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wel
2 years ago
14

b. What would the momentum be if the mass of the bowling ball were doubled and its velocity still was 3 m/s?​

Physics
1 answer:
umka2103 [35]2 years ago
6 0

Answer:

Twice.

Explanation:

The momentum of an object is given by :

p = mv

Where

m is mass and v is the velocity

If the mass of the ball were doubled, m'=2m and v'=v=3 m/s

New momentum,

p'=m'v'

p'=2m × v

p'=2mv

or

p'=2p

So, the new momentum becomes twice the initial momentum.

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Svetllana [295]

If the bag is motionless, then it's not accelerating up or down.
That fact right there tells you that the net vertical force on it
is zero.  So the sum of any upward forces on it is exactly equal
to the downward gravitational force ... the bag's "weight".

If the bag is suspended from a single rope, then the tension
in the rope must be equal to the 100-N weight of the bag.

And if there are four ropes holding it up, then the sum of
the four tensions is 100N.  If the ropes have been carefully
adjusted to share the load equally, then the tension is 25N
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3 years ago
How many wolves are in the Southeast US
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Approximately 100-120 wolves
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3 years ago
Read 2 more answers
A student gives a 5.0 kg box a brief push causing the box to move with an initial speed of 8.0 m/s along a rough surface. The bo
WITCHER [35]

Answer:

The time taken to stop the box equals 1.33 seconds.

Explanation:

Since frictional force always acts opposite to the motion of the box we can find the acceleration that the force produces using newton's second law of motion as shown below:

F=mass\times acceleration\\\\\therefore acceleration=\frac{Force}{mass}

Given mass of box = 5.0 kg

Frictional force = 30 N

thus

acceleration=\frac{30}{5}=6m/s^{2}

Now to find the time that the box requires to stop can be calculated by first equation of kinematics

The box will stop when it's final velocity becomes zero

v=u+at\\\\0=8-6\times t\\\\\therefore t=\frac{8}{6}=4/3seconds

Here acceleration is taken as negative since it opposes the motion of the box since frictional force always opposes motion.

5 0
3 years ago
A train traveled from Station A to Station B at an average speed of 80 kilometers per hour and then from Station B to Station C
Vinil7 [7]

Answer:

1)

75 kmh⁻¹

2)

75 kmh⁻¹

Explanation:

1)

v_{ab} = Speed of train from station A to station B = 80 kmh⁻¹

d_{ab} = distance traveled from station A to station B

t_{ab} = time of travel between station A to station B

we know that

Time = \frac{distance}{speed}

t_{ab} = \frac{d_{ab}}{v_{ab}} = \frac{d_{ab}}{80}

d_{bc} = distance traveled from station B to station C

v_{bc} = Speed of train from station B to station C = 60 kmh⁻¹

t_{bc} = \frac{d_{bc}}{v_{bc}} = \frac{d_{bc}}{60}

Total distance traveled is given as

d = d_{ab} + d_{bc}

Total time of travel is given as

t = t_{ab} + t_{bc}

Average speed is given as

v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{d_{ab} + d_{bc}}{(\frac{d_{ab}}{80} ) + (\frac{d_{bc}}{60} ) }

Given that :

d_{ab} = 4 d_{bc}

So

v_{avg} = \frac{4 d_{bc} + d_{bc}}{(\frac{4 d_{bc}}{80} ) + (\frac{d_{bc}}{60} ) }\\v_{avg} = \frac{4 + 1}{(\frac{4 }{80} ) + (\frac{1}{60} ) }\\v_{avg} = 75 kmh^{-1}

2)

v_{ab} = Speed of train from station A to station B = 80 kmh⁻¹

t_{ab} = time of travel between station A to station B

d_{ab} = distance traveled from station A to station B

we know that

distance = (speed) (time)

d_{ab} = v_{ab} t_{ab}\\d_{ab} = 80 t_{ab}

d_{bc} = distance traveled from station B to station C

v_{bc} = Speed of train from station B to station C = 60 kmh⁻¹

t_{bc} = time of travel for train from station B to station C

we know that

distance = (speed) (time)

d_{bc} = v_{bc} t_{bc}\\d_{bc} = 60 t_{bc}

Total distance traveled is given as

d = d_{ab} + d_{bc}\\d = 80 t_{ab} + 60 t_{bc}

Total time of travel is given as

t = t_{ab} + t_{bc}

Average speed is given as

v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}

Given that :

t_{ab} = 3 t_{bc}

So

v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 (3) t_{bc} + 60 t_{bc}}{(3) t_{bc} + t_{bc}}\\v_{avg} = \frac{(300) t_{bc}}{(4) t_{bc}}\\v_{avg} = 75 kmh^{-1}

4 0
3 years ago
3b. Grace drives her car with a constant speed
miskamm [114]

Time = (distance covered) / (speed)

Time = (224 mi) / (56 mi/hr)

<em>Time = 4 hours</em>

3 0
3 years ago
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