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Pie
4 years ago
12

What is the speed of a high-speed electron traveling at right angles to a 0.80-tesla magnetic field with a force of -7.5 x 10^-1

2 newtons? The charge of an electron is 1.60 x 10^-19
Physics
1 answer:
serg [7]4 years ago
7 0

Explanation:

Answer:

Magnetic field, B = 0.016 Tesla

Explanation:

It is given that,

Velocity of electron, v=9.6\times 10^5\ m/sv=9.6×10

5

m/s

Magnetic force, F=2.6\times 10^{-15}\ NF=2.6×10

−15

N

Charge, q=1.6\times 10^{-19}\ Nq=1.6×10

−19

N

The magnetic force is given by :

F=qvBF=qvB

B=\dfrac{F}{qv}B=

qv

F

B=\dfrac{2.6\times 10^{-15}}{1.6\times 10^{-19}\times 9.6\times 10^5}B=

1.6×10

−19

×9.6×10

5

2.6×10

−15

B=0.016\ TB=0.016 T

So, the magnitude of magnetic field is 0.016 Tesla. Hence, this is the required solution.

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4 0
4 years ago
You lift a 45-newton bag of mulch 1.2 meters and carry it a distance of 10 meters to the garden. How much 
zysi [14]
To lift the bag straight up takes (F · D) = (45 · 1.2) = 54 joules of energy (work).

Moving the bag horizontally 'across' gravity requires no work.
It doesn't matter how far.
4 0
4 years ago
Describe in terms of kinetic and potential energy what happens if an apple falls from a tree and comes to rest on the ground( wr
Lunna [17]

Answer:

An apple hanging at a branch has potential energy due its position. It can be written as PE= mgh where m is the mass of the apple h is the distance between the apple and the ground and g is the acceleration due to gravity.

as the apple falls from the tree it loses its potential energy and gains kinetic energy due to the movement of the apple. Its kinetic energy will be given by KE= 1/2mv²  where m is the mass of the apple and v is the speed with which the apple falls.

As the apple falls the height or the distance reduces and PE becomes reduces. But it gains Kinetic energy due to its speed.

But when the apple falls to the ground and comes to rest its kinetic energy is converted to potential energy.

thus the total energy remains the same. it changes from one form to the other but remains unaltered.

6 0
3 years ago
Suppose high tide is at midnight, the water level at midnight is 3 m, and the water level at low tide is 0.5 m. Assuming the nex
aev [14]

We have that the time, to the nearest minute, when the water level is at 1.125 m for the second time after midnight is

t=10.0hours

From the Question we are told that

Maximum height h_{max}=3m

Minimum height  H_{min}=0.5m

Time for  next high tide will occurT=12 hours =>720 min

Generally Average Height

h_{avg}=\frac{3+0.5}{2}\\\\h_{avg}=1.75

Therefore determine Amplitude to be

A=h_{max}=j_{avg}\\\\A=3-1.75\\\\A=1.25

Generally, the equation for Time is mathematically given by

At t=0

h(x)=Acos(Bx)+h_{avg}

Where

B=\frac{2\pi}{P}\\\\B=\frac{2\pi}{720}\\\\B=8.73*10^{-3}

Therefore

h(t)=Acos8.73*10^{-3}(t)+h_{avg}

Hence the Time at T=1.125 is

1.125(t)=1.25cos(8.73*10^{-3})(t)+1.75

-0.1249t=1.75

t=10.0hours

For more information on this visit

brainly.com/question/22361343

5 0
3 years ago
A rock is tossed straight up from the ground with a speed of 21 m/s . When it returns, it falls into a hole 10 m deep.a.) What i
Arte-miy333 [17]

(a) 25.2 m/s

Let's take the initial vertical position of the rock as "zero" (reference height).

According to the law of conservation of energy, the speed of the rock as it reaches again the position "zero" after being thrown upwards is equal to the initial speed of the rock, 21 m/s (in fact, if there is no air resistance, no energy can be lost during the motion; and since the kinetic energy depends only on the speed of the rock:

K=\frac{1}{2}mv^2

and the gravitational potential energy of the rock has not changed, since the rock has returned into its initial position, it means that the speed of the rock should be the same)

This means that we can only analyze the final part of the motion, the one in which the rock falls into the 10 m hole. Since it is a free fall motion, we can find the final speed by using

v^2 = u^2 + 2gd

where

u = 21 m/s is the initial speed of the rock as it enters the hole

g = 9.8 m/s^2 is the acceleration due to gravity

d = 10 m is the depth of the hole

Substituting,

v=\sqrt{u^2 +2gd}=\sqrt{(21 m/s)^2+2(9.8 m/s^2)(10 m)}=25.2 m/s

(b) 4.72 s

The vertical position of the rock at time t is given by

y(t) = v_y t - \frac{1}{2}gt^2

where

v_y = 21 m/s is the initial vertical velocity

Substituting y(t)=-10 m, we can then solve the equation for t to find the time at which the rock reaches the bottom of the hole:

-10 = 21 t - \frac{1}{2}(9.8)t^2\\10+21 t -4.9t^2 = 0

which has two solutions:

t = -0.43 s --> negative, so we discard it

t = 4.72 s --> this is our solution

7 0
4 years ago
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