A car travels 15 kilometers in 30 minutes. what is the speed of the car

A particle with charge 7.76×10^(−8)C is moving in a region where there is a uniform 0.700 T magnetic field in the +x-direction.

kodGreya [7K]

Answer:

The z-component of the force is $\= F_z = 0.00141 \ N$

Explanation:

From the question we are told that

The charge on the particle is $q = 7.76 *0^{-8} \ C$

The magnitude of the magnetic field is $B = 0.700\r i \ T$

The velocity of the particle toward the x-direction is $v_x = -1.68*10^{4}\r i \ m/s$

The velocity of the particle toward the y-direction is

$v_y = -2.61*10^{4}\ \r j \ m/s$

The velocity of the particle toward the z-direction is

$v_y = -5.85*10^{4}\ \r k \ m/s$

Generally the force on this particle is mathematically represented as

$\= F = q (\= v X \= B )$

So we have

$\= F = q ( v_x \r i + v_y \r j + v_z \r k ) \ \ X \ ( \= B i)$

$\= F = q (v_y B(-\r k) + v_z B\r j)$

substituting values

$\= F = (7.7 *10^{-8})([ (-2.61*10^{4}) (0.700)](-\r z) + [(5.58*10^{4}) (0.700)]\r y)$

$\= F= 0.00303\ \r j +0.00141\ \r k$

So the z-component of the force is $\= F_z = 0.00141 \ N$

Note : The cross-multiplication template of unit vectors is shown on the first uploaded image ( From Wikibooks ).

7 0

3 years ago

A battery with an emf of 24.0 V is connected to a resistive load. If the terminal voltage of the battery is 16.1 V and the curre

lions [1.4K]

Answer:

2.03 Ω

Explanation:

EMF: This can be defined as the potential difference of a cell when it is not delivering any current. The S.I unit of Emf is Volt.

The formula of emf is given as,

E = I(R+r)............................ Equation 1

Where E = Emf, I = current, R = External resistance, r = internal resistance.

Make r the subject of the equation

r = (E-IR)/I........................ Equation 2

Note: From ohm's law, V = IR.

r = (E-V)/I........................ Equation 3

Where V = Terminal voltage

Given: E = 24 V, I = 3.9 A, V = 16.1 V.

Substitute into equation 3

r = (24-16.1)/3.9

r = 7.9/3.9

r = 2.03 Ω

6 0

3 years ago

A proton having an initial velvocity of 20.0i Mm/s enters a uniform magnetic field of magnitude 0.300 T with a direction perpend

Sonja [21]

The time interval for which the proton remains in the field is -

Δt = $$\frac{\pi R}{40}$ .

We have a proton entering a uniform magnetic field which is in a direction perpendicular to the proton's velocity.

We have to determine time interval during which the proton is in the field.

<h3>What is the magnitude of force on the charged particle moving in a uniform magnetic field?</h3>

The magnitude of force on the charged particle moving in a uniform magnetic field is given by -

F = qvB sinθ

According to the question, we have -

Entering Velocity (v) = 20 i m/s

Magnetic field intensity (B) = 0.3 T

Leaving velocity (u) = - 20 j m/s

Now -

The entering and leaving velocity vectors have 90 degrees difference between them. Therefore, only a quarter of distance of the complete circular path of radius 'R' is traced by the proton. Therefore -

d = $$\frac{2\pi r}{4}$ = $$\frac{\pi R}{2}$

Since, the radius of circular path is not given, we will assume it R.

Therefore, time for which proton remained in the field is -

t = $$\frac{\pi R}{2v} = \frac{\pi R}{40}$

Hence, the time interval for which the proton remains in the field is -

Δt = $$\frac{\pi R}{40}$

To solve more questions on Force on charged particle, visit the link below-

brainly.com/question/14597200

#SPJ4

6 0

2 years ago

If I could lift up to ten tons and I threw a ball the size of an orange but weighed a ton, to the ground, how big of an impact w

Anarel [89]

Answer:

The impact force is 98000 N.

Explanation:

mass = 10 tons

The impact force is the weight of the object.

Weight =mass x gravity

W = 10 x 1000 x 9.8

W = 98000 N

The impact force is 98000 N.

5 0

3 years ago

- farmer has recently bought a 40 acre land and wishes to fence the perimeter. given that the Land is square, what lenght of fen

goldfiish [28.3K]

Answer:

Explanation:

4√((40(4046)) = 1,609.17370... = 1609 m

3 0

3 years ago

A car travels 15 kilometers in 30 minutes. what is the speed of the car​

A car travels 15 kilometers in 30 minutes. what is the speed of the car