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Vinil7 [7]
3 years ago
13

Final velocity will be greater than initial velocity of an object is

Physics
1 answer:
Natali [406]3 years ago
4 0

Answer:

accelerating

Explanation:

If we consider(v > u) Acceleration:

final velocity(v)= 14m/s

initial velocity(u)=10m/s

time taken(t)= 2 seconds

a= \frac{(v-u)}{t} =\frac{(14-10)}{2}=2m/s²

If we consider (v<u) Deceleration:

final velocity(v)= 3m/s

initial velocity(u)=9m/s

time taken(t)=2 seconds

a= \frac{(v-u)}{t}=\frac{(3-9)}{2}= -3m/s²

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Explain what is the capacity of a battery in your own words​
Vinil7 [7]

Battery capacity (AH) is defined as a product of the current that is drawn from the battery while the battery is able to supply the load until its voltage is dropped to lower than a certain value for each cell.

4 0
3 years ago
A 18.0-kg rock is sliding on a rough, horizontal surface at 7.10 m/s and eventually stops due to friction. the coefficient of ki
Bond [772]
A = .3*g = 2.94 m/s² 

<span>t = v/a = 9/2.94 = 3.061 sec </span>

<span>W = E/t = ½mv²/t = ½*40*9²/3.061 = 529.2 watts</span>
4 0
3 years ago
How much work does gravity do on the compressor during this displacement?
gayaneshka [121]

Explanation and Examples

let the mass of the compressor be

mass (m):

height in x axis is (h1)

height in y axis be (h2):

Height difference: h2-h1

displacement x force:

mass x gravity x height

(m)*9.8*(height difference) = ___ J

Since gravity is forcing down, it would be negative!

Put the values that you require and get the answer.

6 0
3 years ago
Read 2 more answers
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
A boy drags a suitcase along the ground with a force of 100 N. If the frictional force opposing the motion of the suitcase is 50
stira [4]
Fortunately, 'force' is a vector.  So if you know the strength and direction
of each force, you can easily addum up and find the 'resultant' (net) force.

When we talk in vectors, one newton forward is the negative of
one newton backward.   Hold that thought, while I slog through
the complete solution of the problem.


            (100 N forward) plus (50 N backward)

        =  (100 N forward) minus (50 N forward)

        =           50 N forward .

That's it.
Is there any part of the solution that's not clear ?

4 0
3 years ago
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