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Lesechka [4]
3 years ago
11

1 2 3 4 5 6 7 8 9 10 TIME REMAINING 59:49 Which changes would result in a decrease in the gravitational force between two object

s? Check all that apply.
Physics
1 answer:
Masteriza [31]3 years ago
3 0

Answer:

<em>increasing the distance between the objects</em>

<em>decreasing the mass of one of the objects</em>

<em>decreasing the mass of both objects </em>

Explanation:

The complete question would be:

<em>Which changes would result in a decrease in the gravitational force between two objects? Check all that apply.</em>

<em>increasing the distance between the objects</em>

<em>decreasing the distance between the objects</em>

<em>increasing the mass of one of the objects</em>

<em>increasing the mass of both objects</em>

<em>decreasing the mass of one of the objects</em>

<em>decreasing the mass of both objects </em>

---------------------------------------------------------------------------------------

Gravitational force between two objects can be computed using the formula:

Fg = G\dfrac{Mm}{r^2}

Where:

Fg is the gravitational force (Newtons or N)

G is the gravitational constant (6.674⋅10⁻¹¹N(m/kg)²)

M is the mass of one object (Kilograms or Kg)

m is the mass of the other object (Kilograms or Kg)

r is the distance between the two objects (m)

Now notice the relationship between the mass and gravitational force. Their relationship is direct, meaning as the value of one goes up, the other goes up and the reverse holds true. If the mass of one or both of the objects decreases, the gravitational force also decreases.

Now look at the relationship between the distance between the objects and the gravitational force-- they are indirect. This means that as one increases, the other decreases. So if we increase the distance between the objects, the gravitational force between them decreases as well.

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A long solenoid with 8.22 turns/cm and a radius of 7.00 cm carries a current of 19.4 mA. A current of 3.59 A exists in a straigh
daser333 [38]

Answer:

a. 3.039cm

b.magnetic field is B=2.958\times10^{-5}T

Explanation:

Direction of the solenoid magnetic field is along the axis of the solenoid. and magnetic field due to the wire perpendicular to that due to the solenoid.. Magnetic field at r is given by:

\overrightarrow B = \overrightarrow B_s+ \overrightarrow B_w,\ \ \ \ \  \overrightarrow B_s\perp \overrightarrow B_w

Angle of net magnetic field from axial direction is given by:

tan\  \theta=\frac{B_w}{B_s},

Field due to solenoid:

B_s=\mu_onI_s,  \ \ \ \ n=(8.22 t/cm)(100cm/m)=822turn/m

Field due to wire:

B_w=\frac{\mu_oI_w}{2\pi r}

Therefore, r:

tan\  \theta=\frac{B_w}{B_s}\\\\=\frac{\mu_oI_w}{2\pi r(\mu_o nI_s)}\\\\r=\frac{I_w}{2\pi  nI_stan \ \theta}\\\\r=\frac{3.59A}{2\pi\times822\times19.4\times10^{-3}A \ tan 49.7\textdegree}\\\\r=3.039cm

Hence, the radial distance is 3.039cm

b.The magnetic field strength is given by:

B=\sqrt{B_w^2+B_s^2}\\\\tan 49.7\textdegree=\frac{B_w}{B_s}\\\\1.179=\frac{B_w}{B_s}\\\\B_w=1.179B_s\\\\B=\sqrt{(4\pi\times10^{-7}T.m/A\times 822\times19.4\times10^{3}A)+1.179(4\pi\times10^{-7}T.m/A\times 822\times19.4\times10^{-3}A)}\\\\B=2.958\times10^{-5}T

Hence, the magnetic field is B=2.958\times10^{-5}T

7 0
4 years ago
A 2150 kg satellite used in a cellular telephone network is in a circular orbit at a height of 780 km above the surface of the e
Tom [10]

Answer:

a)F=16741.9N

b)\frac{F}{W}=0.795

Explanation:

The gravitational force on the satellite is calculated with Newton's Gravitation Law:

F=\frac{GMm}{r^2}

Where M=5.97\times10^{24}kg is Earth's mass, m=2150kg is the satellite mass, r=R+h is the distance between their centers, where h=780000m is the height of the satellite (from Earth's surface) and R=6371000m is Earth's radius, and G=6.67\times10^{-11}Nm^2/kg^2 is the gravitational constant.

a) With these values we then have:

F=\frac{GMm}{r^2}=\frac{(6.67\times10^{-11}Nm^2/kg^2)(5.97\times10^{24}kg)(2150kg)}{(6371000m+780000m)^2}=16741.9N

b) And the fraction this force is of the satellite’s weight <em>W=mg</em> is:

\frac{F}{W}=\frac{GMm}{mgr^2}=\frac{GM}{gr^2}=\frac{(6.67\times10^{-11}Nm^2/kg^2)(5.97\times10^{24}kg)}{(9.8m/s^2)(6371000m+780000m)^2}=0.795

5 0
4 years ago
A truck is moving around a circular curve at a uniform velocity of 13 m/s. If the centripetal force on the truck is 3300 N and t
kozerog [31]
Well, first of all, the truck's velocity is constantly changing, not 'uniform'.
Velocity consists of speed and direction.  So, even if the truck's speed is
constant, its direction keeps changing as long as it's on a circular curve,
so its velocity is constantly changing.

The force needed to keep a mass moving in a circle is

                                 F = (mass) x (speed)² / (radius)
                      
                             3300 N  =  (1600 kg) (13 m/s)² / R

                           3300 kg-m/s²  =  (1600 kg) (169 m²/s²) / R

                           R  =  (1600 kg) · (169 m²/s²) / (3300 kg·m/s²)

                               =  (1600 · 169 / 3300)  meters

                               =        81.9  meters     
6 0
3 years ago
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An object, 5 cm high, is placed on the principal axis of a diverging lens of focal length 20 cm. The object is 30 cm from the le
Nata [24]

Answer:

The answer is 70 cm

Explanation:

If you add All the numbers together, you receive an 55 cm then you add 15 because the points on the diagram also count.

3 0
3 years ago
An object can be moving from 10 seconds and still have zero displacement true or false
Tasya [4]

It is True :) :( :) :(


8 0
4 years ago
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