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oee [108]
2 years ago
8

A person tosses a ball from the ground up into the air at an initial speed of 10 m/sec and an initial angle of 43◦ off the groun

d. after the ball is released, what is the total acceleration vector acting on the ball when the ball is at the top of its arc? 1. 9.8 m/s2 , down 2. none of these 3. 9.8 m/s2 , up 4. zero 5. 9.8 m/s2 , in the horizontal direction
Physics
1 answer:
Feliz [49]2 years ago
7 0
Just try your best best friend everyone
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A proton having an initial velvocity of 20.0i Mm/s enters a uniform magnetic field of magnitude 0.300 T with a direction perpend
Sonja [21]

The time interval for which the proton remains in the field is -

Δt = $\frac{\pi R}{40}.

We have a proton entering a uniform magnetic field which is in a direction perpendicular to the proton's velocity.

We have to determine time interval during which the proton is in the field.

<h3>What is the magnitude of force on the charged particle moving in a uniform magnetic field?</h3>

The magnitude of force on the charged particle moving in a uniform magnetic field is given by -

F = qvB sinθ



According to the question, we have -

Entering Velocity (v) = 20 i  m/s

Magnetic field intensity (B) = 0.3 T

Leaving velocity (u) = - 20 j  m/s

Now -

The entering and leaving velocity vectors have 90 degrees difference between them. Therefore, only a quarter of distance of the complete circular path of radius 'R' is traced by the proton. Therefore -

d = $\frac{2\pi r}{4} = $\frac{\pi R}{2}

Since, the radius of circular path is not given, we will assume it R.

Therefore, time for which proton remained in the field is -

t = $\frac{\pi R}{2v} = \frac{\pi R}{40}

Hence, the time interval for which the proton remains in the field is -

Δt = $\frac{\pi R}{40}

To solve more questions on Force on charged particle, visit the link below-

brainly.com/question/14597200

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6 0
1 year ago
What is density?
il63 [147K]
<span>a. the amount of matter in a given volume </span>
8 0
2 years ago
Read 2 more answers
an object is placed 15.8 cm in front of a thin converging lens with an unknown focal length. if a real image forms behind the le
storchak [24]
<span>On what:

f (is the focal length of the lens) = ? 
p (is the distance from the object to the lens) =15.8 cm
p' (is the distance from the image to the spherical lens) = 4.2 cm

</span><span>Using the Gaussian equation, to know where the object is situated (distance from the point).
</span>
\frac{1}{f} = \frac{1}{p} + \frac{1}{p'}
\frac{1}{f} = \frac{1}{15.8} + \frac{1}{4.2}
\frac{1}{f} = \frac{2.1}{33.18} + \frac{7.9}{33.18}
\frac{1}{f} =  \frac{10}{33.18}
Product of extremes equals product of means:
10*f = 1*33.18
10f = 33.18
f =  \frac{33.18}{10}
\boxed{\boxed{f = 3.318\:cm}}\end{array}}\qquad\quad\checkmark

6 0
3 years ago
The forklift exerts a 1,500.0 N force on the box and moves it 3.00 m forward to the stack. How much work does the forklift do ag
allochka39001 [22]
In order to get the propoerty of work you need to use the following formula
 <span>work = force times distance
</span>replacing data you will get:
W = (1.500) (3)
W =  4.500 NM
The answer should be in NM. So it will be 4500 NM againts the force of gravity
4 0
3 years ago
Q011) The Doppler effect a. occurs when the frequency of sound waves received is lower if the wave source is moving toward you t
inn [45]

Answer:

Option (c) is correct.

Explanation:

The apparent change in the frequency of light due to the relative motion between the source and the observer is called Doppler's effect.

When the source is moving towards the observer which is at rest, the apparent frequency increases and if the observer is moving away the frequency of sound decreases.

It occurs for both light and sound.  

So, to explain the blue shift of light in the universe is due to the Doppler's effect of light.

8 0
3 years ago
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