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mamaluj [8]
2 years ago
13

An 16.00-kg object is suspended by a rope at some height, h if the object has a potential energy of 900.01j what is the objects

height ?
Physics
2 answers:
Novosadov [1.4K]2 years ago
6 0

Answer:

\displaystyle 5,7398596939\:m

Step-by-step explanation:

\displaystyle mgh = PE \\ \\ [16][9,8]h = 900,01 → \frac{900,01}{156,8} = \frac{156,8h}{156,8} \\ \\ 5,7398596939 ≈ h

I am joyous to assist you anytime.

qwelly [4]2 years ago
5 0

Answer:

5.740 m

Explanation:

PE = mgh

900.0 J = (16.00 kg) (9.8 m/s²) h

h = 5.740 m

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3 years ago
16x^2y^2-25a^2b^2<br>factorize the expression​
SIZIF [17.4K]

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(4xy+5ab)(4xy-5ab)

Explanation:

16x^{2}y^{2}-25a^{2}b^{2}

4^2 is 16 and 5^2 is 25,

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(4xy+5ab)(4xy-5ab)

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2 years ago
In a chemical equation, the chemicals that react are considered . In a chemical equation, the chemicals that are produced are co
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In a chemical equation, chemicals that react are the reactants, while chemicals that are produced are the products/by products. Both sides of the equation must be balanced.

Explanation

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4 0
3 years ago
Read 2 more answers
A square plate of side 9 m is submerged in water at an incline of 60∘ with the horizontal. Its top edge is located at the surfac
Tpy6a [65]

Answer:

The force on one side of  the plate is 3093529.3 N.

Explanation:

Given that,

Side of square plate = 9 m

Angle = 60°

Water weight density = 9800 N/m³

Length of small strip is

y=\dfrac{\Delta y}{\sin60}

y=\dfrac{2\Delta y}{\sqrt{3}}

The area of strip is

dA=\dfrac{9\times2\Delta y}{\sqrt{3}}

We need to calculate the force on  one side of  the plate

Using formula of pressure

P=\dfrac{dF}{dA}

dF=P\times dA

On integrating

\int{dF}=\int_{0}^{9\sin60}{\rho g\times y\times 6\sqrt{3}dy}

F=9800\times6\sqrt{3}(\dfrac{y^2}{2})_{0}^{9\sin60}

F=9800\times6\sqrt{3}\times(\dfrac{(9\sin60)^2}{2})

F=3093529.3\ N

Hence, The force on one side of  the plate is 3093529.3 N.

7 0
3 years ago
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