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Oliga [24]
2 years ago
8

Given A=125.0=0.4 and 25.0=0.1 calculate A-B​

Physics
1 answer:
kicyunya [14]2 years ago
6 0

Answer: Find an answer to your question given A=125.0=0.4 and 25.0=0.1 calculate A-B​

Explanation:

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How much work does the electric field do in moving a proton from a point with a potential of +145 v to a point where it is -55 v
Aleks04 [339]
The work done by the electric field in moving a charge is the negative of the potential energy difference between the two locations, which is the product between the magnitude of the charge q and the potential difference \Delta V:
W=-\Delta U=-q\Delta V=-q (V_f -V_i)
The proton charge is q=1.6 \cdot 10^{-19}C, and the two locations have potential of V_i = +145 V and V_f=-55 V, therefore the work is
W=-(1.6 \cdot 10^{-19}C)(-55 V-(+145 V))=3.2 \cdot 10^{-17}J
5 0
2 years ago
A crate is pulled with a force of 165 N at an angle 30 ° northwest. What is the resultant horizontal force on the crate?
Ad libitum [116K]

Answer:

Resultant horizontal force = 143 N

Explanation:

Since the a gle is 30° northwest, then it means the resultant force will be horizontal and as such;

Resultant horizontal force = 165 * cos 30

Resultant horizontal force = 142.89

Approximating to a whole number gives;

Resultant horizontal force = 143 N

5 0
1 year ago
You stand on a merry-go-round which is spinning at f = 0.25 revolutions per second. You are R = 200 cm from the center. (a) Find
wariber [46]

Answer:

a) ω = 9.86 rad/s

b) ac = 194. 4 m/s²

c) minimum coefficient of static friction, µs = 19.8

Explanation:

a) angular speed, ω = 2πf, where f is frequency of revolution

1 rps = 6.283 rad/s, π = 3.142

ω = 2 * 3.14 * 0.25 * 6.28

ω = 9.86 rad/s

b) centripetal acceleration, a = rω²

where r is radius in meters; r = 200 cm or 2 m

a = 2 * 9.86²

a = 194. 4 m/s²

c) µs = frictional force/ normal force

frictional force = centripetal force = ma; where a is centripetal acceleration

normal force = mg; where g = 9.8 m/s²

µs = ma/mg = a/g

µs = 194.4 ms⁻²/9.8 ms⁻²

c) minimum coefficient of static friction, µs = 19.8

5 0
2 years ago
How do the tension of the cord and the force of gravity affect a pendulum?
LuckyWell [14K]

Answer:

<em>Force of gravity may not affect a pendulum during its equilibrium state</em>. But  the gravity can affect the pendulum when a force occurs in any direction of the bob connected to the cord that makes a swing sideways. The gravity of pendulum never stops, it always accelerates. So the gravity affects the pendulum acceleration and speed.    

<em>Similarly the tension in the cord will not affect the pendulum</em><em> </em>but if change in the length of the pendulum while keeping other factors constant changes the length of the period of pendulum. longer pendulum swings with lower frequency than shorter pendulums.    


6 0
3 years ago
Read 2 more answers
If a non-rotating object has no acceleration, then we can say for certain that it is:__________
EleoNora [17]

We may be positive that an object is in mechanical equilibrium if it is not rotating and experiences no acceleration.

<h3>What is mechanical equilibrium?</h3>

There are numerous other definitions for mechanical equilibrium that are all mathematically comparable in addition to the definition in terms of force. A system is in equilibrium in terms of momentum if the component motions are all constant. If velocity is constant, the system is in equilibrium in terms of velocity. When an item is in a state of rotational mechanical equilibrium, its angular momentum is preserved and its net torque is zero. More generally, equilibrium is reached in conservative systems at a configuration space location where the gradient of the potential energy concerning the generalized coordinates is zero.

To learn more about mechanical equilibrium, visit:

<u>brainly.com/question/14246949</u>

#SPJ4

6 0
1 year ago
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