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Aleksandr-060686 [28]
3 years ago
12

most of earth water a. contains salt b.is freshwater c. is frozen in glaciers d. is found underground​

Physics
2 answers:
CaHeK987 [17]3 years ago
6 0

Answer:

a. salt

Explanation:

Lyrx [107]3 years ago
4 0

Answer:

a. contains salt

Explanation:

97.5% of Earth's water contains salt, and only 2.5% id fresh water. Only 0.3% is in liquid form on the surface. More than half of Earth's water is stored in glaciers.

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1. What is work done in holding a 15kg suitcase while waiting for a bus for 15 minutes?
Tems11 [23]
Work is (force applied) x (distance through which the force moves).

Since the suitcase doesn't move up or down during the 15 minutes,
no work is done ... zero, zip, nada ... according to the real Physics
definition of 'work'.
6 0
3 years ago
A 500-kilogram sports car accelerates uni-
timama [110]

Answer:

90 meters

Explanation:

Given:

x₀ = 0 m

v₀ = 0 m/s

v = 30 m/s

t = 6 s

Find:

x

x = x₀ + ½ (v + v₀)t

x = 0 + ½ (30 + 0)(6)

x = 90

The car travels 90 meters.

3 0
3 years ago
3.
lozanna [386]

The force of gravity produces acceleration in all C. freely falling objects and this is known as acceleration due to gravity

Explanation:

A body is said to be in free fall when there is only one force acting on the body: the force of gravity.

Gravity is a force that acts downward, i.e. towards the Earth's centre.

If we are near the Earth's surface, the magnitude of the force of gravity on a body is given by

F=mg

where:

m is the mass of the body

g is known as the acceleration of gravity , whose value near the Earth's surface is 9.8 m/s^2).

We can apply Newton's second law on an object in free-fall, to find its acceleration. In fact, we have:

F=ma

where F is the force acting on the body and a is its acceleration.

Solving for the acceleration,

a=\frac{F}{m}

And substituting F,

a=\frac{mg}{m}=g=9.8 m/s^2

Therefore, every object in free-fall accelerates at 9.8 m/s^2 towards the ground.

Learn more about free fall here:

brainly.com/question/1748290

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8 0
3 years ago
What is the total amount of kinetic and potential energy in a system ?
Solnce55 [7]

Answer:

Its the sum of the potential energy and the kinetic energy

7 0
3 years ago
A balloon is rising vertically upwards at a velocity of 10m/s. When it is at a height of 45m from the ground, a parachute bails
harina [27]

(a) 30.9 m

Let's analyze the motion of the parachutist. Its vertical position above the ground is given by

y=h+ut+\frac{1}{2}gt^2

where

h = 45 m is the initial height

u = 10 m/s is the initial velocity (upward)

t is the time

g = -9.8 m/s^2 is the acceleration of gravity (downward)

Substituting t=3 s , we find the height of the parachutist when it opens the parachute:

y=45 m+(10 m/s)(3 s)+\frac{1}{2}(-9.8 m/s^2)(3 s)^2=30.9 m

(b) 44.1 m

Here we have to find first the height of the balloon 3 seconds after the parachutist has jumped off from it. The vertical position of the balloon is given by

y = h + ut

where

h = 45 m is the initial height

u = 10 m/s is the initial velocity (upward)

t is the time

Substituting t = 3 s, we find

y = 45 m + (10 m/s)(3 s) = 75 m

So the distance between the balloon and the parachutist after 3 s is

d = 75 m - 30.9 m = 44.1 m

(c) 8.2 m/s downward

The velocity of the parachutist at the moment he opens the parachute is:

v = u +gt

where

u = 10 m/s is the initial velocity (upward)

t is the time

g = -9.8 m/s^2 is the acceleration of gravity (downward)

Substituting t = 3 s,

v = 10 m/s + (-9.8 m/s^2)(3 s)= -19.4 m/s

where the negative sign means it is downward

After t=3 s, the parachutist open the parachute and it starts moving with a deceleration of

a =+5 m/s^2

where we put a positive sign since this time the acceleration is upward.

The total distance he still has to cover till the ground is

d = 30.9 m

So we can find the final velocity by using

v^2-u^2 = 2ad

where this time we have u = 19.4 m/s as initial velocity. Taking the downward direction as positive, the deceleration must be considered as negative:

a = -5 m/s^2

Solving for v,

v=\sqrt{u^2 +2ad}=\sqrt{(19.4 m/s)^2+2(-5 m/s^2)(30.9 m)}=8.2 m/s

(d) 5.24 s

We can find the duration of the second part of the motion of the parachutist (after he has opened the parachute) by using

a=\frac{v-u}{t}

where

a = -5 m/s^2 is the deceleration

v = 8.2 m/s is the final velocity

u = 19.4 m/s is the initial velocity

t is the time

Solving for t, we find

t=\frac{v-u}{a}=\frac{8.2 m/s-19.4 m/s}{-5 m/s^2}=2.24 s

And added to the 3 seconds between the instant of the jump and the moment he opens the parachute, the total time is

t = 3 s + 2.24 s = 5.24 s

8 0
3 years ago
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