TsAn electric light bulb mixer is used for 19.2 seconds. In that time, it transfers 1
536 J of energy.
Calculate the power output of the cake mixer.
A
I DONT KNOW

Answers

Answer 1

Answer: And electric light bulb mix it is used for nine. Two seconds in that temperature transfers one 536 J of energy calculate shin is 2+2 is 4-1 that’s three quick maths exclamation is calculate the power work output of the kitchen

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Nina and Jon are practicing an ice skating routine. Nina is standing still. Jon, who is twice as heavy as Nina, skates toward her, pushing Nina away with force f. Assuming the system is closed, which statement is correct about this system? a. Nina experiences a force equal to f/2. b. Nina experiences a force equal to f^2. c. Nina experiences a force equal to 2f. d. Nina experiences a force equal to f.

Answers

Answer:

Explanation:

• Nina experiences a force equal to f.

Answer:

Nina experiences a force equal to f

Explanation:

got to get that 2nd answer slot correct too before an abusive expert verifier with an alt comes in and purposely verifies the wrong answer

An F-35 stealth jet takes off from the aircraft carrier Ronald Reagan. Starting from rest, the jet accelerated with a constant acceleration of 55.3 m/s2 along a straight line on the deck. What is the displacement of the jet when it reaches a speed of 181 m/s?

Answers

Answer:

When the jet reaches a speed of 181 m/s, its displacement is 296 m.

Explanation:

Hi there!

The equation of position and velocity of an object traveling with constant acceleration along a straight line are the following:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position of the object at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

v = velocity of the object at time t.

If we place the origin of the frame of reference at the point where the jet starts moving, then, x0 = 0. Since the jet starts from rest, v0 is also zero. Then the equations get reduced to the following:

x = 1/2 · a · t²

v = a · t

We know the acceleration and the final velocity of the jet. So, using the equation of velocity, we can find the time it takes the jet to reach that velocity. Then, we can calculate the position of the jet at that time. Since the initial position is zero, the final position of the jet will be equal to the displacement (because displacement = final position - initial position).

v = a · t

v/a = t

181 m/s / 55.3 m/s² = t

t = 3.27 s

The final position of the jet will be:

x = 1/2 · a · t²

x = 1/2 · 55.3 m/s² · (3.27 s)²

x = 296 m

When the jet reaches a speed of 181 m/s, its displacement is 296 m.

The displacement of the F-35 jet when it reaches a speed of 181 m/s is 16515 m.

To find displacement using constant acceleration,

we can use the following equation:

displacement = (final velocity)^2 - (initial velocity)^2 / 2 * acceleration.

In this case, the initial velocity is 0 m/s and the final velocity is 181 m/s.

The acceleration is given as 55.3 m/s^2.

Plugging in these values, we get:

displacement = (181)^2 - (0)^2 / 2 * 55.3 = 16515 m.

The displacement of the F-35 jet when it reaches a speed of 181 m/s is 16515 m.

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The pendulum consists of two slender rods AB and OC which have a mass of 3 kg/m. The thin plate has a mass of 12 kg/m2 . a) Determine the location ӯ of the center of mass G of the pendulum, then calculate the mass moment of inertia of the pendulum about z axis passing through G. b) Calculate the mass moment of inertia about z axis passing the rotation center O.

Answers

Answer:

The answer is below

Explanation:

a) The location ӯ of the center of mass G of the pendulum is given as:

$y=(0+(\pi*(0.3\ m) ^2*12kg/m^2*1.8\ m-\pi*(0.1\ m) ^2*12kg/m^2*1.8\ m)+0.75\ m*1.5\ m *3\ kg/m)/((\pi*(0.3\ m) ^2*12kg/m^2-\pi*(0.1\ m) ^2*12kg/m^2)+3\ kg/m^2*0.8\ m+3\ kg/m^2*1.5\ m) \n\ny=0.88\ m$

b) the mass moment of inertia about z axis passing the rotation center O is:

$I_G=(1)/(12)*3(0.8)(0.8)^2+ 3(0.8)(0.888)^2-(1)/(2)*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8-\n0.888)^2+(1)/(2)*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8-0.888)^2+(1)/(12)*3(1.5)(1.5)^2+\n3(1.5)(0.888-0.75)^2\n\nI_G=13.4\ kgm^2$

c) The mass moment of inertia about z axis passing the rotation center O is:

$I_o=(1)/(12)*3(0.8)(0.8)^2+ (1)/(3)* 3(1.5)(1.5)^2+(1)/(2)*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8)^2-\n(1)/(2)*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8)^2\n\nI_o=13.4\ kgm^2$

Final answer:

To solve this problem, calculate the mass of each element of the pendulum, use that information to determine the center of mass, and then apply the parallel axis theorem to calculate the two moments of inertia.

Explanation:

To determine the center of mass and the mass moment of inertia of the pendulum, first we calculate the individual masses of the rods: AB and OC, and the plate. Each rod has a mass of 2 kg (given mass per unit length is 3kg/m and length of each rod is 1 m from the first reference paragraph).

The center of mass ӯ can be determined using the formula for center of mass, averaging distances to each mass element weighted by their individual masses. The mass moment of inertia, also known as the angular mass, for rotation about the z axis through G is determined using the parallel axis theorem, which states that the moment of inertia about an axis parallel to and a distance D away from an axis through the center of mass is the sum of the moment of inertia for rotation about the center of mass and the total mass of the body times D squared.

Finally, the moment of inertia about the z axis passing through the center of rotation O can be calculated again using the parallel axis theorem, with distance d being the distance between points G and O.

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A 140 W lightbulb emits 4% of its energy as electromagnetic radiation. What is the radiation pressure (in N/m2) on a perfectly absorbing sphere of radius 14 m that surrounds the bulb

Answers

Answer: 7.578x10^-12

Explanation:

First, we find the power:

Power P = 140x4/100 =5,6W

Distance r = 14m

Then,

Intensity I = P/4πr2

= 5.6/(4π x 14 x 14)

=. 2.27 x10^3 W/m2

Radiation pressure:

P(rad) = I/c =0.00227÷{3 x 10^8)

=7.578x10^-12 N/m2

Answer:

Pr=7.57*10^{11}Pa

Explanation:

We can solve this problem by taking into account the expression

$P_r=(IA)/(c)$

where I is the irradiance, c is the speed of light and A is the area.

We have that the power is 140W, but only 4% is electromagnetic energy, that is

$P=140W=140(J)/(s)\n0.4*140J=56J$

56J is the electromagnetic energy.

The area of the bulb is

$A_b=4\pi r^2=4\pi (14m)^2=2463m^2$

The radiation pressure is

$P_r=(56)/(2463m^2*3*10^8m/s^2)=7.57*10^(-11)Pa$

hope this helps!!

The erg is a unit of work in units of centimeters (cm), grams (g), and seconds (s), and 1 erg=1 g⋅cm^2/s^2 . Recall that the SI unit of force is the newton (N) and is equal to kg⋅m/s^2 . You push an object 0.032 m by exerting 0.010 N of force. The work is the force times the distance.How much work have you done, expressed in ergs?

A: 3,200 ergs

B: 32 ergs

C: 0.32 ergs

D: 0.00032 ergs

Answers

For the given problem, the amount of work done expressed in ergs is 3200 ergs.

Answer: Option A

Explanation:

The work done on an objects are the force acting on it to move the object to a particular distance. So, work done on the object will be directly proportional to the force acting on it and the displacement.

Here, the force acting on the object is given as 0.010 N and the displacement of the object is 0.032 m. So, the work done on the object is

$\text { Work done }=\text { Force } * \text { displacement }$

$\text { Work done }=0.010 \mathrm{N} * 0.032 \mathrm{m}=0.00032 \mathrm{Nm}$

It is known that $1 N=1 \mathrm{kg} \mathrm{ms}^(-2)$

So, the work done can be expressed in $k g m s^(-2)$ as,

$\text { Work done }=0.00032 \mathrm{kgm}^(2) \mathrm{s}^(-2)$

It is known that $1 \mathrm{erg}=1 \mathrm{g} \mathrm{cm}^(2) / \mathrm{s}^(2)$ , so the conversion of units from Nm to erg will be done as follows:

$\text { Work done }=0.00032 \mathrm{kgm}^(2) \mathrm{s}^(-2) * \frac{1000 \mathrm{g}}{1 \mathrm{kg}} * \frac{100 * 100 \mathrm{cm}^(2)}{m^(2)}=3200 \mathrm{g} \mathrm{cm}^(2) \mathrm{s}^(-2)$

Thus, work done in ergs is 3200 ergs.

A wagon is pulled at a speed of 0.40 m/s by a horse exerting 1800 Newtons of horizontal Force. how much work was done by the horse

Answers

The amount of work done per second by the horse exerting a force of 1800 N on a wagon moving with a speed of 0.4 m/s is 720 J/s.

What is power?

Power is the workdone by a body in one second.

To calculate the work done by the horse in one seconds, we use the formula below

Formula:

P = Fv................ Equation 1

Where:

P = work done on the horse in one second
F = Force of the horse
v = Velocity of the wagon

From the question,

Given:

F = 1800 N
v = 0.4 m/s

Substitute these values into equation 1

P = 1800×0.4
P = 720 J/s

Hence, the amount of work done per second by the horse is 720 J/s.

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Complete question: A wagon is pulled at a speed of 0.40 m/s by a horse exerting 1800 Newtons of horizontal Force. how much work was done by the horse per second.

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