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professor190 [17]
2 years ago
7

a proton is moving horizontally at 7.9 x 10^5 m/s. It passes through a vertical magnetic field that points downward. In which di

rection is the proton deflected by the magnetic field? A. up B. to the right of its velocity C. down D. to the left of its velocity

Physics
1 answer:
blondinia [14]2 years ago
7 0
There are four key characteristics of the magnetic force on a moving charge.

First, Its magnitude is proportional to the magnitude of the charge. 
Second, the magnitude of the force is also proportional
Third, the magnetic force depends on the particle’s
velocity. 
Fourth, we find by experiment that the magnetic force \vec{F} does not have the same direction as the magnetic field \vec{F} but instead is always perpendicular to both \vec{B} and the velocity \vec{v} 

So, in a mathematical language this is given by:

\vec{F}=q\vec{v} \times \vec{B}

So, this is a cross product. Therefore, applying the Right-hand rule:

The answer is:

<span>D. to the left of its velocity.

Explaining this in other words:

"The force </span>\vec{F} is directed into the plane of the paper, so this is the direction of the proton"

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The A-string (440 HzHz) on a piano is 38.9 cmcm long and is clamped tightly at both ends. If the string tension is 667N, what's
Mice21 [21]

Answer:

Mass, m = 2.2 kg                                

Explanation:

It is given that,

Frequency of the piano, f = 440 Hz

Length of the piano, L = 38.9 cm = 0.389 m

Tension in the spring, T = 667 N

The frequency in the spring is given by :

f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}

\mu=\dfrac{m}{L} is the linear mass density

On rearranging, we get the value of m as follows :

m=\dfrac{T}{4Lf^2}

m=\dfrac{667}{4\times 0.389\times (440)^2}

m = 0.0022 kg

or

m = 2.2 grams

So, the mass of the object is 2.2 grams. Hence, this is the required solution.

3 0
2 years ago
Read 2 more answers
A child is riding a bike at a speed of 6m/s with a total kinetic energy of 1224J. If the mass of the child is 30kg, what is the
UkoKoshka [18]

Answer:

 Mass of bike = 38 kg.

Explanation:

Kinetic energy is given by the expression, KE = \frac{1}{2} mv^2, where m is the mass and v is the velocity.

Here speed of child riding bike = 6 m/s

Mass of child = 30 kg

Total kinetic energy = 1224 J

Let the mass of bike be, m kg

So, total mass of child and bike = (m + 30) kg

Substituting,

  1224 = \frac{1}{2}* (m+30)*6^2\\ \\ m+30=68\\ \\ m=38kg

So, mass of bike = 38 kg.

3 0
2 years ago
What are the magnitude and direction of the force the pitcher exerts on the ball? (enter your magnitude to at least one decimal
murzikaleks [220]
Details are missing in the question. Complete text of the problem:

"The gravitational force exerted on a baseball is 2.28 N down. A pitcher throws the ball horizontally with velocity 16.5 m/s by uniformly accelerating it along a straight horizontal line for a time interval of 181 ms. The ball starts from rest.

(a) Through what distance does it move before its release? (m)
(b) What are the magnitude and direction of the force the pitcher exerts on the ball? (Enter your magnitude to at least one decimal place.)"


Solution

(a) The pitcher accelerates the baseball from rest to a final velocity of v_f = 16.5 m/s, so \Delta v=16.5 m/s, in a time interval of \Delta t = 181 ms=0.181 s. The acceleration of the ball in the horizontal direction (x-axis) is therefore

a_x =  \frac{\Delta v}{\Delta t}= \frac{16.5 m/s}{0.181 s}=91.2 m/s^2

And the distance covered by the ball during this time interval, before it is released, is:

S= \frac{1}{2} a_x (\Delta t)^2 = \frac{1}{2} (91.2 m/s^2)(0.181 s)^2=1.49 m

(b) For this part we need to consider also the weight of the ball, which is W=mg=2.28 N

From this, we find its mass: m= \frac{W}{g}= \frac{2.28 N}{9.81 m/s^2}=0.23 Kg

Now we can calculate the magnitude of the force the pitcher exerts on the ball. On the x-axis, we have

F_x = m a_x = (0.23 kg)(91.2 m/s^2)=20.98 N

We also know that the ball is moving straight horizontally. This means that the vertical component of the force exerted by the pitcher must counterbalance the weight of the ball (acting downward), in order to have a net force of zero along the y-axis, and so:

F_y=W=mg=2.28 N (upward)

So, the magnitude of the force is

F= \sqrt{F_x^2+F_y^2}=  \sqrt{(20.98N)^2+(2.28N)^2}=21.2 N

To find the direction, we should find the angle of F with respect to the horizontal. This is given by

\tan \alpha =  \frac{F_y}{F_x}= \frac{2.28 N}{20.98 N}=0.11

From which we find \alpha=6.2^{\circ}

7 0
2 years ago
Read 2 more answers
A block rests on a flat plate that executes vertical simple harmonic motion with a period of 0.74 s. What is the maximum amplitu
Mumz [18]

Answer:

maximum amplitude  = 0.13 m

Explanation:

Given that

Time period T= 0.74 s

acceleration of gravity g= 10 m/s²

We know that time period of simple harmonic motion given as

T=\dfrac{2\pi}{\omega}

0.74=\dfrac{2\pi}{\omega}

ω = 8.48 rad/s

ω=angular frequency

Lets take amplitude = A

The maximum acceleration given as

a= ω² A

The maximum acceleration should be equal to g ,then block does not separate

a= ω² A

10= 8.48² A

A=0.13 m

maximum amplitude  = 0.13 m

8 0
2 years ago
A turntable rotates counterclockwise at 76 rpm . A speck of dust on the turntable is at 0.47 rad at t=0. What is the angle of th
icang [17]

To solve this exercise it is necessary to apply the kinematic equations of angular motion.

By definition we know that the displacement when there is constant angular velocity is

\theta= \theta_0 +\omega t

From our given data we know that,

\omega = 76\frac{rev}{min}

\omega = 76\frac{rev}{min}(\frac{2\pi rad}{1rev})(\frac{1 min}{60s})

\omega = 7.958rad/s

Moreover we know that

\theta_0 = 0.47 rad

Therefore for time t=8.1s we have,

\theta= \theta_0+ \omega t

\theta= 0.47+(7.958)(8.1)

\theta = 64.9298rad

That number in revolution is:

\theta = 64.9298rad(\frac{1rev}{2\pi})

\theta = 15.108 Revolutions

Here, we see that there are 15 complete revolutions

And 0.108 revolutions i not complete, so the tunable rotation is

\theta_{net} = 0.108*2\pi=0.216\pi

Therefore the angle of the speck at a time 8.1s is 0.216\pi

4 0
2 years ago
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