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amm1812
3 years ago
6

I need help asap please

Physics
1 answer:
Bad White [126]3 years ago
6 0

Answer:

225√3 N

Explanation:

225√3 N

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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.

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Read 2 more answers
Plzz answer this question ​
Vera_Pavlovna [14]
Mass does not change anywhere in the space .
So the given 30N is equal to the product of mass of the body and the acceleration due to gravity at that place in space.
After finding the mass of the body multiply it by the given acceleration due to gravity.
Mg is the approximate force due to gravity between the mass and the planet.

I hope this may help you.
6 0
4 years ago
A ball weighing 1 lb is attached to a string 2 feet long and is whirled in a vertical circle at a constant speed of 10 ft/sec.
fredd [130]

Explanation:

It is given that,

Mass of the ball, m = 1 lb

Length of the string, l = r = 2 ft

Speed of motion, v = 10 ft/s

(a) The net tension in the string when the ball is at the top of the circle is given by :

F=\dfrac{mv^2}{r}-mg

F=m(\dfrac{v^2}{r}-g)

F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}-1\ lb\times 32\ ft/s^2)

F = 18 N

(b) The net tension in the string when the ball is at the bottom of the circle is given by :

F=\dfrac{mv^2}{r}+mg

F=m(\dfrac{v^2}{r}+g)

F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}+1\ lb\times 32\ ft/s^2)

F = 82 N

(c) Let h is the height where the ball at certain time from the top. So,

T=mg(\dfrac{r-h}{r})+\dfrac{mv^2}{r}

T=\dfrac{m}{r}(g(r-h)+v^2)

Since, v^2=u^2-2gh

T=\dfrac{m}{r}(u^2-3gh+gr)

Hence, this is the required solution.

6 0
3 years ago
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