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Anarel [89]
3 years ago
6

A toaster uses 6700 joules of energy in 45 seconds to toast to a piece of bread what is power of the oven

Physics
1 answer:
just olya [345]3 years ago
8 0
P= w/t and W= Work 
In this case, W= 6,700j, and T= 45 seconds
Power is the ratio of work  per unit time. When you perform a work in a given span of time, the ratio of work performed with respect to time is Called Power. 
si unit for Power is Watt (W) 
so, P= 6,700/45
P= 148 
Final answer is P=148
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What is the ACCELERATION of a 50 kg object pushed with a force of 500 newtons *
Anon25 [30]

Answer:

The answer is

<h2>10 m/s²</h2>

Explanation:

To find the acceleration of an object given the force and mass we use the formula

<h3>acceleration =  \frac{force}{mass}</h3>

From the question

mass of object = 50 kg

force = 500 N

So the acceleration is

<h3>acceleration =  \frac{500}{50}  \\  =  \frac{50}{5}</h3>

We have the final answer as

<h3>10 m/s²</h3>

Hope this helps you

5 0
3 years ago
B. Assuming the acceleration is still -9.81 m/s2, what is the instantaneous velocity of the
goblinko [34]

We have that the instantaneous velocity of the shuttlecock when it hits the ground is

V_{int}=\sqrt{U^2+19.6H}

From the question we are told

Assuming the acceleration is still -9.81 m/s2, what is the instantaneous velocity of the

shuttlecock when it hits the ground? Show your work below.

Generally the equation for acceleration  is mathematically given as

a=v \frac{dv}{dx}\\\\\Therefore\\\\\2ah=v^2-u^2

Where

acceleration is still -9.81 m/s2,

Hence,

V^2-U^2=2(-9.81)*-H

Therefore

V_{int}=\sqrt{U^2+19.6H}

For more information on this visit

brainly.com/question/12319416?referrer=searchResults

7 0
2 years ago
Two point charges of equal magnitude q are held a distance d apart. Consider only points on the line passing through both charge
NemiM [27]

let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.

the mathematical formula for potential is V=\frac{1}{4\pi\epsilon} \frac{q}{d}

for positive charges the potential is positive and is negative for negative charges.

the formula for electric field is given as-E=\frac{1}{4\pi\epsilon} \frac{q}{r^2}

for positive charges,the line filed is away from it and for negative charges the filed is towards it.

we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.

here the net electric field due to the dipole can not be zero  between the two charges,but we can find the points situated on the axial  line but  outside of charges where the electric field is zero.

now let the two charges of same nature.let these are positively charged.

here we can not find a point between two charges and on the line joining  two charges  where the potential is zero.

but at the mid point of the line joining two charges the filed is zero.

5 0
3 years ago
Question is down below need rply fast
My name is Ann [436]

The X and Y components are as follows;

1. X = 35 * cos 57 = 19. 1m/s; Y = 35 * sin 57 = 29.4 m/s

2. X = 12 * -cos 34  = -10 m/s; Y = 12 * -sin 34 = -6.7 m/s

3. X = 8 * -cos 90  = 0 m/s; Y = 12 -sin 90 = -8 m/s

4. X = 20 * cos 75 = 5. 2m/s; Y = 20 * (-sin 75) =  -19.3 m/s

<h3>What are the horizontal and vertical components of the vectors?</h3>

The horizontal and vertical components of the velocities are given as follows:

  • Horizontal component, X = x cos θ
  • Vertical component, Y = y sin θ

1. 35 m/s at  57° from x-axis

X = 35 * cos 57 = 19. 1m/s

Y = 35 * sin 57 = 29.4 m/s

2. 12m/s at 34° S of W

X = 12 * -cos 34  = -10 m/s

Y = 12 * -sin 34 = -6.7 m/s

3. 8 m/s at South

X = 8 * -cos 90  = 0 m/s

Y = 12 -sin 90 = -8 m/s

4. 20 m/s at  275° from x-axis

X = 20 * cos 75 = 5. 2m/s

Y = 20 * (-sin 75) =  -19.3 m/s

In conclusion, the X and Y components are found by taking cosines and sine of the angles.

Learn more about horizontal and vertical components at: brainly.com/question/26446720

#SPJ1

8 0
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A 58 kg boy and a 38 kg girl use an elastic rope while engaged in a tug-of-war on a frictionless icy surface. If the acceleratio
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