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3 years ago

Using clay electric rates, how much does it cost to watch Netflix for 4 hours on a smart tv?

Physics

1 answer:

OverLord2011 [107]3 years ago

7 0

Answer: TVs cost between $0.0015 and $0.0176

Explanation:

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A 0.450 kg soccer ball has a kinetic energy of 119 J.

Anastaziya [24]

Answer:

V is approximately = 23m/s

Explanation:

Kinetic energy = ½ mv²

Where m= mass = 0.450kg

V= velocity =?

K. E = 119J

Therefore

K. E = ½ mv²

Input values given

119= ½ × 0.450 × v²

Multiply both sides by 2

119 ×2 = 2 × 1/2 × 0.450 × v²

238= 0.450v²

Divide both sides by 0.450

238/0.450 = 0.450v²/0.450

v² = 528.89

Square root both sides

Sq rt v² = sq rt 528.89

V = 22.998m/s

V is approximately = 23m/s

I hope this was helpful, please rate as brainliest

8 0

3 years ago

Which type of energy will definitely not be used in the lighting of a match?

In-s [12.5K]

I say that the answere would be B

4 0

3 years ago

An apple dropped from the branch of a tree hits the ground in 0.5 s. If the acceleration of the apple during its motion is 10 ms

Ipatiy [6.2K]

Given that,

Time = 0.5 s

Acceleration = 10 m/s²

(I). We need to calculate the speed of apple

Using equation of motion

$v=u+at$

Where, v = speed

u = initial speed

a = acceleration

t = time

Put the value into the formula

$v=0+10\times0.5$

$v=5\ m/s$

(III). We need to calculate the height of the branch of the tree from the ground

Using equation of motion

$s=ut+\dfrac{1}{2}gt^2$

Put the value into the formula

$s=0+\dfrac{1}{2}\times10\times(0.5)^2$

$s=1.25\ m$

(II). We need to calculate the average velocity during 0.5 sec

Using formula of average velocity

$v_{avg}=\dfrac{\Delta x}{\Delta t}$

$v_{avg}=\dfrac{x_{f}-x_{i}}{t_{f}-t_{0}}$

Where, $x_{f}$ = final position

$x_{i}$ = initial position

Put the value into the formula

$v_{avg}=\dfrac{1.25+0}{0.5}$

$v_{avg}=2.5\ m/s$

Hence, (I). The speed of apple is 5 m/s.

(II). The average velocity during 0.5 sec is 2.5 m/s

(III). The height of the branch of the tree from the ground is 1.25 m.

7 0

3 years ago

At an air show, a stunt pilot performs a vertical loop-the-loop in a circle of radius 3.63 x 103 m. During this performance the

san4es73 [151]

Answer:

189 m/s

Explanation:

The pilot will experience weightlessness when the centrifugal force, F equals his weight, W.

So, F = W

mv²/r = mg

v² = gr

v = √gr where v = velocity, g = acceleration due to gravity = 9.8 m/s² and r = radius of loop = 3.63 × 10³ m

So, v = √gr

v = √(9.8 m/s² × 3.63 × 10³ m)

v = √(35.574 × 10³ m²/s²)

v = √(3.5574 × 10⁴ m²/s²)

v = 1.89 × 10² m/s

v = 189 m/s

5 0

3 years ago

A potential difference of 53 mV is developed across the ends of a 12.0-cm-longwire as it moves through a 0.27 T uniform magnetic

Klio2033 [76]

Answer:

The angle between the magnetic field and the wire’s velocity is 19.08 degrees.

Explanation:

Given that,

Potential difference, V = 53 mV

Length of the wire, l = 12 cm = 0.12 m

Magnetic field, B = 0.27 T

Speed of the wire, v = 5 m/s

Due to its motion, an emf is induced in the wire. It is given by :

$\epsilon=Blv\sin\theta$

Here,

$\theta$ is the angle between magnetic field and the wire’s velocity

$\sin\theta=\dfrac{\epsilon}{Blv}\\\\\sin\theta=\dfrac{53\times 10^{-3}}{0.27\times 0.12\times 5}\\\\\sin\theta=0.327\\\\\theta=19.08^{\circ}$

So, the angle between the magnetic field and the wire’s velocity is 19.08 degrees.

8 0

3 years ago