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3 years ago

The gravitational force between two objects with masses 1kg and 28kg separated by a distance 7m is ____________10-11 N.

Physics

1 answer:

ikadub [295]3 years ago

4 0

Answer:

a. 3.81

Explanation:

F = GMm/r^2

F = (6.67 x 10^-11 x 28 x 1) / 7^2

F = 3.81 x 10^-11 N

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Simple physics question, check the document. Should take about 3-5 minutes.

Ahat [919]

Answer:

The magnitude of the force that the 6.3 kg block exerts on the 4.3 kg block is approximately 41.9 N

Explanation:

Forces on block 4.3 kg are:

63N to the right and R21 (contact force from the 6.3 kg block) to the left

Net force on 4.3 kg block is: 63 N - R21

Forces on the 6.3 kg block are:

R12 to the right (contact force from the 4.3 kg block) and 11 N to the left.

So net force on the 6.3 kg block is: R12 - 11 N

According to the action-reaction principle the contact forces R21 and R12 must be equal in magnitude (let's call them simply "R").

Then, since the blocks are moving with the SAME acceleration, we equal their accelerations:

a1 = (63 N - R)/4.3 = (R - 11 N)/6.3 = a2

solve for R by cross multiplication

6.3 (63 - R) = 4.3 (R - 11)

396.9 - 6.3 R = 4.3 R - 47.3

369.9 + 47.3 = 10.6 R

444.2 = 10.6 R

R = 444.2 / 10.6

R = 41.90 N

5 0

3 years ago

an athlete runs 300 m up a hill at a steady speed of 3.0 m/s. She then immediately runs the same distance at 6.0 m/s . What is h

mina [271]

Answer:

4.0 m/s

Explanation:

In the first part of the run, the athlete runs a distance of

$d_1 = 300 m$

at a speed of

$v_1 = 3.0 m/s$

So, the time he/she takes is

$t_1 = \frac{d_1}{v_1}=\frac{300}{3.0}=100 s$

In the second part of the run, the athlete covers an additional distance of

$d_2 = 300 m$

with a speed

$v_2 = 6.0 m/s$

So, the time taken in this second part is

$t_2 = \frac{d_2}{v_2}=\frac{300}{6.0}=50 s$

So, the total distance covered is

d = 300 m + 300 m = 600 m

And the total time taken

t = 100 s + 50 s = 150 s

Therefore, the average speed for the entire trip is

$v=\frac{d}{t}=\frac{600}{150}=4.0 m/s$

4 0

4 years ago

A projectile fired from a gun has initial horizontal and vertical components of velocity equal to 60 m/s and 80 m/s, respectivel

DerKrebs [107]

This question involves the concepts of projectile motion and launch speed.

(a) The initial launch speed of the projectile is "100 m/s".

(b) The launch angle of the projectile is "53.13°".

<h3>(a) LAUNCH SPEED</h3>

A projectile motion is a motion that takes place on both x and y axes, simultaneously. In this motion the initial launch speed is given by the following formula:

$v_o=\sqrt{v_{ox}^2+v_{oy}^2}$

where,

$v_o$ = initial launch speed = ?
$v_{ox}$ = horizontal component of initial launch speed = 60 m/s
$v_{oy}$ = vertical component of initial launch speed = 80 m/s

Therefore,

$v_o = \sqrt{(60\ m/s)^2+(80\ m/s)^2}\\\\v_o = 100 m/s$

<h3>(b) LAUNCH ANGLE</h3>

Launch angle is given by th following formula:

$\theta = tan^{-1}(\frac{v_{oy}}{v_{ox}})=tan^{-1}(\frac{80\ m/s}{60\ m/s})\\\\\theta=53.13^o$

Learn more about the projectile motion here:

brainly.com/question/11049671

6 0

3 years ago

PLEASE HELP - Your concept map will compare the manner in which energy is transferred by mechanical waves and electromagnetic wa

makkiz [27]

The concept map is given in the attachment:

3 0

3 years ago

Read 2 more answers

A generator rotates at 95 Hz in a magnetic field of 0.025 T. It has 550 turns and produces an rms voltage of 170 V and an rms cu

weeeeeb [17]

Answer:

Peak current= 84.86 A

Area of each turn = 0.029 m^2

Explanation:

The peak value of current can be obtained from Irms= 0.707Io. Where Io is the peak current.

Hence;

Irms= 60.0A

Io= Irms/0.707

Io = 60.0/0.707

Io= 84.86 A

Vrms= 0.707Vo

Vo= Vrms/0.707= 170/0.707 = 240.45 V

From;

V0 = NABω

Where;

Vo= peak voltage

N= number of turns

B= magnetic field

A= area of each coil

ω= angular velocity

But ω= 2πf = 2×π×95= 596.9 rads-1

Substituting values;

A= Vo/NBω

A= 240.45/550×0.025×596.9

A= 0.029 m^2

6 0

4 years ago