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alukav5142 [94]
3 years ago
7

True or false—If a rock is thrown into the air, the increase in the height would increase the rock’s kinetic energy, and then th

e increase in the velocity as it falls to the ground would increase its potential energy.
Physics
1 answer:
nlexa [21]3 years ago
5 0
I think that this is false but I am not sure
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Which example best demonstrates how unbalanced forces change the speed of an object's motion?
exis [7]

I don’t see a picture but unbalanced forces could be two boys pushing with a combined force of 400 Newton’s but the surface of what the box is laying on being 600 meaning since the ground is creating a higher force in the form of friction it will slow the boys down. When forces are unbalanced it means that the object can not be still or moving at a constant speed  when one force is greater by a significant amount the object either slows quickly or accelerates fast depending on which factor is greater.

4 0
3 years ago
Differentiate between a derived quantity and a derived unit.
Inessa [10]

Explanation:

Derived quantities are quantities dependent on fundamental quantities while derived units are the units of these quantities

3 0
3 years ago
slader How much energy is required to move a 1040 kg object from the Earth's surface to an altitude four times the Earth's radiu
andrew-mc [135]

Answer:

ΔU = 5.21 × 10^(10) J

Explanation:

We are given;

Mass of object; m = 1040 kg

To solve this, we will use the formula for potential energy which is;

U = -GMm/r

But we are told we want to move the object from the Earth's surface to an altitude four times the Earth's radius.

Thus;

ΔU = -GMm((1/r_f) - (1/r_i))

Where;

M is mass of earth = 5.98 × 10^(24) kg

r_f is final radius

r_i is initial radius

G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²

Since, it's moving to altitude four times the Earth's radius, it means that;

r_i = R_e

r_f = R_e + 4R_e = 5R_e

Where R_e is radius of earth = 6371 × 10³ m

Thus;

ΔU = -6.67 × 10^(-11) × 5.98 × 10^(24)

× 1040((1/(5 × 6371 × 10³)) - (1/(6371 × 10³))

ΔU = 5.21 × 10^(10) J

7 0
3 years ago
Find the area under the standard normal curve between z=0.19 and z=2.18. round your answer to four decimal places, if necessary.
LiRa [457]

The area under the standard normal curve between z=0.19 and z=2.18 is,

P(0.19<z<2.18)= 0.4101

<h3>How can we calculate the area under the curve?</h3>

To calculate the The area under the standard normal curve  between z=0.19 and z=2.18, we are using two things,

<u>Step 1</u>: The formula,

P(0.19<z<2.18)= P(Z<2.18)- P(Z<0.19)

<u>Step 2</u>: The statistical values of the area under the curve we get from the picture.

Now we put the known values from the picture in the above formula, we get,

P(0.19<z<2.18)= P(Z<2.18)- P(Z<0.19)

Or, P(0.19<z<2.18)=  0.9854-0.5753

Or, P(0.19<z<2.18)=  0.4101

From the above calculation we can easily conclude that,

The area under the standard normal curve between z=0.19 and z=2.18 is,

P(0.19<z<2.18)= 0.4101

Learn more about the standard normal curve:

brainly.com/question/17088387

#SPJ4

5 0
2 years ago
What is the concentration of a solution if 65 g of sugar is dissolved into 200 ml of water?
DIA [1.3K]

Answer:

the concentration in g/L 325 while in molarity is 0.6398(M).

Explanation:

hope this helps

5 0
2 years ago
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