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3 years ago

A 15 kg mass is lifted to a height of 2m. What is gravitational potential energy at this position

Physics

2 answers:

7nadin3 [17]3 years ago

7 0

Answer: 294J

Explanation:

$U_g=mg(y_f-y_i)$

$U_g=(15)(9.8)(2-0)=294J$

JulijaS [17]3 years ago

7 0

Answer:

294.3 J

Explanation:

U = mgh

U = energy

m = mass

g = gravity

h = height

U = 15kg * 9.81 m/s^2 * 2m

294.3 J (joules) [kg *m^2/s^2]

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How do I calculate the tension in the horizontal string?

matrenka [14]

ANSWER

T₂ = 10.19N

EXPLANATION

Given:

• The mass of the ball, m = 1.8kg

First, we draw the forces acting on the ball, adding the vertical and horizontal components of each one,

In this position, the ball is at rest, so, by Newton's second law of motion, for each direction we have,

$\begin{gathered} T_{1y}-F_g=0_{}_{}_{} \\ T_2-T_{1x}=0 \end{gathered}$

The components of the tension of the first string can be found considering that they form a right triangle, where the vector of the tension is the hypotenuse,

$\begin{gathered} T_{1y}=T_1\cdot\cos 30\degree \\ T_{1x}=T_1\cdot\sin 30\degree \end{gathered}$

We have to find the tension in the horizontal string, T₂, but first, we have to find the tension 1 using the first equation,

$T_1\cos 30\degree-m\cdot g=0$

Solve for T₁,

$T_1=\frac{m\cdot g}{\cos30\degree}=\frac{1.8kg\cdot9.8m/s^2}{\cos 30\degree}\approx20.37N$

Now, we use the second equation to find the tension in the horizontal string,

$T_2-T_1\sin 30\degree=0$

Solve for T₂,

$T_2=T_1\sin 30\degree=20.37N\cdot\sin 30\degree\approx10.19N$

Hence, the tension in the horizontal string is 10.19N, rounded to the nearest hundredth.

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1 year ago

Nine steps for developing scientific principles are

xxTIMURxx [149]

Answer:

1)Observe a phenomenon

2)Ask a question/ start inferring

3)Form a hypothesis

4)Create an experiment

5)Collect data

6)Compare results

7)Analyze

8)Report findings

9)Compare with other experiments

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3 years ago

(15 points) (Asap!!) In what two ways can you increase the elastic potential energy of a spring?

adelina 88 [10]

Hello There

Answers: The elastic potential energy can be increased by: 

1) Getting a spring with a higher spring constant

2) Increasing the length at which the spring is compressed.

Reasons: Getting a stronger spring makes it stronger which equals more energy. While increasing the compression on the spring, increases the stored energy which makes it more powerful when its released

I hope this helps
-Chris

5 0

3 years ago

Read 2 more answers

5p - 14 = 8 p + 4 find p

Digiron [165]

Answer:

Explanation:

5p - 14 = 8p + 4

5p = 8p + 18 <-- Moving constants to one side; add the same number of +14 to both sides.

-3p = 18. <-- The same thing with the variable itself.

p = -6 <-- Divide both sides by negative 3.

6 0