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3 years ago

Please help me anyone i will mark brainliest

Physics

1 answer:

Mashcka [7]3 years ago

3 0

Answer:

for 4.567 I can't tell if it x $10^{3}$ or x $10^{5}$ so:

answer using x $10^{3}$ = 0.0000644697N

answer using x $10^{5}$ = 0.00644697

Explanation:

use equation F = GMm/ $R^{2}$

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The gravitational force between two objects that are 2.1 times 10^-1 m apart is 3.2 times 10^-6 N. If the mass of one object is

Shtirlitz [24]

Answer:

Mass of the other object is 38.45 kg.

Explanation:

Given:

The gravitational force between two objects is, $F=3.2\times 10^{-6}$ N

Mass of one object is, $m_{1}=55\ kg$

Distance between the objects is, $r=2.1\times 10^{-1}\ m$

Gravitational constant is, $G =6.674\times 10^{-11}\ m^3 kg^{-1}s^{-2}$

Let the mass of the other object be $m_{2}\ kg$

Gravitational force is given as:

$F=\frac{Gm_1m_2}{r^2}$

Plug in the given values and solve for $m_2$ . This gives,

$3.2\times 10^{-6}=\frac{6.674\times 10^{-11}\times 55\times m_2}{(2.1\times 10^{-1})^2}\\3.2\times 10^{-6}=\frac{6.674\times 10^{-11}\times 55\times m_2}{0.0441}\\3.2\times 10^{-6}=8.3236\times 10^{-8}m_2\\m_2=\frac{3.2\times 10^{-6}}{8.3236\times 10^{-8}}= 38.45\ kg$

Therefore, the mass of the other object is 38.45 kg.

3 0

4 years ago

A metal ring 4.60 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpend

telo118 [61]

Answer:

A. Ein = 8.05*10^-4 V/m

B. Clockwise sense

Explanation:

A. the magnitude of the electric field induced in the ring is obtaind by using the following formula:

$\int E_{in} \cdot ds=-\frac{d\Phi_B}{dt}$ (1)

Ein: induced electric field

ds: differential of a path of the ring

ФB: magnetic flux in the ring

The Ein vector is parallel to ds in the complete ring. Furthermore, the area of the ring is constant, hence, you have in the equation (1):

$\int E_{in}ds=E_{in}(2\pi r)=-A\frac{dB}{dt}\\\\E_{in}=-\frac{A}{2\pi r}\frac{dB}{dt}$ (2)

dB/dt = -0.280T/s (it is decreasing)

A: area of the ring = π(r/2)^2= (π/4) r^2

r: radius of the ring = 4.60/2 = 2.30 cm

Then, you replace the values of all variables in the equation (2):

$E_{in}=-\frac{(\pi/4)r^2}{2\pi r}\frac{dB}{dt}=\frac{r}{8}\frac{dB}{dt}\\\\E_{in}=-\frac{0.0230m}{8}(-0.280T)=8.05*10^{-4}\frac{V}{m}$

hence, the induced electric field is 8.05*10^-4 V/m

B. The induced current in the ring produced a magnetic field that is opposite to the magnetic field of the magnet. The, in this case you have that the induced current is in a clockwise sense.

6 0

4 years ago

Plz help i need help on my homework

tatiyna

The resultant force is 4 N to the right.

6 0

4 years ago

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The force required to maintain an object at a constant velocity in free space is equal to

alex41 [277]

Answer:

The force required to maintain an object at a constant velocity in free space is equal to Zero.