37 microgram to Gigagram

A small sphere with a mass of 441 g is moving upward along the vertical +y-axis when it encounters an electric field of 5.00 N/C

sweet-ann [11.9K]

In your question where the ask is to calculate the charge that the small sphere carries which is the mass of it is 441g moving at an acceleration of 13m/s^2 nad having and electric field of 5N/C. So the formula in getting the charge is mutliply the mass and the quotients of Acceleration and the Electric Field so the answer is 1,146.6

3 0

3 years ago

Under certain circumstances, potassium ions (K+) in a cell will move across the cell membrane from the inside to the outside. Th

choli [55]

Answer:

$1.368\times 10^{-20}\ J$

Explanation:

q = Charge in the potassium ion = $19e-18e$

e = Charge of electron = $1.6\times 10^{-19}\ C$

$V_2-V_1$ = Change in potential = $0-(-85.5\times 10^{-3})$

Change in electric potential is given by

$E=q(V_2-V_1)\\\Rightarrow E=(19e-18e)(0-(-85.5\times 10^{-3})\\\Rightarrow E=1.6\times 10^{-19}\times 85.5\times 10^{-3}\\\Rightarrow E=1.368\times 10^{-20}\ J$

The energy is $1.368\times 10^{-20}\ J$

3 0

3 years ago

An ambulance is traveling east at 61.9 m/s. Behind it there is a car traveling along the same direction at 28.5 m/s. The ambulan

inessss [21]

Answer: 0.4 m

Explanation:

Given

Speed of ambulance, vs = 61.9 m/s

Speed of car = 28.5 m/s

Frequency of ambulance siren, f = 694 Hz

Velocity of sound in air, v = 343 m/s

With speed of ambulance being (61.9 m/s) -> We solve using

fd = f(v + vr) / (v - vs), where vr = 0

fd = 694 * (343 + 0) / (343 - 61.9)

fd = 694 * (343 / 281.1)

fd = 694 * 1.22

fd = 847 Hz

Recall,

λ = v/f

λ = 343/847

λ = 0.4 m

Therefore, the wavelength of the sound of the ambulance’s siren if you are standing at the position of the car is 0.4 m

7 0

3 years ago

For a certain optical medium the speed of light varies from a low value of 1.90 × 10 8 m/s for violet light to a high value of 2

Dmitry_Shevchenko [17]

Answer:

a. The refractive index ranges from 1.5 - 1.56

b. 18.7° for violet light and 19.5° for red light.

c. 33.7° for violet light and 35.3° for red light.

Explanation:

a. The refractive index of an object is the ratio of the speed of light in a vacuum and the speed of light in the object.

Mathematically,

$n = \frac{c}{v}$

The speed of violet light in the object is $1.9 * 10^8 m/s$ .

The speed of red light in the object is $2 * 10^8 m/s$

Hence, the refractive index for violet light is:

$n = \frac{3 * 10^8 }{1.9 * 10^8} \\\\n = 1.56$

and for red light, it is:

$n = \frac{3 * 10^8 }{2 * 10^8} \\\\n = 1.5$

Hence, the refractive index ranges from 1.5 - 1.56.

b. The refractive index is also the ratio of the sine of the angle of incidence to the sine of the angle of refraction.

$n = \frac{sin(i)}{sin(r)}$

The angle of incidence is 30°.

The angle of refraction for violet light will be:

$1.56 = \frac{sin(30)}{sin(r)}\\ \\sin(r) = \frac{sin(30)}{1.56} = \frac{0.5}{1.56} \\\\sin(r) = 0.3205\\\\r = 18.7^o$

And the angle of refraction for red light will be:

$1.5 = \frac{sin(30)}{sin(r)}\\ \\sin(r) = \frac{sin(30)}{1.5} = \frac{0.5}{1.5} \\\\sin(r) = 0.3333\\\\r = 19.5^o$

The angle of refraction for red light is larger than that of violet light when the angle of incidence is 30°.

c. The angle of incidence is 60°.

The angle of refraction for violet light will be:

$1.56 = \frac{sin(60)}{sin(r)}\\ \\sin(r) = \frac{sin(60)}{1.56} = \frac{0.8660}{1.56} \\\\sin(r) = 0.5551\\\\r = 33.7^o$

And the angle of refraction for red light will be:

$1.5 = \frac{sin(60)}{sin(r)}\\ \\sin(r) = \frac{sin(60)}{1.5} = \frac{0.8660}{1.5} \\\\sin(r) = 0.5773\\\\r = 35.3^o$

The angle of refraction for red light is still larger than that of violet light when the angle of incidence is 60°.

6 0

3 years ago

A train has an acceleration of magnitude 0.90 m/s2 while stopping. A pendulum with a 0.55-kg bob is attached to a ceiling of one

anygoal [31]

The angle of the pendulum with the vertical is $5.2^{\circ}$

Explanation:

As the train decelerates, the bob of the pendulum will feel a force given by

$F=ma$

where

m = 0.55 kg is the mass of the bob

$a=0.9 m/s^2$ is the magnitude of the acceleration

In the horizontal direction.

The pendulum will be inclined at an angle $\theta$ from the vertical, so it will be in equilibrium, and therefore the horizontal component of the tension in the string must be equal to the net force F of the previous equation:

$T sin \theta = ma$ (1)

where T is the tension in the string.

We also know that the bob is in equilibrium along the vertical direction: so the vertical component of the tension must be equal to the weight of the bob,

$T cos \theta = mg$ (2)