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nata0808 [166]
2 years ago
11

An object with a mass of 2.0 kg accelerates at 2.0 m/s^2 when an unknown force is applied to it. What is the amount of the force

?
GIVEN:

REQUIRED:

EQUATION:

SOLUTION:

ANSWER:
Physics
1 answer:
lord [1]2 years ago
6 0

ANSWER: 4.0 N

EXPLANATION

The formula for Newton's second law is

F = ma

where:

  • F = force
  • m = mass
  • a = acceleration

<em>Given: </em>

mass (m) = 2.0 kg

acceleration (a) = 2.0 m/s^2

<em>Required:</em>

F = ?

<em>Equation: </em>

F = ma

<em>Solution: </em>

F = 2.0 x 2.0

<em>Answer:</em>

F = 4.0 N

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1. Which of the following is not an acceptable Physical Education activity?
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You have 2 objects flying towards you. One of them is havier with lower velocity and the other one is lighter but has higher vel
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3 years ago
A wheel rotates about a fixed axis with a constant angular acceleration of 3.3 rad/s2. The diameter of the wheel is 21 cm. What
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Answer:

The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

Explanation:

We are given that

Angular acceleration, \alpha=3.3 rad/s^2

Diameter of the wheel, d=21 cm

Radius of wheel, r=\frac{d}{2}=\frac{21}{2} cm

Radius of wheel, r=\frac{21\times 10^{-2}}{2} m

1m=100 cm

Magnitude of total linear acceleration, a=1.7 m/s^2

We have to find the linear speed  of a  at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.

Tangential acceleration,a_t=\alpha r

a_t=3.3\times \frac{21\times 10^{-2}}{2}

a_t=34.65\times 10^{-2}m/s^2

Radial acceleration,a_r=\frac{v^2}{r}

We know that

a=\sqrt{a^2_t+a^2_r}

Using the formula

1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}

Squaring on both sides

we get

2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}

\frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}

v^4=r^2\times 2.7699

v^4=(10.5\times 10^{-2})^2\times 2.7699

v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}

v=0.418 m/s

Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

6 0
3 years ago
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