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mamaluj
1 month ago
5

The image illustrates that as the distance between two objects increases, the force of gravity ____________. A) decreases. B) in

creases. C) remains the same. D) increases then decreases.
Physics
2 answers:
Sav [3.1K]1 month ago
7 0
The image is absent (but it's not essential to resolve the issue).

The right response is A) decreases, as gravitational force is inversely related to the square of the distance. The magnitude of the gravitational force between two masses M and m, separated by a distance d, is expressed as
F=G \frac{Mm}{d^2}
where G is the gravitational constant. The formula demonstrates that as the distance d between the two masses increases, the force magnitude diminishes.
Yuliya22 [3.3K]1 month ago
3 0

B; Decreases

This is because

the image illustrates that with increasing distance between two objects, the gravitational force weakens. The strength of gravity is influenced by both the distance between the two objects and their masses

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The moment the body impacts the ground, two types of Forces are produced: the gravitational pull and the Normal Force. This aligns with Newton's third law, indicating that every action has an equal and opposite reaction. If the downward force of gravity is directed toward the earth, the reactionary force from the block acts upwards, equivalent to its weight:

F = mg

Where,

m = mass

g = gravitational acceleration

F = 5*9.8

F = 49N

Consequently, the answer is E.

5 0
2 months ago
At a given point on a horizontal streamline in flowing air, the static pressure is â2.0 psi (i.e., a vacuum) and the velocity is
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Bernoulli's equation at a point on the streamline is
p/ρ + v²/(2g) = constant
where
p = pressure
v = velocity
ρ = air density, 0.075 lb/ft³ (under standard conditions)
g = 32 ft/s²

Point 1:
p₁ = 2.0 lb/in² = 2*144 = 288 lb/ft²
v₁ = 150 ft/s

Point 2 (stagnation):
The velocity at the stagnation point is zero.

The density stays constant.
Let p₂ denote the pressure at the stagnation location.
Then,
p₂ = ρ(p₁/ρ + v₁²/(2g))
p₂ = (288 lb/ft²) + [(0.075 lb/ft³)*(150 ft/s)²]/[2*(32 ft/s²)
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5 0
1 month ago
What happens when you decrease the thrust on your scooter? A. You stop B. Nothing happens C. You fall over D. You speed up Reset
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Response:

D. You accelerate

Clarification:

Hope this is helpful

4 0
1 month ago
(a) Calculate the absolute pressure at the bottom of a fresh- water lake at a depth of 27.5 m. Assume the density of the water i
kicyunya [3294]

Response:

a) P = 370.993\,kPa, b) F = 25.948\,kN

Clarification:

a) The absolute pressure at a depth of 27.5 meters is:

P = P_{atm} + P_{man}

P = 101.3\,kPa + \left(1000\,\frac{kg}{m^{3}}\right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (27.5\,m)\cdot \left(\frac{1\,kPa}{1000\,Pa} \right)

P = 370.993\,kPa

b) The force applied by the water is:

F = (P - P_{atm})\cdot A

F = (370.993\,kPa-101.3\,kPa)\cdot \left(\frac{\pi}{4} \right)\cdot (0.35\,m)^{2}

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1 month ago
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The kangaroo reaches a maximum vertical altitude of 2.8 m, which can be calculated using the formula 2.8 = 1/2 * 9.8 * t^2. Thus, applying the equation s = ut + 1/2at^2.
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