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rewona [7]
3 years ago
9

The force of Earth's gravity on a capsule in space will lessen as it moves farther away. If the capsule moves to twice its dista

nce, the force toward Earth becomes?
Physics
1 answer:
Bess [88]3 years ago
3 0

Answer: One quarter of the force

Explanation:

According to Newton's law of Gravitation, the force F exerted between two bodies of masses m1 and m2  and separated by a distance r  is equal to the product of their masses and inversely proportional to the square of the distance:

F=G\frac{(m1)(m2)}{r^2}    (1)

Where Gis the gravitational constant

This means that the gravity force decreases when the distance between these two bodies increases.

In this context, if the distance between the capsule and the Earth increases twice, the new distance will be 2r.

Substituting this distance in (1):

F=G\frac{(m1)(m2)}{(2r)^2}    (2)

F=G\frac{(m1)(m2)}{4r^2}    

<u>Finally:</u>

F=\frac{1}{4}G\frac{(m1)(m2)}{r^2} >>>This means the force toward Earth becomes one quarter "weaker"

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Boyle's law says that the volume of a gas varies inversely with the pressure. When the volume of a certain gas is 4l , the pressure is 720 kpa (kilopascals). What is the pressure when the volume is 10l ?

7 0
3 years ago
What direction does the earth rotate?
solniwko [45]
The earth rotates from west to east :)
4 0
3 years ago
There is a parallel plate capacitor. Both plates are 4x2 cm and are 10 cm apart. The top plate has surface charge density of 10C
liberstina [14]

Answer:

1) The total charge of the top plate is 0.008 C

b) The total charge of the bottom plate is -0.008 C

2) The electric field at the point exactly midway between the plates is 0

3) The electric field between plates is approximately 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates is approximately 1.807 × 10⁻⁷ N

Explanation:

The given parameters of the parallel plate capacitor are;

The dimensions of the plates = 4 × 2 cm

The distance between the plates = 10 cm

The surface charge density of the top plate, σ₁ = 10 C/m²

The surface charge density of the bottom plate, σ₂ = -10 C/m²

The surface area, A = 0.04 m × 0.02 m = 0.0008 m²

1) The total charge of the top plate, Q = σ₁ × A = 0.0008 m² × 10 C/m² = 0.008 C

b) The total charge of the bottom plate, Q = σ₂ × A = 0.0008 m² × -10 C/m² = -0.008 C

2) The electrical field at the point exactly midway between the plates is given as follows;

V_{tot} = V_{q1} + V_{q2}

V_q = \dfrac{k \cdot q}{r}

Therefore, we have;

The distance to the midpoint between the two plates = 10 cm/2 = 5 cm = 0.05 m

V_{tot} =  \dfrac{k \cdot q}{0.05} + \dfrac{k \cdot (-q)}{0.05}  = \dfrac{k \cdot q}{0.05} - \dfrac{k \cdot q}{0.05} = 0

The electric field at the point exactly midway between the plates, V_{tot} = 0

3) The electric field, 'E', between plates is given as follows;

E =\dfrac{\sigma }{\epsilon_0 } = \dfrac{10 \ C/m^2}{8.854 \times 10^{-12} \ C^2/(N\cdot m^2)} \approx 1.1294 \times 10^{12}\ N/C

E ≈ 1.1294 × 10¹² N/C

The electric field between plates, E ≈ 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates

The charge on an electron, e = -1.6 × 10⁻¹⁹ C

The force on an electron in the middle of the two plates, F_e = E × e

∴ F_e = 1.1294 × 10¹² N/C ×  -1.6 × 10⁻¹⁹ C ≈ 1.807 × 10⁻⁷ N

The force on an electron in the middle of the two plates, F_e ≈ 1.807 × 10⁻⁷ N

4 0
3 years ago
the height of seven falls in colorado is 5/2 the height of twin falls in idahi. the sum of the two heights is 420 ft. find the h
Masja [62]
Let height of twin falls = x
height of seven falls = 2.5x

x + 2.5x = 420
3.5x = 420
x = 420/3.5 = 120

so twin falls = x = 120 ft
seven falls = 2.5x = 300 ft
6 0
3 years ago
What is the correct answer?
damaskus [11]

Answer:

96 million of dollars

Explanation:

p=-6x^3+72x

p=-6(2)^3+72(2)

p=-6(8)+144

p=-48+144

p=96

8 0
3 years ago
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