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4 years ago

Which statement is true about an object moving in a circular motion due to centripetal force, F, when the mass is doubled?

Physics

1 answer:

marshall27 [118]4 years ago

4 0

Ans) A) Centripetal force will be doubled.

See centripetal force F = mv^2/r

That means centripetal force is directly proportional to the mass of the particle

So, if we double the mass, centripetal force will be increased by twofolds.

So, option A) is correct.

Now, looking at the other options,

B) says centripetal force is unaltered which is incorrect as centripetal force has been altered and increased twofold.

Option C) and D) reduces centripetal force which are also not possible here.

So, only Option A) is correct

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An electron is accelerated by a 5.9 kV potential difference. das (sd38882) – Homework #9 – yu – (44120) 3 The charge on an elect

klio [65]

Complete Question

An electron is accelerated by a 5.9 kV potential difference. das (sd38882) – Homework #9 – yu – (44120) 3 The charge on an electron is 1.60218 × 10−19 C and its mass is 9.10939 × 10−31 kg. How strong a magnetic field must be experienced by the electron if its path is a circle of radius 5.4 cm?

Answer:

The magnetic field strength is $B= 0.0048 T$

Explanation:

The work done by the potential difference on the electron is related to the kinetic energy of the electron by this mathematical expression

$\Delta V q = \frac{1}{2}mv^2$

Making v the subject

$v = \sqrt{[\frac{2 \Delta V * q }{m}] }$

Where m is the mass of electron

v is the velocity of electron

q charge on electron

$\Delta V$ is the potential difference

Substituting values

$v = \sqrt{\frac{2 * 5.9 *10^3 * 1.60218*10^{-19} }{9.10939 *10^{-31]} }$ f

$= 4.5556 *10^ {7} m/s$

For the electron to move in a circular path the magnetic force[ $F = B q v$ ] must be equal to the centripetal force[ $\frac{mv^2}{r}$ ] and this is mathematically represented as

$Bqv = \frac{mv^2}{r}$

making B the subject

$B = \frac{mv}{rq}$

r is the radius with a value = 5.4cm = $= \frac{5.4}{100} = 5.4*10^{-2} m$

Substituting values

$B = \frac{9.1039 *10^{-31} * 4.556 *10^7}{5.4*10^-2 * 1.60218*10^{-19}}$

$= 0.0048 T$

7 0

4 years ago

When a scientific calculator shows the quantity below, what does it mean?<br> 1.5E8

jenyasd209 [6]

1.5 with 8 exponent of 10.
1.5 x 100000000
150000000

6 0

2 years ago

A proton travels at a velocity of 3.0x10^6 m/s in a direction perpendicular to a uniform magnetic field.

Virty [35]

Answer:

0.239 T

Explanation:

Applying,

F = Bvqsin∅................ Equation 1

Where F = magnetic force, B = magnetic Field, q = charge of a proton, v = velocity of proton, ∅ = angle between the velocity and the magnetic field.

make B the subject of the equation

B = F/(vqsin∅)................. Equation 2

From the question,

Given: F = 1.15×10⁻¹³ N, v = 3.0×10⁶ m/s, ∅ = 90°(perpendicular)

Constant: q = 1.602 x 10⁻¹⁹ C

Substitute into equation 2

B = 1.15×10⁻¹³ /(3.0×10⁶×1.602 x 10⁻¹⁹×sin90°)

B = 1.15×10⁻¹³/(4.806×10⁻¹³)

B = 0.239 T.

Hence the magnetic field = 0.239 T

7 0

3 years ago

A 750 g air-track glider attached to a spring with spring constant 14.0 N/m is sitting at rest on a frictionless air track. A 20

alexandr402 [8]

Answer:

the amplitude of the subsequent oscillations is 0.11 m

the period of the subsequent oscillations is 1.94 s

Explanation:

given Information:

the mass of air-track glider, $m_{1}$ = 750 g = 0.75 kg

spring constant, k = 13.0 N/m

the mass of glider, $m_{2}$ = 200 g = 0.2 kg

the speed of glider, $v_{2}$ = 170 cm/s = 1.7 m/s

the amplitude of the subsequent oscillations is A = 0.11 m

according to mechanical enery equation, we have

$A = \sqrt{\frac{m_{1} +m_{2} }{k} }v_{f}$

where

A is the amplitude and $v_{f}$ is the final speed.

to find $v_{f}$ , we can use momentum conservation lwa, where the initial momentum is equal to the final momentum.

$P_{f} = P_{i}$

$(m_{1} +m_{2} )v_{f} = m_{1} v_{1} +m_{2}v_{2}$

$v_{1}$ = 0, thus

(0.75+0.2) $v_{f}$ = (0.75)(0)+(0.2)(1.7)

0.95 $v_{f}$ = 0.34

$v_{f}$ = 0.36 m/s

Now we can calculate the amplitude

$A = \sqrt{\frac{0.75 +0.2 }{10} }0.36$

A = 0.11 m

the period of the subsequent oscillations is T = 1.94 s

the equation for period is

T = 2π $\sqrt{\frac{m_{1}+m_{2} }{k} }$

T = 2π $\sqrt{\frac{0.75+0.2 }{10} }$

T = 1.94 s

7 0

3 years ago

What measures air pressure.?

Alekssandra [29.7K]

That depends on what type of pressure you are attempting to measure, to measure Atmospheric pressure, you would use a Barometer. To measure things like tires, you could use a Tire Pressure Gauge. For Industrial processes and boilers, you would use a Manometer. For pressure vessels, you would use a Bordon Gauge. <span />

8 0

4 years ago

Read 2 more answers