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Natali [406]
3 years ago
15

A electron is released from rest in a uniform electric field oriented from left to right. What happens to the electric potential

energy of the proton-electric field system?A. IncreaseB. DecreaseC. Remains the same
Physics
1 answer:
True [87]3 years ago
4 0

Answer:

Explanation:

When electron is released in an electric field , it will move in the direction opposite to the direction of field. In this process , force is applied by the field and work is done by the field. So its electric potential energy will be decreased because , its energy is converted into kinetic energy of the electron .

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Look at the divisions between 2 and 4. What level of precision does this stopwatch have based on the divisions marked on its fac
klemol [59]

Answer:

10ths of a second

Explanation:

Because I just did it

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Roughly how many stars are in the Milky Way Galaxy?
svetlana [45]

Answer:

there are many stars in the milky way galaky

Explanation:

uncountable stars

7 0
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Read 2 more answers
The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find (a) the tension in the cable a
grandymaker [24]

Answer:

(a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

Explanation:

Given that,

Weight of beam= 190 N

Here, The center of gravity is at its center

According to figure,

The angle is

\sin\theta=\dfrac{3}{5}

The horizontal component is

T_{x}=T\cos\theta

The vertical component is

T_{y}=T\sin\theta

(a). We need to calculate the tension in the cable

Using formula of net torque acting on the pivot

\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'

Put the value into the formula

0=190\times2+300\times 4-T\sin\theta\times 4

T\sin\theta\times 4=380+1200

T=\dfrac{1580\times5}{3\times 4}

T=658.33\ N

(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

F_{x}=T\cos\theta

Put the value into the formula

F_{x}=658.33\times\dfrac{4}{5}

F_{x}=526.66\ N

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

F_{y}=F_{b}+F_{w}-T\sin\theta

Put the value into the formula

F_{y}=190+300-658.33\times\dfrac{3}{5}

F_{y}=95.002\ N

Hence, (a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

3 0
3 years ago
In a collision, a 25.0 kg mass moving at 3.0 m/s transfers all of its momentum to a 5.0 kg mass.
nadezda [96]

Answer:

Explanation:

The momentum of the 25 kg mass is

p=mv

p=25kg*3m/s= 75kg*m/s

If this whole momentum of the object is transferred to the 5.0 kg object then according to the law of conservation of momentum, the momentum of the 25.0 kg object must be transferred to the 5.0 kg object:

75kg*m/s = 5.0kg*v

v=\dfrac{75}{5}

\boxed{v=15m/s}

8 0
3 years ago
Answer the question with step.​
Maru [420]

Answer:

f1/f2 =W1/W2 = 1/3

.0 f2 = 3f1

As ,

1/F= 1/f1 +1/f2

...1/40 = 1/f1 - 1/3f1

f1=> 80/3 cm

... f2 = 2f1 = 3 x 80/3 = 80 cm

7 0
3 years ago
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