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3 years ago

10

A heavy object and a light object are dropped from the same height. If we neglect air resistance, which will hit the ground firs

t?

1 answer:

Maksim231197 [3]3 years ago

6 0

Answer:

None, both objects will hit ground at the same time.

Explanation:

Assuming no air resistance present, and that both objects start from rest, we can apply the following kinematic equation for the vertical displacement:

$\Delta h = \frac{1}{2}*g*t^{2} (1)$

As the left side in (1) is the same for both objects, the right side will be the same also.
Since g is constant close to the surface of the Earth, it's also the same for both objects.
So, the time t must be the same for both objects also.

You might be interested in

Explain how wind erosion changes land forms

valkas [14]

Answer:

the wind carries abrasive materials

Explanation:

such as sand and salt over time theses small particles slowly strip way at the land form sculpting it by eroding the softer layers first

8 0

2 years ago

There are two identical, positively charged conducting spheres fixed in space. The spheres are 44.0 cm apart (center to center)

Aneli [31]

Answer:

q₁ =± 1.30 10⁻⁶ C and q₂ = ± 1.28 10⁻⁶ C

Explanation:

We will solve this problem with Coulomb's law

F = K q₁q₂ / r²

Where the Coulomb constant is value 8.99 10⁹ N m² / C²

Let's apply this equation to our problem

Case 1

F1 = k q₁ q₂ / r₁²

Where r₁ = 0.440 m and F1 = 0.0765 N

Case 2

The charges are the same

F2 = k q q / r₂²

With r₂ = r₁ = 0.440 m, the spheres are fixed and the force is F2 = 0.100 N

When the spheres are joined with the wire, the charge is distributed, distributed and matched in the two spheres

q₁ + q₂ = 2 q

Let's replace

F2 = k ½ (q₁ + q₂) / r²

Let's write the two equations and solve the system of equations

F1 = k q₁ q₂ / r²

F2 = ½ k (q₁ + q₂) / r²

F1 r² / k = (q₁ q₂)

F2 r² / k = (q₁ + q₂)/2

q₁ = 2F2 r² / k - q₂

We substitute in the other equation

F1 r² / k = (2F2 r² / k - q₂) q₂

0 = -F1 r² / k + (2F2 r² / k) q₂ - q₂²

Let's solve the second degree equation

F1 r² / K = 0.0765 0.440² / 8.99 10⁹

F1 r² / K = 1.65 10⁻¹²

(2F2 r² / k) =2 0.10 0.44² / 8.99 10⁹

(2F2 r2 / k) = 4.30 10⁻¹²

q₂² - 4.30 10⁻¹² q₂ + 1.65 10⁻¹² = 0

q₂ = ½ {4.30 10⁻¹² ± √ [(4.30 10⁻¹²)² - 4 1.65 10⁻¹²]}

q₂ = ½ {4.30 10⁻¹² ± 2,569 10⁻⁶}

q₂ = ± 1.2845 10⁻⁶ C

Now we calculate q1

F1 = k q₁ q₂ / r²

q₁ = F1 r² / (k q₂)

q₁ = 0.0765 0.440² / (8.99 10⁹ 1.2845 10⁻⁶)

q₁ = 1.30 10⁻⁶ C

3 0

3 years ago

When the mass of an object decreases, the force of gravity

Viktor [21]

Hello There!

From what i know, gravitational force increases if the mass is increased.
If the mass is being decreased, then i assume it will be B. Decreases.

Hope This Helps You!
Good Luck :)

- Hannah ❤

8 0

3 years ago

A high diver of mass 74.0 kg jumps off a board 9.00 m above the water. If his downward motion is stopped 2.50 seconds after he e

stiv31 [10]

Answer:

1120 N

Explanation:

The velocity with which he hits the water can be found with kinematics:

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (-9.8 m/s²) (-9.00 m)

v = -13.3 m/s

Or it can be found with conservation of energy.

PE = KE

mgh = ½ mv²

v = √(2gh)

v = √(2 × -9.8 m/s² × -9.00 m)

v = -13.3 m/s

Sum of forces on the diver after he hits the water:

∑F = ma

F − mg = m Δv/Δt

F − (74.0 kg) (9.8 m/s²) = (74.0 kg) (0 m/s − (-13.3 m/s)) / (2.50 s)

F = 1120 N

6 0

3 years ago

Suppose a Heel-mobile, moving in a straight line and steadily increases its speed. It moves from 15 m/s to 30 m/s the first seco

Makovka662 [10]

acceleration = change in velocity/change in time = 45-15 / 2

=30/2 = 15 m/s

6 0

3 years ago