we have to use newtons law of gravitation which is
F=GMm/r^2
G=6.67 x 10^-11N kg^2/m^2
M=(15kg)
m=15 kg
r=(3.0m)^2
putting values we have
=(6.67 x 10^-11N kg^2/m^2)(15kg)(15kg)/(3.0m)^2 
=1.67 x 10^-9N

7 0

3 years ago

As shown above an adhesive has been applied to contacting faces of two blocks so that the blocks interact with an adhesive force

rusak2 [61]

The magnitude of the adhesive force allows the top block to remain

attached to the bottom when the blocks and the wire are balanced.

The option that must be true is; $\mathbf{W_{bottom} = F_{adhesion}}$

Reason:

The tension exerted by the wire attached to the top block = T

Magnitude of the adhesive = $F_{adhesion}$

Weight of the top block = $W_{top}$

Weight of the bottom block = $W_{bottom}$

Given that with the exertion of the tension, the two blocks remain at rest, we have;

T = $W_{top}$ + $W_{bottom}$

The adhesive causes the bottom block to remain attached to the top block, we have;

Therefore, the magnitude of the adhesive force adds the bottom weight, to the top, weight, which gives;

The magnitude of the adhesive force = The weight of the bottom block

Therefore;

$W_{bottom}$ = $F_{adhesion}$

Learn more here:

brainly.com/question/18907970

8 0

3 years ago

** URGENT** The voltage across the primary winding is 350,000 V, and the voltage across the secondary winding is 17,500 V. If th

Vlad [161]

As we know that in transformers we have

$\frac{V_s}{V_p} = \frac{N_s}{N_p}$

here we know that

$V_s = 17,500 Volts$

$V_p = 350,000 Volts$

$N_s = 600 coils$

now from above equation we will have

$\frac{17500}{350000} = \frac{600}{N_p}$

$N_p = 600\times \frac{350000}{17500}$

$N_p = 12000 coils$

6 0

3 years ago