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dimulka [17.4K]
2 years ago
7

A 1 kg object sits on the earth’s surface. What is the force of gravity between the object and the earth? (mass of the earth = 5

.97 x 10 24kg, radius of the earth = 6.37 x 10 6m)
Physics
1 answer:
snow_lady [41]2 years ago
4 0

Answer:

9.81N

Explanation:

the force of attraction is given by

F=<u>GmM</u><u>/</u><u>R²</u><u> </u>

where m is mass of the body

M is mass of the earth

R is radius of the earth

G is the universal gravitational constant(6.67x10-¹¹)

hence we substitute the values in the formula.

<em> </em><em>you</em><em> </em><em>can</em><em> </em><em>ask</em><em> </em><em>questions</em>

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NEES HELP!!!!
OLga [1]

Answer:

Doppler Theory

Explanation:

it's a theory regarding the change in wave frequency during the relative motion between a wave source and its observer.

6 0
2 years ago
Meter #1 can measure voltage to within 0.1 volts. Meter #2 can measure voltage to within 0.01 volts.
Ann [662]

Meter #2 is more precise.

There's no information here that tells us which meter is more accurate.

4 0
3 years ago
A pebble is thrown into the air with a velocity of 19/m at an angle of 36 with respect to the horizontal.
kow [346]

Answer:

The maximum height the pebble reaches is approximately;

A. 6.4 m

Explanation:

The question is with regards to projectile motion of an object

The given parameters are;

The initial velocity of the pebble, u = 19 m/s

The angle the projectile path of the pebble makes with the horizontal, θ = 36°

The maximum height of a projectile, h_{max}, is given by the following equation;

h_{max} = \dfrac{\left (u \times sin(\theta) \right)^2}{2 \cdot g}

Therefore, substituting the known values for the pebble, we have;

h_{max} = \dfrac{\left (19 \times sin(36 ^{\circ}) \right)^2}{2 \times 9.8} = 6.3633894140470403035477570509439

Therefore, the maximum height of the pebble projectile, h_{max} ≈ 6.4 m.

3 0
3 years ago
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Dmitrij [34]
Can you translate to english ?
8 0
3 years ago
An isloated point charge produce an electric field with magnitude E at a point 2 m away. At a point 1 m from the charge magnitud
Nina [5.8K]

Answer:

the correct answer is C,      E’= 4E

Explanation:

In this exercise you are asked to calculate the electric field at a given point

         E = k \frac{q}{r^2}

indicates that the field is E for r = 2m

         E = \frac{ k q}{4}                  (1)

the field is requested for a distance r = 1 m

         E ’= k \frac{q}{r'^2}

         E ’= k q / 1

 

from equation 1

         4E = k q

       

we substitute

        E’= 4E

so the correct answer is C

8 0
3 years ago
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