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2 years ago

Rory uses a force of 25 N to lift her grocery bag while doing 50 J of work. How far did she lift the

Physics

1 answer:

Dennis_Churaev [7]2 years ago

8 0

Answer:

$W=Fd \\ 50 = 25d \\ d = \frac{50}{25} \\ \color{yellow} \boxed{d = 2m}$

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3 years ago

Read 2 more answers

I need homework help

KIM [24]

1) the weight of an object at Earth's surface is given by $F=mg$ , where m is the mass of the object and $g=9.81 m/s^2$ is the gravitational acceleration at Earth's surface. The book in this problem has a mass of m=2.2 kg, therefore its weight is
$F=mg=(2.2 kg)(9.81 m/s^2)=21.6 N$

2) On Mars, the value of the gravitational acceleration is different: $g=3.7 m/s^2$ . The formula to calculate the weight of the object on Mars is still the same, but we have to use this value of g instead of the one on Earth: $F=mg=(2.2 kg)(3.7 m/s^2)=8.1 N$

3) The weight of the textbook on Venus is F=19.6 N. We already know its mass (m=2.2 kg), therefore by re-arranging the usual equation F=mg, we can find the value of the gravitational acceleration g on Venus:
$g= \frac{F}{m}= \frac{ 19.6 N}{2.2 kg}=8.9 m/s^2$

4) The mass of the pair of running shoes is m=0.5 kg. Their weight is F=11.55 N, therefore we can find the value of the gravitational acceleration g on Jupiter by re-arranging the usual equation F=mg:
$g= \frac{F}{m} = \frac{11.55 N}{0.5 kg} =23.1 m/s^2$

5) The weight of the pair of shoes of m=0.5 kg on Pluto is F=0.3 N. As in the previous step, we can calculate the strength of the gravity g on Pluto as
$g= \frac{F}{m} = \frac{0.3 N}{0.5 kg} =0.6 m/s^2$

6) On Earth, the gravity acceleration is $g=9.81 m/s^2$ . The mass of the pair of shoes is m=0.5 kg, therefore their weight on Earth is
 $F=mg=(0.5 kg)(9.81 m/s^2)=4.9 N$

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2 years ago

Katelyn (55 kg) is practicing a drop jump in the biomechanics lab. She steps off a plyometrics box, lands on the force plate, an

suter [353]

Answer:

J = 357.5 kg*m/s

Explanation:

The impulse exerted on Katelyn when she was on the force plate, is equal to the change in her momentum, according to Newton's 2nd Law.
Assuming as the positive direction the upward direction (coincident with the positive y-axis) we can express the initial momentum as follows:

$p_{o} = m*v_{o} = 55 kg * (-3.0 m/s) (1)$

By the same token, the final momentum is as follows:

$p_{f} = m*v_{f} = 55 kg * (3.5 m/s) (2)$

As we have already said, the impulse J is just equal to the change in momentum, i.e., the difference between (2) and (1):

$J = p_{f} - p_{o} = m* (v_{f} -v_{o}) = 55 kg* (3.5m/s- (-3.0m/s)) = 357.5 kg*m/s (3)$

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2 years ago