elevator that is moving downwards at a constant speed of </em><em>4.9 m/s. What is the magnitude of the net force acing on the student?</em>

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2 years ago

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A train leaves the station heading south on the tracks. It takes 5 seconds to reach 50 miles per hour. It completes the entire 1

Musya8 [376]

Well, in order to figure out the answer is to divide until you figure out how many miles they went per second. If it takes 5 seconds to reach 50 miles per hour it took 10 seconds per every 10 miles meaning each mile took 1 second. (Not actually possible but the answer) So, If it finished a 100 mile trip in 2 hours it took an hour for 50 miles. If it took 1 hour for 50 miles divide 60/50 which gets you 1.2 so it took 1.2 miles per minute meaning the car went 120 miles per hour I believe. I hope this helps :)

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3 years ago

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Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. A pipe closed at both ends can have standing wav

uranmaximum [27]

Answer:

68.8 Hz

137.6 Hz, 206.4 Hz

Explanation:

L = Length of tube = 2.5 m

v = Velocity of sound in air = 344 m/s

Distance between nodes is given by

$L=\dfrac{\lambda}{2}+m\dfrac{\lambda}{2}\\\Rightarrow \dfrac{\lambda(n+1)}{2}=L\\\Rightarrow \lambda=\dfrac{2L}{n+1}$

Where n = 0, 1, 2, 3, ...

Making n+1 = n

$\lambda=\dfrac{2L}{n}$

where n = 1, 2, 3 .....

For fundamental frequency n = 1

$\lambda=\dfrac{2\times 2.5}{1}\\\Rightarrow \lambda=5\ m$

Frequency is given by

$f=\dfrac{v}{\lambda}\\\Rightarrow f=\dfrac{344}{5}\\\Rightarrow f=68.8\ Hz$

The fundamental frequency is 68.8 Hz

First overtone

$2f=2\times 68.8=137.6\ Hz$

Second overtone

$3f=3\times 68.8=206.4\ Hz$

The overtones are 137.6 Hz, 206.4 Hz

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3 years ago

If the star Sirius emits 23 times more energy than the Sun, why does the Sun appear brighter in the sky?

Ganezh [65]

Answer:

As b ∝ (L/r²) and

the distance of the sun from the earth is 0.00001581 light years

and

the distance of the Sirius from the earth is 8.6 light years

hence,

the Sun appear brighter in the sky

Explanation:

The brightness (b) is directly proportional to the Luminosity of the star (L) and inversely proportional to the square of the distance between the star and the observer (r).

thus, mathematically,

b ∝ (L/r²)

now,

given

L for sirius is 23 times more than the sun i.e 23L

now,

the distance of the sun from the earth is 0.00001581 light years

and

the distance of the Sirius from the earth is 8.6 light years

thus,

using the the relation between conclude that the value of brightness for the Sirius comes very very low as compared to the value for brightness for the Sun.

hence, the sun appears brighter

5 0

3 years ago

Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=m_{1} V_{o} + m_{2} U_{o}$ (2)

$p_{f}=(m_{1} + m_{2}) V_{f}$ (3)

$m_{1}=110 kg$ is the mass of the first football player

$V{o}=-7 m/s$ is the velocity of the first football player (to the south)

$m_{2}=75 kg$ is the mass of the second football player

$U_{o}=4 m/s$ is the velocity of the second football player (to the north)

$V_{f}$ is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

$m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}}$ (5)

$V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg}$ (6)

$V_{f}=-2.54 m/s$ (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy $\Delta K$ is defined as:

$\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2}$ (8)

Simplifying:

$\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2})$ (9)

$\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2})$ (10)

Finally:

$\Delta K=-2351.25 J$ (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

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2 years ago