Please give the answer
please give the answer - 1

Answers

Answer 1
Answer:

Answer:

75

Explanation:

just took it e2020

Answer 2
Answer:

Answer:

60%

Explanation:

efficiency= useful/input x 100%,

Here, kinetic energy is useful for food processor (i.e. spinning blades)

600J/1000J=60%


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You hit a hockey puck and it slides across the ice at nearly a constant speed.Is a force keeping it in motion?Explain.
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A 1000-kg car rolling on a smooth horizontal surface ( no friction) has speed of 20 m/s when it strikes a horizontal spring and is brought to rest in a distance of 2 m What is the spring’s stiffness constant?

Answers

Explanation:

kinetic energy was converted to potential energy in the spring.

the answer is in the above image

Which of the following is a description of the Remote Associates Test (RAT)?

Answers

Answer:

The description is outlined throughout the clarification section following, and according to the given word.

Explanation:

  • Throughout the 1960s, Sarnoff Mednick created the RAT as a tool used for testing imaginative convergent thought. Through each RAT test query lists a set of terms, which demands that we have a single additional term that will tie any of the others around.  
  • Those other words may also be related in something like a variety of ways, such as through creating a compound word or perhaps a semantic connexon.

A radio station broadcasts its electromagnetic (radio)waves at a frequency of 9.05 x 107 Hz.
These radio waves travel at a speed of 3.00 x 108 m/s.
What is the wavelength of these radio waves?

Answers

Wavelength = speed/frequency
Wavelength = 3.00x108/9.05x107=
3.3x10 risen to the power of -1

The particle with charge q is now released and given a quick push; as a result, it acquires speed v. Eventually, this particle ends up at the center of the original square and is momentarily at rest. If the mass of this particle is m , what was its initial speed v ?.

Answers

The given situation is illustrated below. A particle is released and given a quick push. As a result, it acquires a speed v. Eventually, this particle ends up at the center of the original square and is momentarily at rest. If the mass of this particle is m, the initial speed of the particle is √((2qV/m).)


To solve the problem, we need to apply the law of conservation of energy, which states that energy can neither be created nor destroyed; it can only be transformed from one form to another.
Initial potential energy = Final kinetic energy
The initial potential energy of the particle is given by
U = qV
where V is the potential difference between the corner and the center of the square.
At the center of the square, the potential energy is zero.
The final kinetic energy of the particle is given by
K = (1/2) mv^2
where m is the mass of the particle and v is its final velocity.
Since the particle is momentarily at rest at the center of the square, its final kinetic energy is zero.
Therefore, we have
qV = (1/2) mv^2
Solving for v, we get
v = √((2qV/m).)

for such more question on speed

brainly.com/question/13943409

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Two coils are wound around the same cylindrical form. When the current in the first coil is decreasing at a rate of -0.245 A/s , the induced emf in the second coil has a magnitude of 1.60×10−3 V . Part A What is the mutual inductance of the pair of coils? MM = nothing H Request Answer Part B If the second coil has 22 turns, what is the flux through each turn when the current in the first coil equals 1.25 A ? ΦΦ = nothing Wb

Answers

To solve this problem we need to use the emf equation, that is,

E=m(dI)/(dT)

Where E is the induced emf

I the current in the first coil

M the mutual inductance

Solving for a)

M=(E)/((dI)/(dT))\nM=(1.6*10^(-3))/(0.245)=6.53*10^(-3)H

Solving for b) we need the FLux through each turn, that is

\Phi=(MI)/(N)

Where N is the number of turns in the second coil

\Phi=(6.53*10^(-3)*1.25)/(22)=3.71*10^(-4)Wb

The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in meters, t in seconds, and c and b are positive constants. What are the units of (a) constant c and (b) constant b? Find a formula in terms of c, b, and t of the (c) velocity v and (d) acceleration a. (e) At what time t ≥ 0 does the particle reach its maximum x value?

Answers

Answer:

(a):  \rm meter/ second^2.

(b):  \rm meter/ second^3.

(c):  \rm 2ct-3bt^2.

(d):  \rm 2c-6bt.

(e):  \rm t=(2c)/(3b).

Explanation:

Given, the position of the particle along the x axis is

\rm x=ct^2-bt^3.

The units of terms \rm ct^2 and \rm bt^3 should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of \rm ct^2=meter

Therefore, unit of \rm c= meter/ second^2.

(b):

Unit of \rm bt^3=meter

Therefore, unit of \rm b= meter/ second^3.

(c):

The velocity v and the position x of a particle are related as

\rm v=(dx)/(dt)\n=(d)/(dx)(ct^2-bt^3)\n=2ct-3bt^2.

(d):

The acceleration a and the velocity v of the particle is related as

\rm a = (dv)/(dt)\n=(d)/(dt)(2ct-3bt^2)\n=2c-6bt.

(e):

The particle attains maximum x at, let's say, \rm t_o, when the following two conditions are fulfilled:

  1. \rm \left ((dx)/(dt)\right )_(t=t_o)=0.
  2. \rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)<0.

Applying both these conditions,

\rm \left ( (dx)/(dt)\right )_(t=t_o)=0\n2ct_o-3bt_o^2=0\nt_o(2c-3bt_o)=0\nt_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = (2c)/(3b).

For \rm t_o = 0,

\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)=2c-6bt_o = 2c-6\cdot 0=2c

Since, c is a positive constant therefore, for \rm t_o = 0,

\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)>0

Thus, particle does not reach its maximum value at \rm t = 0\ s.

For \rm t_o = (2c)/(3b),

\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)=2c-6bt_o = 2c-6b\cdot (2c)/(3b)=2c-4c=-2c.

Here,

\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)<0.

Thus, the particle reach its maximum x value at time \rm t_o = (2c)/(3b).

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