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2 years ago

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Den pushes a desk 400 cm across the floor. He exerts a force of 10 N for 8 s to move the desk. What is his power output? (Power:

P = W/t) 1.25 W 5 W 40 W 500 W

1 answer:

White raven [17]2 years ago

7 0

Answer:

5 W

Explanation:

The formula of the power is:

● P = W/t

W is the work and t is the time needed to do it(in seconds)

Let's calculate first the work that the force exerced:

W = Vector F . Vector d

D is the distance ( here 400 cm wich is 4 m)

Make a representation to see how are the vectors F and V.(picture below)

The vector F and d are colinear since Den is pushing the desk on the ground.

● W = 4 × 10 = 40 J

J is Joule

■■■■■■■■■■■■■■■■■■■■■■■■■■

● P = W / t

● P = 40/ 8

● P = 5 W

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$\frac{Pressure_{1} *Volume_{1}}{Temperature_{1}} = \frac{Pressure_{2} *Volume_{2}}{Temperature_{2}}$ with 1 and 2 both being anything with pressure, volume, and temperature. Since pressure is on the top and temperature is on the bottom, they are inversely related, meaning that when the temperature gets high the pressure goes low. In addition, since pressure and volume are both on the top, when volume goes down pressure goes down too. Therefore, since A and C are right, D is the correct answer

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The velocity of the mass at time, t = 1 s, is determined as 20 m/s.

<h3>Magnitude of the force experienced by the body</h3>

The magnitude of the force experienced by the body is calculated as follows;

|F| = √(8² + 6²)

|F| = 10 N

<h3>Acceleration of the mass</h3>

From Newton's second law of motion;

F = ma

where;

m is mass
a is acceleration

a = F/m

a = 10 /0.5

a = 20 m/s²

<h3>Velocity of the mass after 1 second</h3>

v = at

v = 20 x 1

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1 year ago

Explain what matter is

djverab [1.8K]

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2 years ago

Read 2 more answers

A particle with mass 1.09 kg oscillates horizontally at the end of a horizontal spring. A student measures an amplitude of 0.985

e-lub [12.9K]

Answer:

a) $f = 0.598\,hz$ , b) $v_{max} = 3.701\,\frac{m}{s}$ , c) $k = 15.385\,\frac{N}{m}$ , d) $U = 1.081\,J$ , e) $K = 6.382\,J$ , f) $v\approx 3.422\,\frac{m}{s}$

Explanation:

a) The frequency of oscillation is:

$f = \frac{76}{127\,hz}$

$f = 0.598\,hz$

b) The angular frequency is:

$\omega = 2\pi \cdot f$

$\omega = 2\pi \cdot (0.598\,hz)$

$\omega = 3.757\,\frac{rad}{s}$

Lastly, the speed at the equilibrium position is:

$v_{max} = \omega \cdot A$

$v_{max} = (3.757\,\frac{rad}{s} )\cdot (0.985\,m)$

$v_{max} = 3.701\,\frac{m}{s}$

c) The spring constant is:

$\omega = \sqrt{\frac{k}{m}}$

$k = \omega^{2}\cdot m$

$k = (3.757\,\frac{rad}{s} )^{2}\cdot (1.09\,kg)$

$k = 15.385\,\frac{N}{m}$

d) The potential energy when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

$U = \frac{1}{2}\cdot (15.385\,\frac{N}{m} )\cdot (0.375\,m)^{2}$

$U = 1.081\,J$

e) The maximum potential energy is:

$U_{max} = \frac{1}{2}\cdot (15.385\,\frac{N}{m} )\cdot (0.985\,m)^{2}$

$U_{max} = 7.463\,J$

The kinetic energy when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

$K = U_{max} - U$

$K = 7.463\,J - 1.081\,J$

$K = 6.382\,J$

f) The speed when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

$K = \frac{1}{2}\cdot m \cdot v^{2}$

$v = \sqrt{\frac{2\cdot K}{m} }$

$v = \sqrt{\frac{2\cdot (6.382\,J)}{1.09\,kg} }$

$v\approx 3.422\,\frac{m}{s}$

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2 years ago

Q/C An undersea earthquake or a landslide can produce an ocean wave of short duration carrying great energy, called a tsunami. W

iragen [17]

The amplitude $A_{2}$ is 8.307 m

When describing the peak value of a quantity, such as the level of sound waves or the power and voltage in electrical and electronic systems, the term amplitude is employed. A louder sound has a bigger amplitude, and a softer sound has a smaller amplitude.

As we previously discussed in the concept session, the square root of the water's speed is inversely related to the amplitude of its waves.

$A\alpha\frac{1}{\sqrt{v} }$

Solve for first case:

$A_{1}=\frac{1}{\sqrt{v_{1} } }$

Solve for second case:

$A_{2}=\frac{1}{\sqrt{v_{2} } }$

So, the two equations is equal

$A^{2} _{1} v_{1} = A^{2} _{2} v_{2}$

$=A^{2} _{2} \sqrt{g d_{2} }$

Rearrange and solve for the amplitude $A_{2}$ :

$A_{2} =\sqrt{\frac{A^{2}_{1} v_{1} }{v_{2} } }$

$=\sqrt{\frac{A^{2}_{1} v_{1} }{\sqrt{g d_{2} } } }$

$=\sqrt{\frac{(1.8m)^{2} X200 m/s }{\sqrt{9.8m/s^{2}X9 m } } }$

= 8.307 m

So, The amplitude $A_{2}$ is 8.307 m.

Learn more about amplitude here:

brainly.com/question/3613222

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1 year ago