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saveliy_v [14]
1 year ago
6

• An object of mass 2.0 kg moves in uniform circular motion on a horizontal friction less table. The radius of the circle is 0.7

5 m and the centripetal force is 10.0 N. The work done by this force when the object moves through one-half of a complete revolution is​
Physics
1 answer:
Gekata [30.6K]1 year ago
4 0

Answer:

23.562

Explanation:

First it is good to work out the circumference:

2 pi r

2 pi .75

circumference = 4.7124

Next we need half of this because it is half a resolution. This will be our distance travelled.

half = 2.3562

word done = force x distance

work done = 10 x 2.3562

work done = 23.562

(I didn't see the need for mass...)

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Name the physical quantity which is expressed in the unit atm state its value in pascal​
olasank [31]

Answer:

The unit of measurement called standard atmosphere (atm) is defined as 101,325 Pa. Common multiple units of the pascal are the hectopascal (1 hPa = 100 Pa), which is equal to one millibar, and the kilopascal (1 kPa = 1000 Pa), which is equal to one centibar.

4 0
3 years ago
Calculate the net force on the right charge due to the other two. Enter a positive value if the force is directed to the right a
lbvjy [14]

Answer:

Answer:

A. - 0.017N. It acts to the left.

B. - 0.043N. It acts to the left.

C. 0.060N. It acts to the right.

Explanation:

A. For the +65μC charge, we consider it to be the origin. Hence, the two other charges are on the +x axis.

The net coulombs force on the charge is

F = [KQ(1)Q(2)]/(r^2) + [KQ(1)Q(3)]/(r^2)

Where K = Coloumbs constant =

Q(1) = charge on the leftmost side.

Q(2) = charge in the middle.

Q(3) = charge on the rightmost side.

F = [(8.988 × 10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(-95×10^-6)×(65×10^-6)]/(40^2)

F = 0.01753 - 0.03469

F = -0.017N

It has a negative sign, hence, it acts to the left.

B. For the +48μC charge, we consider it to be the origin. Hence, the leftmost charge is on the - x axis and the rightmost charge is on the +x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) + [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = -0.017 - 0.02562

F = - 0.043N

It has a negative sign, hence, it acts to the left.

C. For the -95μC charge, we consider it to be the origin. Hence, the two other charges are on the - x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) - [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(-95×10^-6)]/(40^2) - [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = +0.03469 + 0.02562

F = +0.060N

It has a positive sign, hence, it acts to the right.

Read more on Brainly.com - brainly.com/question/14592748#readmore

Explanation:

5 0
4 years ago
You are helping your friend move a new refrigerator into his kitchen. You apply a horizontal force of 275 N in the positive x di
Volgvan

Answer:

f = 347.08 N

Explanation:

The frictional force exerted by the floor on the refrigerator is given as follows:

f = \mu R = \mu W

where,

f = frictional force = ?

μ = coefficient of static friction = 0.58

W = Weight of refrigerator = mg

m = mass of refrigerator = 61 kg

g = acceleration due to gravity = 9.81 m/s²

Therefore,

f = \mu mg\\f = (0.58)(61\ kg)(9.81\ m/s^2)\\

<u>f = 347.08 N</u>

8 0
3 years ago
The entropy of an isolated system must be conserved, so it never changes.a. Trueb. Fasle
Snowcat [4.5K]

Answer:

B: False

Explanation:

The second law of thermodynamics states that: the entropy of an isolated system will never decrease because isolated systems always tend to evolve towards thermodynamic equilibrium which is a state with maximum entropy.

Thus, it means that the entropy change will always be positive.

Therefore, the given statement in the question is false.

6 0
3 years ago
In a process called pair production, an energetic
vredina [299]

Answer:

(3) mass-energy must be conserved

Explanation:

As we know that gamma rays are mass less and charge less photons which will have sufficient energy.

Now in case of Pair production the gamma photons convert its whole energy into  mass by the law of Einstein's mass energy equivalence relation.

As per his relation we can say

E = \Delta m c^2

here we will have

[tex]\Delta m[\tex] = mass produced

now we also have to think that here the two particles must have a pair of particles and antiparticles so that the combined mass system will have energy equivalent to the energy of gamma photons and also it must follow the conservation of charge

So here it will form an electron and a positron such that total charge will be zero and the energy will be same as energy of gamma photon.

3 0
4 years ago
Read 2 more answers
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