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mars1129 [50]
3 years ago
10

Can anyone help ????

Physics
1 answer:
user100 [1]3 years ago
5 0

Explanation:

1. serie

2.Red

3.16 ohms

4.

I=V/R

6=V/16

V=16×6

V=96 volts

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where would the spaceprobe experience the strongest net (or total) gravitional force exerted on it by Earth and Mars
Arte-miy333 [17]

Answer:

r = 41.1 10⁹ m

Explanation:

For this exercise we use the equilibrium condition, that is, we look for the point where the forces are equal  

                  ∑ F = 0

                  F (Earth- probe) - F (Mars- probe) = 0

                  F (Earth- probe) = F (Mars- probe)

Let's use the equation of universal grace, let's measure the distance from the earth, to have a reference system

the distance from Earth to the probe is        R (Earth-probe) = r

the distance from Mars to the probe is        R (Mars -probe) = D - r

where D is the distance between Earth and Mars

                   

                 G  \ \frac{m \ M_{Earth}}{r^2} = G  \ \frac{m \ M_{Mars}}{(D-r)^2}

                 M_earth (D-r)² = M_Mars r²

                 (D-r) = \sqrt{ \frac{M_{Mars}}{ M_{Earth}} }    r

                  r ( 1 + \sqrt{ \frac{M_{Mars}}{M_{Earth}} }) = D

                  r = \frac{D}{ 1+ \sqrt{\frac{M_{Mars}}{ M_{Earth}} } }

We look for the values ​​in tables

                  D = 54.6 10⁹ m (minimum)

                  M_earth = 5.98 10²⁴ kg

                  M_Marte = 6.42 10²³ kg = 0.642 10²⁴ kg

                   

let's calculate

                  r = 54.6 10⁹ / (1 + √(0.642/5.98)  )

                  r = 41.1 10⁹ m

5 0
2 years ago
A woman exerts a horizontal force of 113 N on a crate with a mass of 31.2 kg.
sergeinik [125]

Answer:

a) 113N

b) 0.37

Explanation:

a) Using the Newton's second law:

\sum Fx =ma

Since the crate doesn't move (static), acceleration will be zero. The equation will become:

\sum Fx = 0

\sumFx = Fm - Ff = 0.

Fm is the applied force

Ff is the frictional force

Since Fm - Ff = 0

Fm = Ff

This means that the applied force is equal to the force of friction if the crate is static.

Since applied force is 113N, hence the magnitude of the static friction force will also be 113N

b) Using the formula

Ff = nR

n is the coefficient of friction

R is the reaction = mg

R = 31.2 × 9.8

R = 305.76N

From the formula

n = Ff/R

n = 113/305.76

n = 0.37

Hence the minimum possible value of the coefficient of static friction between the crate and the floor is 0.37

8 0
2 years ago
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GarryVolchara [31]
Diverge is the right answer
8 0
2 years ago
Water can’t stick to itself. True or false?
astra-53 [7]

Answer:ture

Explanation:

6 0
2 years ago
How much work is done in lifting a 6 kg object from the ground to a height of 4m?
Gekata [30.6K]
W=mgh W=(6)(9.8)(4) W= 235.2J
8 0
3 years ago
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