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4 years ago

11

Suzie Spacewalker hovers in space beside a rotating space station in outer space. Both she and the center of mass of the space s

tation are at relative rest. If the space station is in Earth orbit, then Suzie____________

1 answer:

Inga [223]4 years ago

3 0

Answer:

is in the earths orbit

Explanation:

for Suzie to hover in space beside the rotating space station, she and the center of mass of the space station are at relative rest which happens when space station is in Earth orbit, hence she is in the earths orbit.

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URGENT. Physics quiz on force, distance, etc. will reward brainliest.

goblinko [34]

13a) 9 J

The work done is equal to the area under the curve between x=0 cm and x=30 cm. However, first we should find the magnitude of the force for x=30 cm. If we notice that the force is proportional to the stretching x, we can set the following proportion to find the value of F for x=30 cm:

$10 N : 5 cm = x : 30 cm$

$x=\frac{30 cm \cdot 10 N}{5 cm}=60 N$

And so, the work done is

$W=Area=\frac{1}{2}(base)(height)=\frac{1}{2}(0.30 m)(60 N)=9 J$

13b) 24.5 m/s

The kinetic energy gained by the arrow is equal to the work done in stretching the bow:

$K=W=9 J$

Given the formula for the kinetic energy:

$K=\frac{1}{2}mv^2$

we can find the speed v of the arrow:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2\cdot 9J}{0.030 kg}}=24.5 m/s$

13c) 30.6 m

If shot vertically upward, at the point of maximum height all the initial kinetic energy of the arrow is converted into gravitational potential energy:

$\frac{1}{2}mv^2 = mgh$

Re-arranging the formula and using the initial speed of the arrow, we can find its maximum height h:

$h=\frac{v^2}{2g}=\frac{(24.5 m/s)^2}{2(9.81 m/s^2)}=30.6 m$

14) 20 m/s

We can solve the problem by using the work-energy theorem. In fact, the work done by the frictional force of the brake is equal to the change in kinetic energy of the car:

$W=\Delta K=K_f -K_i$

$Fd=\frac{1}{2}mv^2-\frac{1}{2}mu^2$

where

$F=-2500 N$ is the force applied by the brakes (with a negative sign, since it is opposite to the displacement of the car)

$d=100 m$ is the displacement of the car

$m=1000 kg$ is the car's mass

$v$ is the final speed of the car

$u=30 m/s$ is the initial speed of the car

By re-arranging the equation, we can find v:

$v=\sqrt{\frac{2(Fd+\frac{1}{2}mu^2)}{m}}=20 m/s$

15) 5.0 m/s

We can solve the problem by using the law of conservation of energy:

$U_i + K_i = U_f + K_f\\mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2$

where

m is the mass of the pendulum

$h_i=1.2 m$ is the initial height of the pendulum

$u=3 m/s$ is the initial speed of the pendulum

$h_f=0.4 m$ is the final height of the pendulum

$v$ is the final speed of the pendulum

Re-arranging the equation, we can find v:

$v=\sqrt{2gh_i + u^2 - 2gh_f}=5.0 m/s$

16) Point B (at the top of the loop)

Gravitational potential energy is defined as:

$U=mgh$

where m is the mass, g is the gravitational acceleration and h is the height above the ground. Therefore, we see that the potential energy is proportional to h: the higher the ball above the ground, the greater its potential energy. In this example, the point of maximum height is point B, therefore it is the point where the ball has the largest potential energy.

17) Law of conservation of energy: the total mechanical energy of an isolated object is conserved (if no frictional force act on it)

Example: A stone left falling from rest from a cliff. Let's call h the height of the cliff, m the mass of the stone. The mechanical energy of the stone is constant, and it is sum of the potential energy and kinetic energy:

$E=U+K$

At the top of the cliff, the kinetic energy is zero (the stone is at rest), so all its energy is potential energy:

$E_i = U_i = mgh$

When the stone falls, its energy is converted into kinetic energy. Just before hitting the ground, the height has become zero, h=0, so the potential energy is zero and all the mechanical energy is now kinetic energy:

$E_f=K_f=\frac{1}{2}mv^2$

since the mechanical energy must be conserved, we can write

$E_i=E_f\\mgh = \frac{1}{2}mv^2\\2gh=v^2$

6 0

4 years ago

A block of mass m = 2.0 kg lies on a rough ramp that is inclined at an angle θ = 20oto the horizontal. A force F of magnitude 5.

Marina86 [1]

Answer:

a) 0.64 b) 2.17m/s^2 c) 8.668joules

Explanation:

The block was on the ramp, the ramp was inclined at 20degree. A force of 5N was acting horizontal to the but not parallel to the ramp,

Frictional force = horizontal component of the weight of the block along the ramp + the applied force since the block was just about move

Frictional force = mgsin20o + 5N = 6.71+5N = 11.71

The force of normal = the vertical component of the weight of the block =mgcos20o = 18.44

Coefficient of static friction = 11.71/18.44= 0.64

Remember that g = acceleration due to gravity (9.81m/s^2) and m = mass (2kg)

b) coefficient of kinetic friction = frictional force/ normal force

Fr = 0.4* mgcos 20o = 7.375N

F due to motion = ma = total force - frictional force

Ma = 11.71 - 7.375 = 4.335

a= 4.335/2(mass of the block) = 2.17m/s^2

C) work done = net force *distance = 4.335*2= 8.67Joules

8 0

3 years ago

After your school's team wins the regional championship, students go to the dorm roof and start setting off fireworks rockets. T

Ugo [173]

Answer:

the distance is 315.3696 m

Explanation:

The computation of the distance is given below:

Given that

Sound intensity = 1.67 × 10^-6 W/m^2

And, the distance = 233 m

Now as we know that

Power = Intensity × surface area

1.67 × 10^-6 × 4π(233)^2 = 1.67 × 10^-6 ÷ 2× 4π × d^2

d^2 = 2 × (223)^2

= √2 × 223

= 315.3696 m

Hence, the distance is 315.3696 m

5 0

3 years ago

If a car's mass is 439 kg, what is its weight on earth?

trasher [3.6K]

The correct answer is B.

8 0

4 years ago

Describe how the motion of a planet (like Mars) among the stars of the zodiac on the sky is qualitatively different from the mot

Advocard [28]

Explanation:

In the plane of the Solar System, called Ecliptic plane, Sun is at the center and all the planets including Earth are going around the Sun in their respective orbits. As seen form the Earth, Sun appear to go around it and in this process it crosses through 12 constellations. These constellations are called as Zodiacs. Now since all the planets move in almost the same plane, they will appear in the sky to be lying near the ecliptic.

Sun's own motion is not that evident to us because of which the impact of that motion is nullified. But, motion of planet is clearly visible and it also shows in their annual round around the Sun. Here, Mars also shows retrograde motion which means it will show a back and forth motion in the sky. Because of which Mars might be visible in the same zodiac for a longer Duration as compared to the Sun.

7 0

3 years ago