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stiv31 [10]
2 years ago
6

6. A warehouse employee is pushing a 15.0 kg desk across a floor at a constant speed of 0.50 m/s. How much work must the employe

e do on the desk to change the speed to 1.00 m/s?​
Physics
1 answer:
MariettaO [177]2 years ago
3 0

Answer:

7.5 J

Explanation:

To answer the question given above, we need to determine the energy that will bring about the speed of 1 m/s. This can be obtained as follow:

Mass (m) = 15 Kg

Velocity (v) = 1 m/s

Energy (E) =?

E = ½mv²

E = ½ × 15 × 1²

E = ½ × 15 × 1

E = ½ × 15

E = 7.5 J

Therefore, to change the speed to 1 m/s, the employee must do a work of 7.5 J.

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the radius of planet is drecresing by 50%without changing its mass . the acceleration due to gravity will be?​
tigry1 [53]

4 times more then real

Explanation:

if the radius of the earth decrease by 50% then the acceleration due to gravity increases by 4 times.

4 0
2 years ago
How can I solve the following statement?
Firlakuza [10]

Answer:

The net electric field at the midpoint is 6.85 x 10^7 N/C.

Explanation:

q = − 8.3 μC

q' = + 7.8 μC

d =  9.2 cm

d/2 = 4.6 cm

The electric field due to the charge q at midpoint is

E = \frac{k q}{r^2}\\\\E = \frac{9\times 10^9\times 8.3\times 10^{-6}}{0.046^2}\\\\E = 3.53\times 10^7 N/C leftwards

The electric field due to the charge q' at midpoint is

E' = \frac{k q}{r^2}\\\\E' = \frac{9\times 10^9\times 7.8\times 10^{-6}}{0.046^2}\\\\E' = 3.32\times 10^7 N/C

The resultant electric field at mid point is

E'' = E + E' = (3.53 + 3.32) x 10^7 = 6.85 x 10^7 N/C

7 0
2 years ago
What happens to the pressure in a tire if air is slowly leaking out of the tire? explain your answer?
creativ13 [48]
It's pretty simple. When air is leaking out of a tire, like a tiny hole or something, the pressure in the tire decreases, because without air in the tire, there is no pressure.
3 0
3 years ago
Two electrons with a charge of magnitude 1.6×10-19 C in an atom are separated by 1.5×10-10 m, the typical size of an atom. What
vesna_86 [32]

Answer:

1.02\cdot 10^{-8} N, repulsive

Explanation:

The magnitude of the electric force between two charged particles is given by Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the two charges of the two particles

r is the separation between the two charges

The force is:

- repulsive if the two charges have  same  sign

- Attractive if the two charges have opposite signs

In this problem, we have two electrons, so:

q_1=q_2=1.6\cdot 10^{-19}C is the magnitude of the two electrons

r=1.5\cdot 10^{-10} m is their separation

Substituting into the formula, we find the electric force between them:

F=(8.99\cdot 10^9)\frac{(1.6\cdot 10^{-19})^2}{(1.5\cdot 10^{-10})^2}=1.02\cdot 10^{-8} N

And the force is repulsive, since the two electrons have same sign charge.

4 0
2 years ago
The force of gravitation between two spherical bodies is Gm1 m2 /r2, where r is separation between their dash
AnnZ [28]

Explanation:

F = Gm1m2/r^2

kya nikalna hai bhai isme

6 0
3 years ago
Read 2 more answers
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