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Bess
1 month ago
9

Consider what happens when you jump up in the air. Which of the following is the most accurate statement? Consider what happens

when you jump up in the air. Which of the following is the most accurate statement? Since the ground is stationary, it cannot exert the upward force necessary to propel you into the air. Instead, it is the internal forces of your muscles acting on your body itself that propels your body into the air. You are able to spring up because the earth exerts a force upward on you that is greater than the downward force you exert on the earth. It is the upward force exerted by the ground that pushes you up, but this force cannot exceed your weight. When you jump up the earth exerts a force F1 on you and you exert a force F2 on the earth. You go up because F1 > F2. When you push down on the earth with a force greater than your weight, the earth will push back with the same magnitude force and thus propel you into the air.
Physics
1 answer:
ValentinkaMS [3.4K]1 month ago
6 0
The most accurate choice is E. When downward force exceeds your weight, pushing down on the earth, it responds with an equal force, which lifts you into the air. This follows Newton's third law: every action has an equal and opposite reaction.
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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
Softa [3030]
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)
     = 314.37 lb/ft²
     = 314.37/144 = 2.18 lb/in²

Thus, the answer is 2.2 psi

5 0
2 months ago
A 20.00-kg lead sphere is hanging from a hook by a thin wire 2.80 m long and is free to swing in a complete circle. Suddenly it
Keith_Richards [3271]
Objects in vertical motion are an illustration of non-uniform motion. At the peak of the circle, centripetal force is balanced by the object's weight. Therefore, the minimum speed required at this top point is given by v = \sqrt{rg} = \sqrt{2.80 \times 9.8} = 5.23 m/s. As the sphere descends from the top to the bottom of the circle, according to the law of conservation of energy, potential energy can be expressed as

P.E_{highest} = mgh

, where h signifies the diameter of the circle (2r). Hence, the expression will be written as P.E_{highest} = mg(2r)

where u is the velocity at the lowest point. Consequently, the modified equation is

= \sqrt{v^{2} + 4gr}

= \sqrt{(5.23)^{2} + (4 \times 9.8 \times 2.80)}

= 11.71 m/s. The collision of the dart with the bullet is an inelastic one. According to the conservation of momentum: v = \frac{(m_{1} + m_{2})u}{m_{2}}

= \frac{(20 + 5) \times 11.71}{5}

= \frac{292.75}{5}

= 58.55 m/s. Thus, the dart's minimum initial speed for the combined system to complete a circular loop post-collision is 58.55 m/s.

3 0
2 months ago
Two very small 8.55-g spheres, 15.0 cm apart from center to center, are charged by adding equal numbers of electrons to each of
ValentinkaMS [3465]

Answer:

1.43 x 10¹⁷.

They will move away from each other.

Explanation:

The force acting on each charged sphere is determined as F = mass x acceleration

= 8.55 x 10⁻³ x 25 x 9.8

= 2.095 N

Assuming Q is the charge on each sphere

F = \frac{9\times10^9\times Q^2}{(15\times10^{-2})^2}

Using the values, 2.095 = \frac{9\times10^9\times Q^2}{(15\times10^{-2})^2}

We find that Q² = \frac{2.095\times(15)^2\times10^{-4}}{9\times10^9}

Thus, Q = 2.289 X 10⁻⁶

The quantity of electrons = Charge / charge of a single electron

= \frac{2.289\times10^{-6}}{1.6\times10^{-19}}

=1.43 x 10¹³.

They will accelerate away from each other.

4 0
2 months ago
Read 2 more answers
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