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3 years ago

11. A student lifts a 25 kg mass a vertical distance of 1.6 m in a time of 2.0 seconds.

Physics

1 answer:

goldenfox [79]3 years ago

4 0

Answer:

a) 245 N

b) 392 J

c) 196 W

Explanation:

a) Force needed to lift the 25 kg object is the same as the weight of the object:

$Force= m * g = 25kg * 9.8 \frac{m}{s^2} = 245 N$

b) Work is the force exerted times the distance moved:

$Work = F * d = 245N * 1.6 m = 392 Joules = 392 J$

c) The power exerted is the quotient of the work done over the amount of time it took to move the object:

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A package of mass 5 kg sits on an airless asteroid of mass 7.6 × 1020 kg and radius 8.0 × 105 m. We want to launch the package i

Effectus [21]

Answer:

s = 1.7 m

Explanation:

from the question we are given the following:

Mass of package (m) = 5 kg

mass of the asteriod (M) = 7.6 x 10^{20} kg

radius = 8 x 10^5 m

velocity of package (v) = 170 m/s

spring constant (k) = 2.8 N/m

compression (s) = ?

Assuming that no non conservative force is acting on the system here, the initial and final energies of the system will be the same. Therefore

• Ei = Ef

• Ei = energy in the spring + gravitational potential energy of the system

• Ei = \frac{1}{2}ks^{2} + \frac{GMm}{r}

• Ef = kinetic energy of the object

• Ef = \frac{1}{2}mv^{2}

• \frac{1}{2}ks^{2} + (-\frac{GMm}{r}) = \frac{1}{2}mv^{2}

• s = $\sqrt{\frac{m}[k}(v^{2}+\frac{2GM}{r})}$

s = $\sqrt{\frac{5}[2.8 x 10^5}(170^{2}+\frac{2 x 6.67 x10^{-11} x 7.6 x 10^{20}}{8 x 10^5})}$

s = 1.7 m

7 0

3 years ago

A rope is tied to a tree 4.5 feet from the ground and then run through a pulley hooked to a vehicle 33 feet from the tree. If a

Angelina_Jolie [31]

Answer:

The vehicle displacement is 9.90 feet.

Explanation:

Given that,

Height of tree = 4.5 feet

Distance = 33 feet

According to figure,

We need to calculate the value of l

Using Pythagorean theorem

$l=\sqrt{(33)^2+(4.5)^2}$

We need to calculate the vehicle displacement

Using horizontal component

Vehicle displacement =horizontal component of pulled rope

$Vehicle\ displacement= d\cos\theta$

Where, $\theta$ is angle between rope and ground

d = pulled length of rope

$Vehicle\ displacement=10\times\dfrac{33}{\sqrt{(33)^2+(4.5)^2}}$

$Vehicle\ displacement=9.90\ feet$

Hence, The vehicle displacement is 9.90 feet.

3 0

3 years ago

A uniform-density 7 kg disk of radius 0.21 m is mounted on a nearly frictionless axle. Initially it is not spinning. A string is

GREYUIT [131]

Answer:

$\omega = 22.67 rad/s$

Explanation:

Here we can use energy conservation

As per energy conservation conditions we know that work done by external source is converted into kinetic energy of the disc

Now we have

$W = \frac{1}{2}I\omega^2$

now we know that work done is product of force and displacement

so here we have

$W = F.d$

$W = (44 N)(0.9 m) = 39.6 J$

now for moment of inertia of the disc we will have

$I = \frac{1}{2}mR^2$

$I = \frac{1}{2}(7 kg)(0.21^2)$

$I = 0.154 kg m^2$

now from above equation we will have

$39.6 = \frac{1}{2}(0.154)\omega^2$

$\omega = 22.67 rad/s$

5 0

3 years ago

Read 2 more answers

Two horses, Thunder and Misty, are accelerating a wagon 1.3 m/s2. The force of friction is 75 N. Thunder is pulling with a force

just olya [345]

Answer:

1327 kg

Explanation:

So the net force exerted on the wagon would be the sum of forces from 2 horses subtracted by friction force

$F = 1000 + 800 - 75 = 1725 N$

This force results in an acceleration of a = 1.3 m/s2. We can use Newton's 2nd law to calculate the mass of the wagon

$F = ma$

$m = F / a = 1725 / 1.3 \approx 1327 kg$

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3 years ago

In which of these examples is the greatest movement occurring?

Elenna [48]

Answer:

not clear pic...but it's definitely not A)

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3 years ago