A car travels from boston

An upward force act on a proton as it moves with a speed of 2.0 x 10^5 meters/seconds through a magnetic field of 8.5 x 10^2

Gekata [30.6K]

Force on a moving charge is given by formula

$\vec F = q(\vec v \times \vec B)$

here we know that this force will be maximum when velocity is perpendicular to magnetic field

$\vec F = qvB$

here we know that

$v = 2.0 \times 10^5 m/s$

$q = 1.6 \times 10^{-19} C$

$B = 8.5 \times 10^2 T$

now we have

$F = (1.6 \times 10^{-19})(2 \times 10^5)(8.5 \times 10^2)$

$F = 2.72 \times 10^{-11} N$

7 0

3 years ago

A ball is thrown upward. At a height of 10 meters above the ground, the ball has a potential energy of 50 joules (with the poten

lions [1.4K]

Answer:

h = 20 m

Explanation:

given.

height, h = 10 m

Potential energy at 10 m = 50 J

Kinetic energy at 10 m = 50 J

maximum height the ball will reach, H = ?

Total energy of the system

T E = 50 J + 50 J

T E = 100 J

now,

A h = 10 m

P E = m g h

50 = m g x 10

mg = 5 ..............(1)

at the top most Point the only Potential energy will be acting on the body.

now, TE = Potential energy

100 = m g h

5 h = 100

h = 20 m

hence, the maximum height reached by the ball is equal to 20 m.

5 0

3 years ago

Two point charges, with charge magnitudes q and ????, are placed a distance r apart. In this arrangement, each point charge expe

sammy [17]

Answer:

1) Q ’= 8 Q , 2) q ’= 16 q , 3) r ’= ¾ r

Explanation:

For this exercise we will use Coulomb's law

F = k q Q / r²

It asks us to calculate the change of any of the parameters so that the force is always F

Original values

q, Q, r

Scenario 1

q ’= 2q

r ’= 4r

F = k q ’Q’ / r’²

we substitute

F = k 2q Q ’/ (4r)²

F = k 2q Q '/ 16r²

we substitute the value of F

k q Q / r² = k q Q '/ 8r²

Q ’= 8 Q

Scenario 2

Q ’= Q

r ’= 4r

we substitute

F = k q ’Q / 16r²

k q Q / r² = k q’ Q / 16 r²

q ’= 16 q

Scenario 3

q ’= 3/2 q

Q ’= ⅜ Q

we substitute

k q Q r² = k (3/2 q) (⅜ Q) / r’²

r’² = 9/16 r²

r ’= ¾ r

6 0

3 years ago

What happens to light when it travels from air into water

spin [16.1K]

Answer:

Water is more dense than air. When water goes through a denser thing, the light is "bent" more towards the "normal" which is a straight, vertical line.

Follow me on i n s t a g r a m, message me and I'll follow you back: ifaulkner24

8 0

2 years ago

suppose a car manufacturer tested its cars for front end collsion by hauling them up on a crane and dropping them from a certain

IRINA_888 [86]

Initial height: 66.5 m

Explanation:

The problem can be solved by using the principle of conservation of energy.

If we neglect air resistance, the total mechanical energy of the car is conserved during the fall, therefore we can write:

$K_i + U_i = K_f + U_f$

where :

$K_i = 0$ is the kinetic energy of the car at the top (it starts from rest)

$U_i = mgh$ is the gravitational potential energy of the car at the top, with:

m = the mass of the car

g = the acceleration of gravity

h = the heigth of the car

$K_f = \frac{1}{2}mv^2$ is the kinetic energy of the car just before hitting the ground, with

v = 130 km/h final speed of the car

$U_f = 0$ is the gravitational potential energy of the car at the bottom

Re-arranging the equation, we find

$mgh=\frac{1}{2}mv^2$

and we have:

$g=9.8 m/s^2\\v = 130 km/h = 36.1 m/s$

Solving for h, we find the initial height of the car:

$h=\frac{v^2}{2g}=\frac{36.1^2}{2(9.8)}=66.5 m$

Learn more about kinetic energy and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

5 0

3 years ago