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Alborosie
3 years ago
6

A proton moving at 8.9 × 106 m/s through a magnetic field of 0.96 T experiences a magnetic force of magnitude 3.8 × 10−13 N. Wha

t is the angle between the proton’s velocity and the field? The charge on a proton is 1.60218 × 10−19 C and its mass is 1.67262 × 10−27 kg.
Physics
1 answer:
goblinko [34]3 years ago
8 0

Answer: 15.66 °

Explanation: In order to solve this proble we have to consirer the Loretz force for charge partcles moving inside a magnetic field. Thsi force is given by:

F=q v×B = qvB sin α where α is teh angle between the velocity and magnetic field vectors.

From this expression and using the given values we obtain the following:

F/(q*v*B) = sin α

3.8 * 10^-13/(1.6*10^-19*8.9*10^6* 0.96)= 0.27

then  α =15.66°

You might be interested in
What would the weight of an astronaut be on Saturn if his mass is 68 kg and acceleration of gravity on Saturn is 10.44 m/s2? Ple
alex41 [277]

Here's the part you need to know:

       (Weight of anything) =

                 (the thing's mass)
times
                 (acceleration of gravity in the place where the thing is) .

                 Weight = (mass ) x (gravity) .

That's always true everywhere.
You should memorize it.

For the astronaut on Saturn . . .

                   Weight = (mass ) x (gravity) .

                
  Weight =  (68 kg) x (10.44 m/s²)

                        
      =    709.92 newtons .
__________________________________

On Earth, gravity is only  9.8 m/s².
So as long as the astronaut is on Earth, his weight is only

                                   (68 kg) x (9.8 m/s²)

                               =    666.4 newtons .

Notice that his mass is his mass ... it doesn't change
no matter where he goes. 

But his weight changes in different places, because
it depends on the gravity in each place.

4 0
3 years ago
5- A 2500g object is pushed with 55N for 12m in 11s, there was a force of friction of 30N.
Assoli18 [71]

Answer:

1kg =1000g

2.5kg

D=12m

t=11s

F=2.5KG

Explanation:

work done =f.d

=2.5×12

=30Nm

55-30

average speed

final - initial

divide by time t(s)

3 0
2 years ago
While driving, your car has an initial position of 3.2 m, an initial velocity of -8.4 m/s, and
KIM [24]

Answer:

The position of the car at t = 1.5 s is at -8.1625 meters

Explanation:

The initial position of the car is 3.2 meters

The initial velocity is -8.4 m/s

The constant acceleration is 1.1 m/s²

We need to find the final position of the car at the time t = 1.5 seconds

The displacement <em>s</em> = final position - initial position

s=ut+\frac{1}{2}at^{2}, where <em>u</em> is the initial velocity, <em>a</em> is the

constant acceleration and <em>t</em> is the time

So we can find the final velocity by using the rule:

final position - initial position = ut+\frac{1}{2}at^{2}

initial position = 3.2 meters , u = -8.4 m/s , a = 1.1 ²m/s , t = 1.5 s

Substitute these values in the rule

final position - 3.2 = (-8.4)(1.5)+\frac{1}{2}(1.1)(1.5)^{2}

final position - 3.2 = -12.6 + 1.2375

final position - 3.2 = -11.3625

add 3.2 for both sides

final position = -8.1625

<em>That means the car is at 8.1625 meters in opposite direction</em>

<em>The position of the car at t = 1.5 s is at -8.1625 meters </em>

4 0
3 years ago
Give an example of a normal force. type below.
Oksi-84 [34.3K]

Normal force is the force exerted when an object is on an surface. So an example could be a pile of books on top of a table.

8 0
3 years ago
If a trapezoidal channel has side slopes of 1:1, hydraulic depth is 5 feet, the bottom width is 8 feet, flow is 2,312 cubic feet
Evgesh-ka [11]

Answer:

S = \dfrac{1}{2.5}

Explanation:

given,

side slope = 1 : 1

hydraulic depth(y) = 5 ft

bottom width (b)= 8 ft

x = 1

Q = 2,312 ft³/s

n = 0.013

slope of channel = ?

R = \dfrac{A}{P}

R = \dfrac{y(b + xy)}{b+2y\sqrt{1+x^2}}

R = \dfrac{5(8+ 5)}{8+2\times 5\sqrt{1+1^2}}

R = 4.69 m

using manning's equation

Q = \dfrac{1}{n}AR^{2/3} S^{1/2}

2312= \dfrac{1}{0.013}\times (5(8+5))\times 4.69^{2/3} S^{1/2}

2312=14009.37\times S^{1/2}

S = 0.406

S = \dfrac{1}{2.5}

5 0
3 years ago
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