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CaHeK987 [17]
3 years ago
12

0.45 kg soccer ball changes its velocity by 20.0 m/s due to a force applied to it in 0.10 seconds. What force was necessary for

this change in velocity?
Physics
1 answer:
Paladinen [302]3 years ago
8 0

Assuming the accleration applied was constant, we have

v=v_0+at\implies v_0+20.0\,\dfrac{\mathrm m}{\mathrm s}=v_0+a(0.10\,\mathrm s)

\implies20.0\,\dfrac{\mathrm m}{\mathrm s}=a(0.10\,\mathrm s)

\implies a=200\,\dfrac{\mathrm m}{\mathrm s^2}

Then the force applied to the ball is given by

F=ma=(0.45\,\mathrm{kg})\left(200\,\dfrac{\mathrm m}{\mathrm s^2}\right)

\implies F=90\,\dfrac{\mathrm{kg}\,\mathrm m}{\mathrm s^2}=90.\,\mathrm N

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Codominence i think, its both parents genes are used

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3 years ago
The Earth revolves around the Sun once a year at an average distance of 1.50×1011m. Find the orbital radius that corresponds to
DedPeter [7]

Answer:

9.4\cdot 10^{10} m

Explanation:

We can solve the problem by using Kepler's third law, which states that the ratio between the cube of the orbital radius and the square of the orbital period is constant for every object orbiting the Sun. So we can write

\frac{r_a^3}{T_a^2}=\frac{r_e^3}{T_e^2}

where

r_o is the distance of the new object from the sun (orbital radius)

T_o=180 d is the orbital period of the object

r_e = 1.50\cdot 10^{11} m is the orbital radius of the Earth

T_e=365 d is the orbital period the Earth

Solving the equation for r_o, we find

r_o = \sqrt[3]{\frac{r_e^3}{T_e^2}T_o^2} =\sqrt[3]{\frac{(1.50\cdot 10^{11}m)^3}{(365 d)^2}(180 d)^2}=9.4\cdot 10^{10} m

3 0
3 years ago
A wheel starts from rest and has an angular acceleration that is given by α (t) = (6.0 rad/s4)t2. After it has turned through 10
marissa [1.9K]

Answer:

75 rad/s

Explanation:

The angular acceleration is the time rate of change of angular velocity. It is given by the formula:

α(t) = d/dt[ω(t)]

Hence: ω(t) = ∫a(t) dt

Also, angular velocity is the time rate of change of displacement. It is given by:

ω(t) = d/dt[θ(t)]

θ(t) = ∫w(t) dt

θ(t) = ∫∫α(t) dtdt

Given that: α (t) = (6.0 rad/s4)t² = 6t² rad/s⁴. Hence:

θ(t) = ∫∫α(t) dtdt

θ(t) = ∫∫6t² dtdt =∫[∫6t² dt]dt

θ(t) = ∫[2t³]dt = t⁴/2 rad

θ(t) = t⁴/2 rad

At θ(t) = 10 rev = (10 *  2π) rad = 20π rad, we can find t:

20π = t⁴/2

40π = t⁴

t = ⁴√40π

t = 3.348 s

ω(t) = ∫α(t) dt = ∫6t² dt = 2t³

ω(t) = 2t³

ω(3.348) = 2(3.348)³ = 75 rad/s

7 0
3 years ago
Is the answer C or B​
inessss [21]

Answer:

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6 0
3 years ago
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Two astronauts of mass 100 kg are 2 m apart in outer space. What is the
fredd [130]

The force of gravity between the astronauts is 1.67\cdot 10^{-7}N

Explanation:

The magnitude of the gravitational force between two objects is given by:

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In this problem, we have two astronauts, whose masses are:

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Substituting into the equation, we can find the gravitational force between the two astronauts:

F=\frac{(6.67\cdot 10^{-11})(100)(100)}{2^2}=1.67\cdot 10^{-7}N

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

4 0
3 years ago
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