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3 years ago

At what speed must the electron revolve round the nucleus of

Physics

1 answer:

EleoNora [17]3 years ago

6 0

Explanation:

I think this is it, give it a try

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A roller coaster has a "hump" and a "loop" for riders to enjoy (see picture). The top of the hump has a radius of curvature of 1

aivan3 [116]

Answer:

Part a)

$F_n = 306 N$

Part b)

$v = 12.1 m/s$

So this speed is independent of the mass of the rider

Explanation:

Part a)

By force equation on the rider at the position of the hump we can say

$mg - F_n = ma_c$

now we will have

$mg - F_n = \frac{mv^2}{R}$

$F_n = mg - \frac{mv^2}{R}$

now we have

$F_n = 100(9.81) - \frac{100(9^2)}{12}$

$F_n = 981 - 675$

$F_n = 306 N$

Part b)

At the top of the loop if the minimum speed is required so that it remains in contact so we will have

$F_n + mg = ma_c$

$F_n = 0$ at minimum speed

$mg = \frac{mv^2}{R}$

$v = \sqrt{Rg}$

$v = \sqrt{15 \times 9.81}$

$v = 12.1 m/s$

So this speed is independent of the mass of the rider

5 0

4 years ago

50 kg of water at 75o C is cooled to 25o C. How much heat was given off?

attashe74 [19]

Answer:

b the answer is b

Explanation:

b is the awnser because it cools after the heat on the water witch lets the steam out

8 0

2 years ago

A certain corner of a room is selected as the origin of a rectangular coordinate system. If a fly is crawling on an adjacent wal

Helga [31]

Answer:

2.59 m

Explanation:

Coordinates of origin = (0, 0)

Coordinates of Point p where the fly reach = (2.3 m, 1.2 m)

Use the distance formula of coordinates to find the distance between the origin and the point P.

$d=\sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}$

$d=\sqrt{\left ( 2.3- 0 \right )^{2}+\left ( 1.2-0 \right )^{2}}$

d = 2.59 m

Thus, the distance between the origin and the point P is 2.59 m.

5 0

3 years ago

The critical angle for a special type of glass in air is 30.8 ◦ . the index of refraction for water is 1.33. what is the critica

Alina [70]

When light moves from a medium with higher refractive index to a medium with lower refractive index, the critical angle is the angle above which there is no refracted light, and all the light is reflected. The value of this angle is given by
$\theta_c = \arcsin ( \frac{n_2}{n_1} )$
where n2 and n1 are the refractive indices of the second and first medium, respectively.

In the first part of the problem, light moves from glass to air ( $n_a=1.00$ ) and the critical angle is $\theta_c = 30.8^{\circ}$ . This means that we can find the refractive index of glass by re-arranging the previous formula:
$n_g=n_1 = \frac{n_2}{\sin \theta_c}= \frac{1.00}{\sin 30.8^{\circ}}=1.95$

Now the glass is put into water, whose refractive index is $n_w = 1.33$ . If light moves from glass to water, the new critical angle will be
$\theta_c = \arcsin ( \frac{n_2}{n_1} )=\arcsin( \frac{n_w}{n_g} )=\arcsin( \frac{1.33}{1.95} )=\arcsin(0.68)=43.0^{\circ}$

6 0

3 years ago

Dog starts at position X equals 2.50 m and undergoes displacement of 8.25 m what is its final position

BabaBlast [244]

Answer:

displacement is how far the dog is from the original position so use addition

Explanation:

2.50+8.25= 10.75

4 0

3 years ago