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3 years ago

You push a table 8 meters for 16 minuutes and do 6720 j of work how much power do ypu use

Physics

1 answer:

Yuri [45]3 years ago

3 0

P=W/t

P=Power
W=Work
t=Time

Convert 16 minutes in seconds:
16 mins = 960 secs

P=6720/960=7.23 W [Watt]

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A pendulum is formed by taking a 2.0 kg mass and hanging it from the ceiling using a steel wire with a diameter of 1.1 mm. it is

Lera25 [3.4K]

Answer: 1.39 s

Explanation:

We can solve this problem with the following equations:

$\frac{\Delta l}{l_{o}}=\frac{F}{AY}$ (1)

$T=2 \pi \sqrt{\frac{l_{o}}{g}}$ (2)

Where:

$\Delta l=0.05 mm=5(10)^{-5} m$ is the length the steel wire streches (taking into account 1mm=0.001 m)

$l_{o}$ is the length of the steel wire before being streched

$F=mg=(2 kg)(9.8 m/s^{2})=19.6 N$ is the force due gravity (the weight) acting on the pendulum with mass $m=2 kg$

$A$ is the transversal area of the wire

$Y=2(10)^{11} Pa$ is the Young modulus for steel

$T$ is the period of the pendulum

$g=9.8 m/s^{2}$ is the acceleration due gravity

Knowing this, let's begin by finding $A$ :

$A=\pi r^{2}=\pi (\frac{d}{2})^{2}=\pi \frac{d^{2}}{4}$ (3)

Where $d=1.1 mm=0.0011 m$ is the diameter of the wire

$A=\pi \frac{(0.0011 m)^{2}}{4}$ (4)

$A=9.5(10)^{-7}m^{2}$ (5)

Knowing this area we can isolate $l_{o}$ from (1):

$l_{o}=\frac{\Delta l AY}{F}$ (6)

And substitute $l_{o}$ in (2):

$T=2 \pi \sqrt{\frac{\frac{\Delta l AY}{F}}{g}}$ (7)

$T=2 \pi \sqrt{\frac{\frac{(5(10)^{-5} m)(9.5(10)^{-7}m^{2})(2(10)^{11} Pa)}{2(10)^{11} Pa}}{9.8 m/s^{2}}}$ (8)

Finally:

$T=1.39 s$

3 0

3 years ago

A block of wood is floating in water; it is depressed slightly and then released to oscillate up and down. Assume that the top a

Marysya12 [62]

Explanation:

Equilibrium position in y direction:

W = Fb (Weight of the block is equal to buoyant force)

m*g = V*p*g

V under water = A*h

hence,

m = A*h*p

Using Newton 2nd Law

$-m*\frac{d^2y}{dt^2} = Fb - W\\\\-m*\frac{d^2y}{dt^2} = p*g*(h+y)*A - A*h*p*g\\\\-A*h*p*\frac{d^2y}{dt^2} = y *p*A*g\\\\\frac{d^2y}{dt^2} + \frac{g}{h} * y =0$

Hence, T time period

T = 2*pi*sqrt ( h / g )

4 0

3 years ago

The ability to adjust to an artificially displaced or even inverted visual field is called

egoroff_w [7]

Answer:

méthode if séparation muddy water

3 0

2 years ago

The gravitational force between two objects that

leonid [27]

Answer:

The answer to your question is m₂ = 38.5 kg

Explanation:

Data

distance = d = 2.1 x 10⁻¹ m

Force = 3.2 x 10⁻⁶ N

m₁ = 55 kg

m₂ = ?

G = 6.67 x 10 ⁻¹¹ Nm²/kg²

Process

1.- To solve this problem use Newton's law of Universal Gravitation.

F = G m₁m₂ / r²

-Solve for m₂

m₂ = Fr² / Gm₁

2.- Substitution

m₂ = (3.2 x 10⁻⁶)(2.1 x 10⁻¹)² / (6.67 x 10⁻¹¹)(55)

3.- Simplification

m₂ = 1.411 x 10⁻⁷ / 3.669 x 10⁻⁹

4.- Result

m₂ = 38.5 kg

5 0

3 years ago

a worker uses a board that is 7 m long to pry up a bolder A small rock is used for the fulcrum and is placed 2.5 m from the resi

e-lub [12.9K]

Answer:

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2 years ago