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Doss [256]
3 years ago
7

An electron is traveling with initial kinetic energy K in a uniform electric field. The electron comes to rest momentarily after

traveling a distance d. (a) What is the magnitude of the electric field? (Use any variable or symbol stated above along with the following as necessary: e for the charge of the electron.)
Physics
1 answer:
geniusboy [140]3 years ago
3 0

Answer:

E=\frac{K}{ed}

Explanation:

We are given that

Initial kinetic energy of an electron=K

Distance=d

Final velocity=v=0

Charge,q=-1e

We have to find the magnitude of electric field.

Work done=Force\times displacement

Using the formula

Work done=qE\times d=-eEd

Using work energy theorem

Work done=Final K.E-Initial K.E=0-K

Work done=-K

Substitute the values

-K=-eEd

K=eEd

E=\frac{K}{ed}

Hence, the magnitude of the electric field=E=\frac{K}{ed}

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The wind pushes a paper cub along the sand at a beach. The cup has a mass of 25 grams (= ? Kg's) and accelerates at a rate of 5
bixtya [17]

Force = (mass) x (acceleration)

= (0.025 kg) x (5 m/s²)

=   0.125 Newton
7 0
3 years ago
At summer camp, the swimming course runs the length (L) of a small lake. To determine the length of the course, the camp counsel
sleet_krkn [62]

Answer:

47 m

Explanation:

Data obtained from the question include the following:

Length of dry leg 1 (L1) = 40 m

Length of dry leg 2 (L2) = 25 m

Length of swimming course (L) =..?

The length of the swimming course can be obtained by using pythagoras theory as shown below:

L² = L1² + L2²

L² = 40² + 25²

L² = 1600 + 625

L² = 2225

Take the square root of both side.

L = √2225

L = 47.1 ≈ 47 m

Therefore, the length of the swimming course is approximately 47 m.

7 0
3 years ago
A dentist uses a concave mirror (focal length 2.5 cm) to examine some teeth. If the distance from the object to the mirror is 1.
AysviL [449]

Answer:

2.28

Explanation:

From mirror formula,

1/f = 1/u+1/v .......... Equation 1

Where f = focal length of the mirror, v = image distance, u = object distance.

Note: The focal length mirror is positive.

make v the subject of the equation,

v = fu/(u-f)............ Equation 2

Given: f = 2.5 cm, u = 1.4 cm

Substitute into equation 2

v = 2.5(1.4)/(1.4-2.5)

v = 3.5/-1.1

v = -3.2 cm.

Note: v is negative because it is a virtual image.

But,

Magnification = image distance/object distance

M = v/u

Where M = magnification.

Given: v = 3.2 cm, u = 1.4 cm

M = 3.2/1.4

M = 2.28.

Thus the magnification of the tooth = 2.28.

3 0
4 years ago
**100 points** PLEASE ANSWER IN 3 PARAGRAPHS
Deffense [45]

Answer:

In the previous section, we defined circular motion. The simplest case of circular motion is uniform circular motion, where an object travels a circular path at a constant speed. Note that, unlike speed, the linear velocity of an object in circular motion is constantly changing because it is always changing direction. We know from kinematics that acceleration is a change in velocity, either in magnitude or in direction or both. Therefore, an object undergoing uniform circular motion is always accelerating, even though the magnitude of its velocity is constant.

You experience this acceleration yourself every time you ride in a car while it turns a corner. If you hold the steering wheel steady during the turn and move at a constant speed, you are executing uniform circular motion. What you notice is a feeling of sliding (or being flung, depending on the speed) away from the center of the turn. This isn’t an actual force that is acting on you—it only happens because your body wants to continue moving in a straight line (as per Newton’s first law) whereas the car is turning off this straight-line path. Inside the car it appears as if you are forced away from the center of the turn. This fictitious force is known as the centrifugal force. The sharper the curve and the greater your speed, the more noticeable this effect becomes.

Figure 6.7 shows an object moving in a circular path at constant speed. The direction of the instantaneous tangential velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity; in this case it points roughly toward the center of rotation. (The center of rotation is at the center of the circular path). If we imagine Δs becoming smaller and smaller, then the acceleration would point exactly toward the center of rotation, but this case is hard to draw. We call the acceleration of an object moving in uniform circular motion the centripetal acceleration ac because centripetal means center seeking.

hope it helps! stay safe and tell me if im wrong pls :D

(brainliest if you want, or if its right pls) :)

4 0
2 years ago
Tina's calculations of the tarantula found that the spider was able to cover 20 centimeters in five seconds what was the average
Sergeu [11.5K]
<span>(20 cm)/(5 sec) = (0.20 meters)/(5 seconds) 
</span>
8 0
4 years ago
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