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navik [9.2K]
3 years ago
9

2. What is the smallest particle into which an

Physics
2 answers:
just olya [345]3 years ago
7 0

Explanation:

atommmmmmmmmmmmmmmmmmmmm

kiruha [24]3 years ago
3 0
Atom hope this helps!
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What is the length of the hypotenuse of a right triangle, whose legs have lengths of 12 meters and 35 meters?​
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c

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How does the resistance of the following change with rhe rise in temperature ? pure metals and German silver and carbon​
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resistance in the increasing order from least to maximum:

German silver, pure metals, carbon

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3 years ago
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An object of mass m is traveling in a circle with centripetal force f c. If the velocity of the object is v, what is the radius
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Answer:

r = mv^2/ Fc

Explanation:

Got it right

7 0
3 years ago
The position of a ball is given by x(t) (2.3 m/s) .t + (5.3m/s3) t3. What is the position 348 m at time t 4 s? You are correct Y
Georgia [21]

Answer:

(a) 348.4 m

(b) 256.7 m/s

(c)  127.2 m/s^2

Explanation:

x = 2.3 t + 5.3 t^{3}

(a) at t = 4 s

x = 2.3 x 4 + 5.3 x 4 x 4 x 4

x = 348.4 m

(b) The derivative of displacement function gives the value of instantaneous velocity.

So, v = dx / dt = 2.3 + 5.3 x 3 x t^2

v = 2.3 + 15.9 t^2

Put t = 4 s

So, v = 2.3 + 15.9 x 4 x 4

v = 256.7 m/s

(c) The derivative of velocity function with respect to time gives the value of instantaneous acceleration.

So, a = dv / dt = 5.3 x 3 x 2 x t

a = 31.8 t

Put t = 4 s

a = 31.8 x 4 = 127.2 m/s^2

4 0
4 years ago
A rubber ball with a mass of 0.145 kg is dropped from rest. From what height (in m) was the ball dropped, if the magnitude of th
Elena-2011 [213]

Answer:

1.55 m

Explanation:

Momentum: This can be defined as the product of  mass of a body and it velocity. the S.I unit of momentum is kgm/s.

Mathematically,

Momentum can be represented as,

M = mv................................. Equation 1

Where m = mass of the body, v = velocity of the body, M = momentum.

Making v the subject of the equation,

v = M/m........................................... Equation 2

Given: M = 0.80 kg.m/s, m = 0.145 kg.

Substituting into equation 2,

v = 0.8/0.145

v = 5.52 m/s.

Using the equation of motion,

v² = u² + 2gs ....................... Equation 3.

Where v = final velocity of the rubber ball, u = initial velocity of the rubber ball, s = distance, g = acceleration due to gravity.

Given: v = 5.52 m/s, u = 0 m/s, g = 9.81 m/s².

Substituting into equation 2

5.52² = 0² + 2(9.81)s

30.47 = 19.62s

s = 30.47/19.62

s = 1.55 m.

Thus the ball was dropped from a height of 1.55 m

8 0
4 years ago
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