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Shkiper50 [21]
2 years ago
15

If you drop an object, it accelerates downward at 9.8 m/s2 (in the absence of air resistance). If, instead, you throw it downwar

d, its downward acceleration after release is?
Physics
1 answer:
Korvikt [17]2 years ago
4 0

If you drop an object, it accelerates downward at 9.8 m/s2 (in the absence of air resistance). If instead, you throw it downward, its downward acceleration after release is 9.8 m/s2.

Acceleration is the rate at which an object's velocity with respect to time changes. They are vector quantities and accelerations. The direction of the net force acting on an object determines the direction of its acceleration. Uniform acceleration, non-uniform acceleration, and average acceleration are the three different forms of accelerated motions.

A free-falling object experiences a downward acceleration of 9.8 m/s/s (on Earth). This specific designation is given to the numerical value for an object in free fall because it is such an essential value. The longer an object is in free fall, the faster it descends toward the ground due to gravity. In actuality, an object's velocity rises by 9.8 m/s2, so it reaches 9.8 m/s by the time it begins to fall.

To know more about acceleration refer to: brainly.com/question/14468548

#SPJ4

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VMariaS [17]

Answer:

Ted is correct

Explanation:

The equation for gravitational potential energy is PE = m·g·h

The equation for gravitational kinetic energy is KE = 1/2·m·v²

Where:

m = Mass of the object (The racing car)

g = Acceleration due to gravity

h = The height to which the object is raised

v = Velocity of motion of the object

From the principle of conservation of energy, energy can neither be created nor destroyed but changes from one form to another, we have;

Potential energy gained from location at height h = Kinetic energy gained as the object moves down the level ground

m·g·h = 1/2·m·v² canceling like terms gives

g·h = 1/2·v²

v = (√2·g·h)

If the speed is doubled, we have

2·v = 2× (√2·g·h) =  (√2·g·4·h)

Therefore, if 2·v = v₂ then v₂ =  (√2·g·4·h)

Since g, the acceleration due to gravity, is constant, it means that the initial  height must be multiplied or increased 4 times to get the new height, that is we have;

v₂ =  (√2·g·4·h) = (√2·g·h₂)

Where:

4·h = h₂

Which gives;

v₂² = 2·g·h₂

1/2·v₂² = g·h₂

1/2·m·v₂² = m·g·h₂ Just like in the first relation

Therefore, Ted is correct s they need to go up four times the initial height to double the speed.

5 0
3 years ago
What is the distance and the displacement of the race car drivers in the Indy 500?
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Answer here
The displacement of the cars is somewhere near 0 miles since they virtually finish where they started. Yet the successful cars have covered a distance of 500 miles
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3 years ago
An elf pushes a sleigh with force of 9N over a distance of 11m. How much work did the elf. Do on the sleigh
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Answer:

<h2>99 J</h2>

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question we have

workdone = 9 × 11

We have the final answer as

<h3>99 J</h3>

Hope this helps you

5 0
3 years ago
A spherical balloon has a radius of 7.15 m and is filled with helium. The density of helium is 0.179 kg/m^3, and the density of
tekilochka [14]

The largest mass of cargo the balloon can lift is 791.06 kg

First, we need to calculate the mass of helium.

Since the radius of the spherical balloon is r = 7.15 m, its volume is V = 4πr³/3.

The volume of the balloon also equals the volume of helium present.

Now, the mass of helium m = density of helium, ρ × volume of helium, V

m = ρV

Since ρ = 0.179 kg/m³

m = ρV

m = ρ4πr³/3.

m = 0.179 kg/m³ × 4π(7.15 m)³/3

m = 0.179 kg/m³ × 4π(365.525875 m³)/3

m = 0.179 kg/m³ × 1462.1035π m³/3

m = 261.7165265π/3 kg

m = 822.207/3 kg

m = 274.07 kg

Since the mass of the skin and structure of the balloon is 910 kg, the total mass, M of the balloon = mass of skin and structure + mass of helium gas is 910 kg + 274.07 kg = 1184.07 kg.

The weight of this mass W = Mg where g = acceleration due to gravity.

The buoyant force on the balloon due to the air is the weight of air displaced, W' = mass of air, m' × acceleration due to gravity, g.

W' = m'g

Now, the mass of air m' = density of air, ρ' × volume of air displaced, V'

We know that the volume of air displaced, V' = volume of balloon, V

So, V' = V = 4πr³/3.

Since the density of air, ρ' = 1.29 kg/m³,

m' = ρ'V

m = 1.29 kg/m³ × 4π(7.15 m)³/3

m = 1.29 kg/m³ × 4π(365.525875 m³)/3

m = 1.29 kg/m³ × 1462.1035π m³/3

m = 1886.113515π/3 kg

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So, the net weight W" that the balloon can lift is W" = W' - W = m'g - Mg = (m' - M )g = (1975.13 kg - 1184.07 kg)g = 791.06g.

So, the net mass m" = W"/g = 791.06g/g = 791.06 kg

This net mass is the largest mass of cargo that the balloon can lift.

Thus, the largest mass of cargo the balloon can lift is 791.06 kg

Learn more about balloons here:

brainly.com/question/21890581

8 0
3 years ago
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