How many joules of energy are required to accelerate a 1.0-kg mass from rest to a speed of 86.6% the speed of light?

Consider two parallel plate capacitors. The plates on Capacitor B have half the area as the plates on Capacitor A, and the plate

vichka [17]

Answer:

CB = 4.45 x 10⁻⁹ F = 4.45 nF

Explanation:

The capacitance of a parallel plate capacitor is given by the following formula:

C = ε₀A/d

where,

C = Capacitance

ε₀ = Permeability of free space

A = Area of plates

d = Distance between plates

FOR CAPACITOR A:

C = CA = 17.8 nF = 17.8 x 10⁻⁹ F

A = A₁

d = d₁

Therefore,

CA = ε₀A₁/d₁ = 17.8 x 10⁻⁹ F ----------------- equation 1

FOR CAPACITOR B:

C = CB = ?

A = A₁/2

d = 2 d₁

Therefore,

CB = ε₀(A₁/2)/2d₁

CB = (1/4)(ε₀A₁/d₁)

using equation 1:

CB = (1/4)(17.8 X 10⁻⁹ F)

5 0

3 years ago

A stationary sub uses sonar to send a 1.18x10^3 hertz sound wave through ocean water. The reflected sound wave from the flat oce

vampirchik [111]

Answer:

a) v = 1524.7 m/s

b) T = 8.47*10^-4 s

λ = 1.29 m

Explanation:

a) First, in order to calculate the speed of the sound wave, you take into account that the velocity is constant, then, you use the following formula:

$v=\frac{d}{t}$

d: distance traveled by the sound wave, which is twice the distance to the ocean bottom = 2*324 m = 648 m

t: time that sound wave takes to return to the sub = 0.425

$v=\frac{648m}{0.425s}=1524.7\frac{m}{s}$

hence, the speed of the sound wave is 1524.7 m/s

b) Next, with the value of the velocity of the wave you can calculate the wavelength of the wave, by using the following formula:

$v=\lambda f\\\\\lambda=\frac{v}{f}$

f: frequency = 1.18*10^3 Hz

$\lambda=\frac{1524.7m/s}{1.18*10^3s^{-1}}=1.29m$

And the period is:

$T=\frac{1}{f}=\frac{1}{1.18*10^3s^{-1}}=8.47*10^{-4}s$

hence, the wavelength and period of the sound wave is, respectively, 1.29m and 8.47*10^-4 s

5 0

3 years ago

A box of mass 20kg is pulled up an inclined plane by a force of 285N. Given that the value of the incline angle is 30 degrees an

docker41 [41]

Given :

Mass of box , m = 250 kg.

Force applied , F = 285 N.

The value of the incline angle is 30°.

the coefficient of dynamic friction is $\mu=0.72$ .

To Find :

The speed with which the box is moving with, assuming it takes 4 seconds to reach the top of the incline.

Solution :

Net force applied in box is :

$F=285 - mgsin\ \theta - \mu mg cos \ \theta\\ \\F=285-mg( sin \ \theta - \mu cos\ \theta)\\\\F=285 - 20\times 10( \dfrac{1}{2}+0.72\times \dfrac{\sqrt{3}}{2})\\\\F=60.29\ N$

Acceleration , $a=\dfrac{F}{m}=\dfrac{60.29}{20}=3.01\ m/s^2$ .

By equation of motion :

$v=u+at\\\\v=0+3.01\times 4\\\\v=12.04\ m/s$

Therefore, the speed of box is 12.04 m/s.

Hence, this is the required solution.

6 0

3 years ago

Someone help please.....

Westkost [7]

Answer:

0.0928km/min (4dp)

Explanation:

To find the jogger's speed in km per minute, we just need to divide the number of km jogged by the time in minutes it took to jog that distance. This will give us the distance they jogged every minute which is their speed.

4km in 32 minutes:

4/32 = 0.125km/min

2km in 22 minutes:

2/22 = 0.091 (3dp)km/min

1km in 16 minutes:

0.0625km/min

Now to find the average speed of these 3 speeds, we just add them all together and divide by how many values there are (3 values).

Average (mean) = $\frac{0.125+0.091+0.0625}{3}$

Average = 0.2785/3

Average speed of jogger = 0.0928 (4dp) km/min

Hope this helped!

8 0

3 years ago

Why do blades come in different lengths in a jig saw sander?

trasher [3.6K]

Answer:

To determine the minimum blade length, add 1" to the workpiece thickness. One type of material, and some materials can be cut by more than one type of blade. No matter the material, there's likely a jigsaw blade designed specifically for. Armed with the right blade, follow these pointers to make your work go (and cut) .

Explanation:

6 0

3 years ago