If john takes 45 minutes to bicycle a total of four kilometers to his grandmother’s house, what is his velocity in km/hr

A positively charged particle of mass 7.2 x 10-8 kg is traveling due east with a speed of 88 m/s and enters a 0.6-T uniform magn

Marianna [84]

Answer:

q = 8.57 10⁻⁵ mC

Explanation:

For this exercise let's use Newton's second law

F = ma

where force is magnetic force

F = q v x B

the bold are vectors, if we write the module of this expression we have

F = qv B sin θ

as the particle moves perpendicular to the field, the angle is θ= 90º

F = q vB

the acceleration of the particle is centripetal

a = v² / r

we substitute

qvB = m v² / r

qBr = m v

q = $\frac{m\ v}{B\ r}$

The exercise indicates the time it takes in the route that is carried out with constant speed, therefore we can use

v = d / t

the distance is ¼ of the circle,

d = $\frac{1}{4} \ 2\pi r$

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

we substitute

v = $\frac{\pi r}{2t}$

r = $\frac{2 \ t \ v}{\pi }$

let's calculate

r = $\frac{2 \ 2.2 \ 10^{-3} \ 88}{\pi }$ 2 2.2 10-3 88 /πpi

r = 123.25 m

let's substitute the values

q = $\frac{ 7.2 \ 10^{-8} \ 88}{ 0.6 \ 123.25}$ 7.2 10-8 88 / 0.6 123.25

q = 8.57 10⁻⁸ C

Let's reduce to mC

q = 8.57 10⁻⁸ C (10³ mC / 1C)

q = 8.57 10⁻⁵ mC

4 0

3 years ago

At what angle do the cars move after the collision?

liubo4ka [24]

Answer:

his results in the final angle after the collision of 37.2 degrees basically what we did there is turn the vector into a right triangle. We use sohcahtoa to solve for the angle. Being.

Explanation:

6 0

2 years ago

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

vlada-n [284]

Answer:

$\frac{1}{10}$ M

Explanation:

To apply the concept of angular momentum conservation, there should be no external torque before and after

As the asteroid is travelling directly towards the center of the Earth, after impact ,it does not impose any torque on earth's rotation, So angular momentum of earth is conserved

⇒ $I_{1} \times W_{1} =I_{2} \times W_{2}$

$I_{1}$ is the moment of interia of earth before impact
$W_{1}$ is the angular velocity of earth about an axis passing through the center of earth before impact
$I_{2}$ is moment of interia of earth and asteroid system
$W_{2}$ is the angular velocity of earth and asteroid system about the same axis

let $W_{1}=W$

since $\text{Time period of rotation}∝$ $\frac{1}{\text{Angular velocity}}$

⇒ if time period is to increase by 25%, which is $\frac{5}{4}$ times, the angular velocity decreases 25% which is $\frac{4}{5}$ times

therefore $W_{1}$ $=$ $\frac{4}{5} \times W_{1}$

$I_{1}=\frac{2}{5} \times M\times R^{2}$ (moment of inertia of solid sphere)

where M is mass of earth

R is radius of earth

$I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}$

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where $M_{1}$ is mass of asteroid