Ask question

Ask question

All categories

3 years ago

15

A proton moves at a speed of 1,140 m/s in a direction perpendicular to a magnetic field with a magnitude of 0.78 T. If the proto

n is replaced with an electron, how will the magnitude of the force change?

1 answer:

Yakvenalex [24]3 years ago

3 0

Answer:

1.42×10⁻¹⁶ N.

Explanation:

The force on a charge moving in a magnetic field is given as

F = qBvsin∅..................... Equation 1

Where F = Force on the charge, q = charge, B = Magnetic Field, v = speed, ∅ = angle between the magnetic field and the speed

Given: B = 0.78 T, v = 1140 m/s, ∅ = 90° ( Perpendicular) q = 1.60 × 10⁻¹⁹ C.

Substitute into equation 1

F = 0.78(1140)(1.60 × 10⁻¹⁹)sin90°

F = 1.42×10⁻¹⁶ N.

Hence the force on the charge = 1.42×10⁻¹⁶ N.

You might be interested in

The logarithmic damping decrement of a mathematical pendulum is DeltaT=0.5. How will the amplitude of oscillations decrease duri

Alika [10]

Explanation:

Finding the (maximum) respective prime powers would yield the answer. Also we need not ... Is perfectly divisible by 720^n? ... So we can say that for any positive value of n it not divisible.

3 0

3 years ago

a body of a mass 1kg suspended from a spring is found to stretch the spring by 10cm. (A) what is the spring constant? what is pe

Nikitich [7]

Answer:

(A) The spring constant is <u>98 N/m.</u>

(B) The period of oscillation is <u>0.635 s.</u>

Explanation:

Given:

Mass of the body is, $m=1\ kg$

Extension length of the spring is, $x=10\ cm=0.1\ m$

Now, let 'k' be the spring constant.

The force acting on the body is due to gravity only and is equal to its weight.

So, weight of the body is given as:

$F_g=mg\\F_g=1\times 9.8\\F_g=9.8\ N$

Now, we know that for a spring-mass system, the net force acting on the body is equal to the product of the spring constant and extension length. Therefore,

$F_g=kx\\9.8=k(0.1)\\k=\frac{9.8}{0.1}=98\ N/m$

Hence, the spring constant is 98 N/m.

(B)

Period of oscillation of a body of spring-mass system in SHM is given as:

$T=2\pi\sqrt{\frac{m}{k}}$

Plug in the given values and solve for period 'T'. This gives,

$T=2\pi\sqrt{\frac{1}{98}}\\T=2\pi\times 0.101\\T=0.635\ s$

Therefore, the period of oscillation is 0.635 s.

5 0

4 years ago

An 88 kg person steps into a car of mass 2002 kg, causing it to sink 5.36 cm on itssprings. Assuming no damping, with what fre-q

Marina CMI [18]

Answer:

The required frequency = 0.442 Hz

Explanation:

Frequency $f = ( \dfrac{1}{2 \pi}) \omega$

where;

$\omega = \sqrt{\dfrac{k}{m} }$

Then;

$f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{k}{m} } \Bigg )$

However;

$k = \dfrac{F}{x}$ and;

mass $m = m_{car } + m_{person}$

$f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{\dfrac{F}{x}}{m_{car}+m_{person}} } \Bigg )$

$f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{{F}}{x(m_{car}+m_{person})} } \Bigg )$

where;

$F = m_{person}g$

Then;

$f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{ {m_{person}g }}{x(m_{car}+m_{person})} } \Bigg )$

replacing the values;

$f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{ {(88 \ kg)* (9.81 \ m/s^2) }}{(5.36 \times 10^{-2} \ m) (2002 \ kg +88 \ kg)} } \Bigg )$

$\mathbf{f = 0.442 \ Hz}$

8 0

3 years ago

A tennis player tosses a tennis ball straight up and then catches it after 2.00 s at the same height as the point of release. (a

Readme [11.4K]

Answer:

Part a)

a = -9.81 m/s/s

Part b)

v = 0

Part c)

v = 9.81 m/s

Part d)

$H = 4.905 m$

Explanation:

Part a)

During the motion of ball it will have only gravitational force on the ball

so here the acceleration of the ball is only due to gravity

so it is given as

$a = g = 9.81 m/s^2$

Part b)

As we know that ball is moving against the gravity

so here the velocity of ball will keep on decreasing as the ball moves upwards

so at the highest point of the motion of the ball the speed of ball reduce to zero

$v_f = 0$

Part c)

We know that the total time taken by the ball to come back to the initial position is T = 2 s

so in this time displacement of the ball will be zero

$\Delta y = 0 = v_y t + \frac{1}{2} at^2$

$0 = v_y (2) - \frac{1}{2}(9.81)(2^2)$

$v_y = 9.81 m/s$

Part d)

at the maximum height position we know that the final speed will be zero

so we will have

$v_f^2 - v_i^2 = 2 a d$

here we have

$0 - (9.81^2) = 2(-9.81)H$

$H = 4.905 m$

4 0

3 years ago

What is the wavelength of an fm radio signal that has a frequency of 91.7 mhz?

Digiron [165]

Answer:

The wavelength of the radio wave will be 3.271 m

Explanation:

We have given frequency of the radio signal $f=91.7MhZ=91.7\times 10^6Hz$

Speed of the light $c=3\times 10^8m/sec$

We have to find the wavelength of the radio signal

We know that wavelength is given by $\lambda =\frac{c}{f}=\frac{3\times 10^8}{91.7\times 10^6}=3.271m$

So the wavelength of the radio wave will be 3.271 m

3 0

3 years ago