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mafiozo [28]
3 years ago
10

g Two identical metal spheres (small enough to be treated as particles) each with a mass of 3.1 × 10−3 kg are hung from separate

identical strings. The strings are 0.70 m long and are attached to the same nail in the ceiling. Surplus electrons are added to each sphere, and then the spheres are brought in contact with each other and released. Their equilibrium position is such that each string makes an angle of 15.0° with the vertical. How many surplus electrons are on each sphere?
Physics
1 answer:
gogolik [260]3 years ago
8 0

Answer:

Explanation:

Let q be the charges on each spheres , 2d be the   distance between them in equilibrium , T be tension in the string and F be the force of repulsion between them

F = k q² /4d²

For equilibrium

T sin15 = F

T cos 15 = mg

tan15 = F / mg

F = mg tan15

k q² /4d² = mg tan15

k q² = 4d² x mg tan15

= 4 x( .7 sin15)² x 3.1 x 10⁻³ x 9.8 x .2679

= 1.068 x 10⁻³

q = .3444 x 10⁻⁶ C

no of electrons

=  .3444 x 10⁻⁶  / 1.6 x 10⁻¹⁹

= 215.25 x 10¹⁰

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Vika [28.1K]
A=Fh
A - work
F - force
h - distance

F=mg
m - mass (god+basket)

so
A=mgh
187 = m*10*4
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6 0
3 years ago
In trampoline competitions, a good jump is one that lasts about 1.8 seconds. (A) How high can an athlete who stays in the air 1.
KonstantinChe [14]

Answer:

3.97305 m

Explanation:

a = Acceleration due to gravity = 9.81 m/s²

If a jump lasts for 1.8 seconds this means that from the moment when the person leaves the ground till the person touches the ground again it takes 1.8 seconds. So, maximum height reached will be at half the time of the jump i.e., 0.9 seconds.

u = Initial velocity = 0

Equation of motion

s=ut+\frac{1}{2}at^2\\\Rightarrow s=0t+\frac{1}{2}9.81\times 0.9^2\\\Rightarrow s=3.97305\ m

So, height of the jump is 3.97305 m.

4 0
3 years ago
The force that one massive object excerts to attract another object is called
Bogdan [553]
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5 0
3 years ago
A 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the cable for the following case
Murljashka [212]

Answer:

The answer is below

Explanation:

A 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the

cable for the following cases:

a. The load moves downward at a constant velocity

b. The load accelerates downward at a rate 0.4 m/s??

C. The load accelerates upward at a rate 0.4 m/s??

Solution:

Acceleration due to gravity (g) = 10 m/s²

a) Given that the mass of the crane (m) is 140 kg. If the load moves downward, the tension (T) is given by:

mg - T = ma

Since the load has a constant velocity, hence acceleration (a) = 0. Therefore:

mg - T = m(0)

mg - T = 0

T = mg

T = 140(10) = 1400 N

T = 1400 N

b)  If the load moves downward, the tension (T) is given by:

mg - T = ma

T = mg - ma = m(g - a)

T = 140(10 - 0.4) = 140(9.96) = 134.4

T = 134.4 N

c)  If the load moves upward, the tension (T) is given by:

T - mg = ma

T = ma + mg = m(a + g)

T = 140(0.4 + 10) = 140(10.4)

T = 145.6 N

2) To find the distance (s) if the load move from rest (u= 0) and accelerates for 20 seconds (t = 20). We use:

s = ut + (1/2)gt²

s = 0(20) + (1/2)(10)(20)²

s = 2000 m

7 0
3 years ago
Four forces of magnitude 10N,15N, 4N, 6N act on on object in the direction North West, East and South respectively, find their m
maria [59]

Magnitude 17 root

Hope it Helps! Please mark as Brainliest!

7 0
3 years ago
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