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3 years ago

A 2MeV proton is moving perpendicular to a uniform magnetic field of 2.5 T.the force on a proton is

Physics

2 answers:

Snowcat [4.5K]3 years ago

5 0

Answer:

7.8x10-12N

Explanation:

We know that

Magnetic force = F = qVB

And

Also Kinetic energy K.E is

E = (1/2)mV²

So making v subject

V = √(2E / m)

And

E = KE = 2MeV

= 2 × 106 eV

= 2 × 106 × 1.6 × 10–19 J

= 3.2 × 10–13 J

And then

V= √2x3.2E-13/1.6E-27

1.9E7m/s

Given that

mass of proton = 1.6 × 10–27 kg,

Magnetic field strength B = 2.5 T.

So F= qBv sinစ

So F = 1.6 × 10–19 × 2.5 × 1.9 x10^7 x sin 90°

= 7.8 x 10^-12N

larisa86 [58]3 years ago

5 0

Answer:

8*10^-12

Explanation:

Given that

Energy of proton, K = 2 MeV = 2 * 1.6*10^-19 *10^6 = 3.2*10^-13

magnetic field strength, B = 2.5 T

mass of proton, m = 1.67*10^-27 kg

K = ½mv², making v² the subject of formula by rearranging, we have

v² = 2k/m

v² = (2 * 3.2*10^-13) / 1.67*10^-27

v² = 6.4*10^-13 / 1.6*10^-27

v² = 4*10^14

v = √4*10^14

v = 2*10^7 m/s

f = qvbsinθ, where

θ = 90

v = 2*10^7 m/s

b = 2.5 T

q = 1.6*10^-19

f = 1.6*10^-19 * 2*10^7 * 2.5 sin 90

f = 8*10^-12 N

thus, the force on the proton is 8*10^-12

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An object with a mass M = 250 g is on a plane inclined at 30º above the horizontal and is attached by a string to a mass m = 150

malfutka [58]

Answer:

0.495 m/s

Explanation:

$T$ = tension force in the string connecting the two objects

$M$ = Mass of the object on inclined plane = 250 g = 0.250 kg

$m$ = Mass of the hanging object = 150 g = 0.150 kg

$a$ = acceleration of each object

From the force diagram, force equation for the motion of the object on the inclined plane is given as

$T - Mg Sin30 = Ma\\T = Mg Sin30 + Ma$

From the force diagram, force equation for the motion of the hanging object on the inclined plane is given as

$mg - T = ma\\T = mg - ma$

Using the above two equations

$Mg Sin30 + Ma = mg - ma$

$(0.250)(9.8) Sin30 + (0.250) a = (0.150) (9.8) - (0.150)a$

$a = 0.6125 ms^{-2}$

$h$ = height dropped by the hanging object = 10 cm = 0.10 m

$v$ = Speed gained by the object

Speed gained by the object can be given as

$v = sqrt(2ah)\\v = sqrt(2(0.6125)(0.20))\\v = 0.495 ms^{-1}$

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3 years ago

The Jamaican bobsled team hit the brakes on their sled so that it decelerates at a uniform rate of 0.43 m/s^2. How long does it

topjm [15]

Answer: The distance covered by a certain object which is travelling at a certain speed is calculated through the equation,

d = (V₀)t + 0.5at²

where d is the distance, V₀ is the initial speed, a is deceleration, and t is the time. Substituting the known values,

85 = (V₀)(t) + (0.5)(-0.43 m/s)(t²)

Because we are not given with the initial velocity, our answer would remain as the equation which is written above.

Explanation: Hope this helps

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Suppose there is a uniform magnetic field, B, pointing into the page (so your index finger will point into the page). If the vel

Naily [24]

Answer:

Explanation:

We shall show all given data in vector form and calculate the direction of force with the help of following formula

force F = q ( v x B )

q is charge , v is velocity and B is magnetic field.

Given B = - Bk ( i is right , j is upwards and k is straight up the page )

v = v j

F = q ( vj x - Bk )

= -Bqvi

The direction is towards left .

a ) If velocity is down

v = - v j

F = q ( - vj x - bk )

= qvB i

Direction is right .

b ) v = v i

F = q ( vi x - Bk )

= qvB j

force is upwards

c ) v = - vi

F = q ( -vi x - Bk )

= -qvBj

force is downwards

d ) v = - v k

F = q( - vk x -Bk )

= 0

No force will be created

e ) v = v k

F = q( vk x -Bk )

= 0

No force will be created

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3 years ago

A bullet of mass 0.017 kg traveling horizontally at a high speed of 290 m/s embeds itself in a block of mass 5 kg that is sittin

rodikova [14]

Answer:

(a) vf = 0.98 m/s

(b) K₁ = 714.85 J : Total translational kinetic energy before the collision.

K₂= 2.41 J : Total translational kinetic energy after the collision.

Explanation:

Theory of collisions

Linear momentum is a vector magnitude (same direction of the velocity) and its magnitude is calculated like this:

p=m*v

where

p:Linear momentum

m: mass

v:velocity

There are 3 cases of collisions : elastic, inelastic and plastic.

For the three cases the total linear momentum quantity is conserved:

P₀ = Pf Formula (1)

P₀ :Initial linear momentum quantity

Pf : Final linear momentum quantity

Data

m₁ = 0.017 kg : mass of the bullet

m₂ = 5 kg : mass of the block

v₀₁ = 290 m/s : initial velocity of the bullet

v₀₂ = 0 : initial velocity of the block₂

(a) Speed of the block after the bullet embeds itself in the block

We appy the formula (1):

P₀ = Pf

m₁*v₀₁ + m₂*v₀₂ = (m₁+ m₂)vf

vf: final velocity of the block

( 0.017)*( 290) + (5)*(0) = ( 0.017 + 5 )*vf

4.93+ 0 = ( 5.017 )*vf

vf = 4.93 / 5.017

vf = 0.98 m/s

b) Total translational kinetic energy before (K₁) and after the collision(K₂).

K₁ = 1/2(m₁*v₀₁² + m₂*v₀₂²)

K₁ = 1/2(0.017*(290)² + 5*(0)²) = 714.85 J

K₂= 1/2(m₁+ m₂)*vf²

K₂= 1/2(0.017+ 5)*(0.98)² = 2.41 J

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3 years ago

If fuel consumption is 80 pounds per hour and groundspeed is 180 knots, how much fuel is required for an airplane to travel 477

lisov135 [29]

Answer:

212 pounds

Explanation:

477 nm / 180 nm/hr * 80 #/hr = 212 #

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