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Bond [772]
4 years ago
11

Which is NOT true related to the following equation?

Physics
1 answer:
Harrizon [31]4 years ago
4 0

Answer:

B will Be Your Best Choice Have a nice day

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A 2 kilogram bowling ball is 2.5 meters off the ground on a post when it falls just before it reaches the ground it is traveling
s2008m [1.1K]
I need the answer too
7 0
3 years ago
Two small metal spheres are 25 cm apaft.The spheres have equal amount of negative charge and repel each other with a force of 0.
Mars2501 [29]

Answer:

0.5\times 10^{-6}C

Explanation:

According to coulombs law force between two charges is given by F=\frac{1}{4\pi \epsilon _0}\frac{Q_1Q_2}{R^2} here R is the distance between both the charges which is given as 25 cm

We have given force F =0.036 N

So  0.036=\frac{1}{4\pi \times 8.85\times 10^{-12}}\frac{Q^2}{(0.25^2)} As \epsilon _0 is constant which value is 8.85\times 10^{-12}

Q^2=0.250\times 10^{-12}

Q=0.5\times 10^{-6}C

8 0
4 years ago
Arcsin 0.9331 in degrees​
lisabon 2012 [21]

Answer:

68.9233231661

Explanation:

Just put it into your calculator, shift sin should do it but it will come up like this: sin^{-1} which is the same as arcsin

6 0
3 years ago
In a laboratory experiment, a diffraction grating produces an interference pattern on a screen. If the number of slits in (inclu
Darya [45]

Answer:

Explanation:

When the number of slits increases, the intensity of fringes increases.

So, the fringes appear to be more bright.

As we know that the fringe width is inversely proportional to the number of slits, so as the number of slits increases, the fringe width decreases, hence the fringes are narrower, bright and close together.

7 0
3 years ago
En la Tierra un volcán puede expulsar rocas verticalmente hasta una altura máxima H. A) ¿A qué altura (en términos de H) llegarí
Nonamiya [84]

A) 2.64 H

The maximum height that the expelled rock can reach can be found by using the equation:

v^2-u^2 = 2gd

where

v = 0 is the velocity at the maximum height

u is the initial velocity

g is the acceleration of gravity

d is the maximum height

Solving for d,

d=\frac{-u^2}{2g}

We see that the maximum heigth is inversely proportional to g. On the Earth,

d=H and g=g_e = -9.81 m/s^2

So we can write:

\frac{H}{H'}=\frac{g_m}{g_e}

where H' is the maximum height reached on Mars, and g_m = -3.71 m/s^2 is the acceleration of gravity on Mars. Solving for H',

H' = \frac{g_e}{g_m}H = \frac{-9.81}{3.71}H=2.64 H

B) 2.64T

The time after which the rock reaches the maximum height can be found by using

v=u+gt

where

v = 0 is the velocity at the maximum height

u is the initial velocity

Solving for t,

t=\frac{v-u}{g}

The total time of the motion is twice this value, so:

t=2\frac{v-u}{g}

So we see that it is inversely proportional to g.

On the Earth, t = T. So we can write:

\frac{T}{T'}=\frac{g_m}{g_E}

where T' is the total time of the motion on Mars. Solving for T',

T' = \frac{g_e}{g_m}T=\frac{-9.81}{-3.71}T=2.64T

4 0
3 years ago
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