What do objects in space that are moving at a constant velocity in a straight line do?

Calculate the potential V(r) for r>rb. (Hint: The net potential is the sum of the potentials due to the individual spheres.)

Marat540 [252]

Answer:

The potential for r > rb is equal to zero.

Explanation:

For r > rb, the potential is:

$V=\frac{Kq}{r}$

Then, the net potential is:

$V_{(r)} =\frac{K(+\epsilon )}{r} +\frac{K(-\epsilon )}{r}$

$K=\frac{1}{4\pi \epsilon _{o} }$

$V_{(r)} =\frac{K(+\epsilon )}{r} -\frac{K(\epsilon )}{r}\\V_{(r)}=0$

8 0

4 years ago

If the average pitcher is releasing the ball from a height of 1.8 m above the ground, and the pitcher's mound is 0.2 m higher th

mina [271]

The catcher can catch the ball at a height of 0.96 m from the ground.

The distance between the pitcher's mound and the catcher's box is about 60'6", which translates to 18.44 m. An average pitcher can pitch with speeds ranging from 88 mph to 97 mph, which is from 39.3 m/s to 43.4 m/s.

Assume the pitcher pitches a ball horizontally with a speed of 40 m/s. If the catcher catches the ball in a time t, then the ball travels a horizontal distance x of 18.44 m and at the same time falls through a height y.

The horizontal motion of the ball is uniform motion since no force acts on the ball ( assuming no air resistance) and hence the acceleration of the ball along the horizontal direction is zero.

Therefore,

$x=ut$

Calculate the time t by substituting 18.44 m for x and 40 m/s for u.

$t=\frac{x}{u} \\ =\frac{18.44 m}{40 m/s} \\ =0.461s$

The ball is acted upon by the earth's gravitational attraction and hence it accelerates downwards with an acceleration equal to the acceleration due to gravity g.

Since a horizontal projection is assumed, the ball has no component of velocity in the downward direction.

Therefore, for vertical motion, which is an accelerated motion, the distance y, the ball falls in the time t taken by it to reach the catcher's box is given by the equation,

$y=\frac{1}{2} gt^2$

Substitute 9.8 m/s² for g and 0.461 s for t.

$y=\frac{1}{2} gt^2\\ y=\frac{1}{2}(9.8 m/s^2)(0.461s)^2=1.04 m$

The pitcher releases the ball at a height of 1.8 m from a mound which is at a height of 0.2 m. Thus, the ball is released at a height of 2.0 m from the ground. It falls through a distance of 1.04 m in the time it takes to reach the catcher.

Therefore, the height at which the catcher needs to keep his glove so as to catch the ball is given by, $(2.0 m)-(1.04 m)=0.96 m$

The catcher needs to hold his glove at a height of <u>0,96 m from the ground.</u>

8 0

3 years ago

a refridgerator has a coefficient of performance equal to 4.2 how much work must be done on the refridgerator in order to remove

anastassius [24]

Answer:

60 J

Explanation:

Given That

Coefficient of performance, CoP = 4.2

Quantity of heat of the refrigerator, Q = 250 J

Work done by the refrigerator, W = ?

First, it should be noted that the Coefficient of Performance can be said to be, the ratio of the Heat required to the Work done by the system. Mathematically written as,

CoP = Q / W

Since we are already given the values from our question, we can plug it in as it's a pretty straightforward question

4.2 = 250 / W, making W subject of formula, we have

W = 250 / 4.2

W = 60J

Thus, the Work done by the refrigerator is 60 J

5 0

3 years ago

A car is moved 26km/hr due east for 4minute.what is its average velocity in m/s?

fenix001 [56]

Answer:

25m/sec

Explanation:

Speed of car in first 15 minutes = 40 km/h

Distance covered = speed × Time taken

60 minutes = 1 hour

15 minutes =

60

15

hour

Distance covered = 240×

60

15

=10km

Speed of car in next 15 minutes = 60 km/h

Distance covered = 60×

60

15

=15km

∴ Total distance covered = (10+15)km = 25 km

6 0

3 years ago

I need help with this please

Firdavs [7]

C is the answer to the question

6 0

3 years ago