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Viefleur [7K]
3 years ago
12

A plant box falls from the windowsill 25.0 m above the sidewalk and hits the cement 3.0 s later. What is the box's velocity when

it hits the ground?
24 m/s
22 m/s
120 m/s
Physics
1 answer:
mamaluj [8]3 years ago
8 0

Answer:

22m/s

Explanation:

To find the velocity we employ the equation of free fall: v²=u²+2gh

where u is initial velocity, g is acceleration due to gravity h is the height, v is the velocity the moment it hits the ground, taking the direction towards gravity as positive.

Substituting for the values in the question we get:

v²=2×9.8m/s²×25m

v²=490m²/s²

v=22.14m/s which can be approximated to 22m/s

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If the density of a diamond is 3.5 /cm^ 3 what would be the mass of a diamond whose volume is 0.5 cm ^ 3 ?
Elan Coil [88]

Answer:

1.75g

Explanation:

Given parameters:

Density  = 3.5g/cm³

Volume  = 0.5cm³  

Unknown:

Mass of the diamond  = ?

Solution:

Density is the mass per unit volume of a substance

  Density  = \frac{mass}{volume}  

So;

  Mass  = 3.5 x 0.5  = 1.75g

7 0
3 years ago
A worker at the top of a 588-m-tall television transmitting tower accidentally drops a heavy tool. If air resistance is negligib
liq [111]

The final velocity of the tool is 107.4 m/s

Explanation:

We can solve this problem by using the principle of conservation of energy.

In fact, if air resistance is negligible, the total mechanical energy of the tool is conserved during the fall, so we can write:

K_i + U_i = K_f + U_f

where

K_i = 0 is the kinetic energy of the tool at the top (zero since it is at rest)

U_i = mgh is the gravitational potential energy of the tool at the top, where

m is the mass of the tool

g is the acceleration of gravity

h is the heigth of the tool

K_f = \frac{1}{2}mv^2 is the kinetic energy of the tool just before hitting the ground, where

v is the final speed of the tool

U_f = 0 is the gravitational potential energy of the tool at the bottom (zero since the height is zero)

Re-arranging the equation,

mgh=\frac{1}{2}mv^2

where we have

g=9.8 m/s^2\\h=588 m

And solving for v,

v=\sqrt{2gh}=\sqrt{2(9.8)(588)}=107.4 m/s

Learn more about kinetic energy and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

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3 0
3 years ago
Determine la fuerza aplicada sobre un émbolo en una prensa hidráulica si se quiere cargar una lanza de 2970kg; toma en cuenta qu
Artist 52 [7]
I’m not sure sorry I really wish I could help
4 0
3 years ago
If all objects that have mass also have gravity, why doesn't your pencil get pulled towards you while it sits on your desk?
Lemur [1.5K]

Answer:

The pencil is not pulled towards a person due to a very small magnitude of force between them, due to lighter masses.

Explanation:

Let us apply Newton's Law of Gravitation between a person and pencil.

Average Mass of a Normal Pencil = m₁ = 10 g = 0.01 kg

Average Mass of a Person = m₂ = 80 kg

Distance between both = r = 1 cm = 0.01 m (Taking minimal distance)

Gravitational Constant = G = 6.67 x 10⁻¹¹ N.m²/kg²

So,

F = Gm₁m₂/r²

F = (6.67 x 10⁻¹¹ N.m²/kg²)(0.01 kg)(80 kg)/(0.01 m)²

<u>F = 5.34 x 10⁻⁷ N</u>

This Force is very small in magnitude due to the light masses of both objects.

<u>Therefore, the pencil is not pulled towards a person due to a very small magnitude of force between them, due to lighter masses.</u>

5 0
3 years ago
How does reverberation explain why your voice sounds different in a gym than it does in your living room
TEA [102]
Sound can reach the inner ear by way of two separate paths, and those paths in turn affect what we perceive. Air-conducted sound is transmitted from the surrounding environment through the external auditory canal, eardrum and middle ear to the cochlea, the fluid-filled spiral in the inner ear. Bone-conducted sound reaches the cochlea directly through the tissues of the head.
When you speak, sound energy spreads in the air around you and reaches your cochlea through your external ear by air conduction. Sound also travels from your vocal cords and other structures directly to the cochlea, but the mechanical properties of your head enhance its deeper, lower-frequency vibrations. The voice you hear when you speak is the combination of sound carried along both paths. When you listen to a recording of yourself speaking, the bone-conducted pathway that you consider part of your “normal” voice is eliminated, and you hear only the air-conducted component in unfamiliar isolation. You can experience the reverse effect by putting in earplugs so you hear only bone-conducted vibrations.
Some people have abnormalities of the inner ear that enhance their sensitivity to this component so much that the sound of their own breathing becomes overwhelming, and they may even hear their eyeballs moving in their sockets.
6 0
3 years ago
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