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3 years ago

A 65 kg student sits 3 m away from a 70 kg student. What is the magnitude of the gravitational force between the two students?

Physics

1 answer:

LiRa [457]3 years ago

3 0

Answer:

$F=3.37\times 10^{-8}\ N$

Explanation:

Given that,

Mass of student 1, m₁ = 65 kg

Mass of student 2, m₂ = 70 kg

The distance between the students, d = 3 m

We need to find the magnitude of the gravitational force between the two students. The formula for the magnitude of the gravitational force between two masses is given by :

$F=G\dfrac{m_1m_2}{d^2}\\\\F=6.67\times 10^{-11}\times \dfrac{65\times 70}{(3)^2}\\F=3.37\times 10^{-8}\ N$

So, the gravitational force between the two students is $3.37\times 10^{-8}\ N$ .

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Calculate the current through a 3.0 Ď‰ resistor with a voltage of 9.0 v across it.

gulaghasi [49]

Answer: 3 A

Explanation:

According to<u> Ohm's law</u>:

$V=R.I$

Where:

$V=9 V$ is the voltage

$R=3\Omega$ is the resistance of the resistor

$I$ is the electric current (the value we want to find)

Isolating $I$ :

$I=\frac{V}{R}$

$I=\frac{9 V}{3\Omega}$

Finally:

$I=3 A$

7 0

3 years ago

a woman walks south at a speed of 2.0mph for 60 minutes. She then turns around and walks north at a distance of 3000m in 25 minu

TiliK225 [7]

When a woman walks south at a speed of 2.0mph for 60 minutes. She then turns around and walks north at a distance of 3000m in 25 minutes. then the woman's average speed during her entire motion would be 73.15 meters /minute.

<h3>What is speed?</h3>

The total distance covered by any object per unit of time is known as speed.

the mathematical expression for speed is given by

speed = total; distance /total time

As given in the problem a woman walks south at a speed of 2.0mph for 60 minutes

60 min = 1 hour

1 mile = 1.60934 km

The distance covered by her southwards = speed ×time

=2 mph × 60 minutes

= 3.218 km

She then turns around and walks north at a distance of 3000m in 25 minutes

The distance covered northward is 3000m

speed = total distance /total time

=(3218 +3000) /(60+25)

=73.15 meters /minutes

Thus, The average speed of the woman would be 73.15 meters /minute.

Learn more about speed from here

brainly.com/question/13263542

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8 0

1 year ago

What is the torque τa about axis a due to the force f⃗ ? express the torque about axis a at cartesian coordinates (0,0)?

pentagon [3]

Answer:

$\tau_a = F a sin \theta$

Explanation:

The torque of a force is given by:

$\tau = F d sin \theta$

where

F is the magnitude of the force

d is the distance between the point of application of the force and the centre of rotation of the system

$\theta$ is the angle between the direction of the force and d

In this problem, we have:

$F$ , the force

$a$ , the distance of application of the force from the centre (0,0)

$\theta$ , the angle between the direction of the force and a

Therefore, the torque is

$\tau_a = F a sin \theta$

5 0

3 years ago

block of mass 5kgriding on a horizontal frictionlessxy-plane surface is subjected tothree applied forces:→F1= 12√2N[ 45◦]→F2= (8

dsp73

Answer:

(i) See attached image for the drawing

(ii) net force given in component form: (20, 20)N with magnitude: $\sqrt{800} \,\,\,N$

Explanation:

First try to write all forces in vector component form:

The force F1 acting at 45 degrees would have multiplication factors of $\frac{\sqrt{2} }{2}$ on both axes, to take care of the sine and cosine projections. Therefore, the:

x-component of F1 is $F1_x=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N$

y-component of F1 is $F1_y=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N$

As far as force F2, it is given already in x and y components, then:

x-component of F2 = 8 N

y-component of F2 = -6 N (negative meaning pointing down the y-axis)

Force F3 has only component (upwards) in the y-direction

x-component of F3 = 0 N

y-component of F3 =14 N

The additions of all these component by component, gives the resultant force (R) acting on the 5 kg mass:

x-component of R = 12 + 8 = 20 N

y-component of R = 12 + 14 - 6 = 20 N

Therefore, the acceleration that the mass receives due to this force is given in component form as:

x-component of acceleration: 20 N / 5 kg = $4\,\,\,m/s^2$

y-component of acceleration: 20 N / 5 kg = $4\,\,\,m/s^2$

Now we can calculate the components of the velocity of this mass after 2 seconds of being accelerated by this force, using the formula of acceleration times time:

x-component of the velocity is: $v_x=4\,*\,2=8\,m/s$