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DaniilM [7]
3 years ago
11

A 3-kg rock is thrown upward with a force of 200 N at a location where the local gravitational acceleration is 9.79 m/s2 . Deter

mine the acceleration of the rock, in m/s
Physics
1 answer:
expeople1 [14]3 years ago
3 0

Answer: 56.87m/s^{2}

Explanation:

If we make an analysis of the net force F_{net} of the rock that was thrown upwards, we will have the following:

F_{net}=F_{up}-W  (1)

Where:

F_{up}=200N is the force with which the rock was thrown

W is the weight of the rock

Being the weight the relation between the mass m=3kg of the rock and the acceleration due gravity g=9.79m/s^{2} :

W=m.g=(3kg)(9.79m/s^{2}) (2)

W=29.37 N (3)

Substituting (3) in (1):

F_{net}=200N-29.37 N  (4)

F_{net}=170.63 N  (5) This is the net Force on the rock

On the other hand, we know this force is equal to the multiplication of the mass with the acceleration, according to Newton's 2nd Law:

F_{net}=m.a  (6)

Finding the acceleration a:

a=\frac{F_{net}}{m}  (7)

a=\frac{170.63 N}{3kg} (8)

Finally:

a=56.87m/s^{2}

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A skateboarder wants to cross a large playground and notices that there are large shapes painted on its asphalt surface. One sha
Elden [556K]
<h2>Diagonal of circle </h2>

Explanation:

As the skateboarder wants to cross the play ground . The surface is rough .

As we know , the force of friction is non-conservative force . Thus work is required against this force .

We have formula:

work done = Force x distance (in one direction )

Te force applied cannot be changed , so he is to decrease the distance .

In case of circle , diameter is the minimum distance . Thus he is supposed to move along it .

6 0
3 years ago
GIVING BRAINLIEST PLEASE HELP!!
Marrrta [24]
The answer is D hope it helps
8 0
3 years ago
A cable is 100-m long and has a cross-sectional area of 1.0 mm2. A 1000-N force is applied to stretch the cable. Young's modulus
Blizzard [7]

Answer:

1 m

Explanation:

L = 100 m

A = 1 mm^2 = 1 x 10^-6 m^2

Y = 1 x 10^11 N/m^2

F = 1000 N

Let the cable stretch be ΔL.

By the formula of Young's modulus

Y=\frac{F\times L}{A\times\Delta L}

\Delta L=\frac{F\times L}{A\times\Y}

\Delta L=\frac{1000\times 100}{10^{-6}\times10^{11}}

ΔL = 1 m

Thus, the cable stretches by 1 m.

4 0
3 years ago
At t=0, a block A of mass 8 kg and block B of mass 16 kg are both at position x=0 . Block A is at rest, and block B is moving at
love history [14]

The center of mass of the two objects is the average position of the parts of the two object system

The center of mass of block <em>A</em>, and block <em>B</em>  after displacement of block <em>B</em> is at <u>20 m from block </u><u><em>A</em></u>

<em />

Reason:

The given parameters are;

The position of block A and block B at t = 0 is x = 0

The mass of block A, m₁ = 8 kg

Mass of block B, m₂ = 16 kg

Speed of block <em>A</em> = 0 m/s

Speed of block <em>B</em>, v₂ = 10 m/s

Location of the center of mass of the two object at t = 3 s; Required

Solution;

The location of block <em>A</em>, after 3 s is x₁ = 0 (block A is at rest)

The location of block <em>B</em>, = v₂ × t

The location of block <em>B</em>, after 3 s is x₂ = 10 m/s × 3 s = 30 m

The center of mass of two masses are given as follows;

x_{cm} = \dfrac{m_1 \cdot x_1 +m_2\cdot x_2}{m_1 + m_2}

x_{cm} = \dfrac{8  \times0 + 16 \times  30}{8 + 16} = 20

The center of mass of the two objects is at at the position x = <u>20 m</u> (from block <em>A</em>)

Learn more about the center of mass here:

brainly.com/question/18557256

brainly.com/question/20714030

brainly.com/question/17088562

4 0
2 years ago
The fulcrum of a first-class lever divides its 9.0 m arm into two sections—a 6.0 m arm and a 3.0 m arm. You place a rock weighin
nexus9112 [7]
For balancing the lever, force on both the sides shall be equal. so,
Force on 3 m end = m × a = 3 × 98.1 = 294.3

Now, on 6 m end, it would be: = 294.3/6 = 49.05
After rounding-off to the nearest hundredth value, it would be: 49 N

Finally, Option A would be your correct answer.

Hope this helps!
6 0
2 years ago
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