I really need help with have no clue

At a particular instant, a proton at the origin has velocity < 5e4, -2e4, 0> m/s. You need to calculate the magnetic field

vesna_86 [32]

Answer:

$9.7\times 10^{-5} T$

Explanation:

Velocity = $5\times 10^4i-2\times 10^4j$

r= $0.03i+0.05j$

r= $\mid r\mid=\sqrt{(0.03)^2+(0.05)^2}=0.058$

v= $\mid V\mid=\sqrt{(5\times 10^4)^2+(-2\times 10^{4})^2}=5.39\times 10^{2}$

We know that

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

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

Mass of proton= $1.67\times 10^{-27} kg$

Using the formula

$B=\frac{1.67\times 10^{-27}\times 5.39\times 10^2}{1.6\times 10^{-19}\times 0.058}$

$B=9.7\times 10^{-5} T$

3 0

3 years ago

A block of mass M slides down an inclined plane that makes an angle θ with the horizontal. The coefficient of kinetic friction b

Llana [10]

Answer:

Mg Cosθ

Explanation:

mass of block = M

Angle of inclination = θ

coefficient of friction = μk

The force of gravity acts of the block is Mg

there are two components of the weight

the component parallel to the inclined plane is Mg Sinθ

the component perpendicular to the plane of inclined is Mg Cosθ

So, the normal force exerted by the inclined is Mg Cosθ

5 0

4 years ago

A car accelerates from rest at a constant rate of 2 m/s^2 for 5 s. what is the speed of the car at the end of that time? g

lara31 [8.8K]

Hope this helps you! This is my step-by-step work. Lemme know if you have any questions!

5 0

3 years ago

Use the slider to apply a force of about 400 N. After 2 s have elapsed in the simulation, decrease the Applied Force (force exer

nikklg [1K]

Answer:

<em>c. decreasing.</em>

Explanation:

Force produces acceleration or deceleration. Force is the product of a body's mass and its acceleration. When a force is applied to an object, the force tends to cause the body to move if the body was originally stagnant, cause the body to accelerate if applied in the direction of the body's velocity, or decelerate the body if applied in opposite direction to the velocity of the body. <em>When the force that is exerted on a moving body is slowly reduced to zero, frictional forces between the body and the floor surface gradually decelerates the body. When this deceleration occurs, the velocity of the body gradually decreases t a stop.</em>

6 0

4 years ago

A newspaper delivery boy throws a newspaper onto a balcony 1.25 m above the

velikii [3]

Answer:

(a) 3.22 m

(b) The vertical velocity, $v_y$ , at maximum height is 0 m/s, the horizontal velocity, vₓ, is 12.72 m/s

(c) The acceleration at maximum height = g = 9.81 m/s²

(d) The time it takes for the paper to reach the balcony is 1.212 seconds

(e) The horizontal range, of the paper is 15.42 m.

Explanation:

(a) Given that we re given a projectile motion, we have the following governing equations;

y = y₀ + v₀·sin(θ₀)·t - 0.5×g·t²

$v_y$ = v₀·sin(θ₀) - g·t

Where:

y = Height of the paper

y₀ = Initial height of the paper = Ground level = 0

v₀ = Inititial velocity of the paper = 15.0 m/s

θ₀ = Angle in which the paper is thrown = 32° above the horizontal

g = Acceleration due to gravity = 9.81 m/s²

t = Time taken to reach the height h

$v_y$ = Vertical velocity of the paper

At maximum height, $v_y$ = 0, therefore;

$v_y$ = v₀·sin(θ₀)·t - g·t = 0

v₀·sin(θ₀) = g·t

t = v₀·sin(θ₀)/g = 15×sin(32°)/9.81 = 0.81 seconds

y = y₀ + v₀·sin(θ₀)·t - 0.5×g·t² = 0 + 15×sin(32°)×0.81-0.5×9.81×0.81² = 3.22 m

(b) The vertical velocity, $v_y$ , at maximum height = 0 m/s, the horizontal velocity, vₓ, = 15×cos(32°) = 12.72 m/s

(c) The acceleration at maximum height = g = 9.81 m/s²

(d) The time it takes to maximum height = 0.81 seconds

The time the paper will take to fall to 1.25 m above the ground, which is 3.22 - 1.25 = 1.97 meters below maximum height is therefore given as follows;

y = y₀ + v₀·sin(θ₀)·t - 0.5×g·t²

Where:

v₀ = 0 m/s at maximum height

y = -1.97 m downward motion

y₀ = 0 starting from maximum height downwards

1.97 = 0 + 0·sin(θ₀)·t - 0.5×9.81×t²

-1.97 = - 0.5×9.81×t²

t = (-1.97)/(-0.5*9.81) = 0.402 seconds

The time the paper will take to fall to 1.25 m above the ground = 0.81+0.402 = 1.212 seconds

Therefore, the time it takes for the paper to reach the balcony = 1.212 seconds

(e) The horizontal range, x, is given by the relation;

x = x₀ + v₀·cos(θ₀)· $t_{tot}$

x₀ = Starting point of throwing the paper = 0

$t_{tot}$ = Total time of flight of the paper

∴ x = x₀ + v₀·cos(θ₀)· $t_{tot}$ = 0 + 15×cos(32°)×1.212 = 15.42 m

The horizontal range, of the paper = 15.42 m.

4 0

4 years ago