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3 years ago

What is the angle of reflection

Physics

2 answers:

Naddika [18.5K]3 years ago

7 0

The angle of reflection of light is equal to 90 degrees

Pachacha [2.7K]3 years ago

7 0

The answer of edge is 45 degrees.

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(a) What is the cost of heating a hot tub containing 1440 kg of water from 10.0°C to 40.0°C, assuming 75.0% efficiency to take h

BaLLatris [955]

Answer:

a) $E = 6.024\,USD$ , For m kilograms, it is 4184m J., 3600000 joules, b) $i = 88.200\,A$

Explanation:

a) The amount of heat needed to warm water is given by the following expression:

$Q_{needed} = m_{w}\cdot c_{w}\cdot (T_{f}-T_{i})$

Where:

$m_{w}$ - Mass of water, measured in kilograms.

$c_{w}$ - Specific heat of water, measured in $\frac{J}{kg\cdot ^{\circ}C}$ .

$T_{f}$ , $T_{i}$ - Initial and final temperatures, measured in $^{\circ}C$ .

Then,

$Q_{needed} = (1440\,kg)\cdot \left(4184\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (40^{\circ}C - 10^{\circ}C)$

$Q_{needed} = 180748800\,J$

The energy needed in kilowatt-hours is:

$Q_{needed} = 180748800\,J\times \left(\frac{1}{3600000}\,\frac{kWh}{J} \right)$

$Q_{needed} = 50.208\,kWh$

The electric energy required to heat up the water is:

$E = \frac{50.208\,kWh}{0.75}$

$E = 66.944\,kWh$

Lastly, the cost of heating a hot tub is: (USD - US dollars)

$E = (66.944\,kWh)\cdot \left(0.09\,\frac{USD}{kWh} \right)$

$E = 6.024\,USD$

The heat needed to raise the temperature a degree of a kilogram of water is 4184 J. For m kilograms, it is 4184m J. Besides, a kilowatt-hour is equal to 3600000 joules.

b) The current required for the electric heater is:

$i = \frac{Q_{needed}}{\eta \cdot \Delta V \cdot \Delta t}$

$i = \frac{180748800\,J}{0.75\cdot (220\,V)\cdot (3.45\,h)\cdot \left(3600\,\frac{s}{h} \right)}$

$i = 88.200\,A$

7 0

3 years ago

A 2.0 kg block is released from rest at the top of a curved incline in the shape of a quarter of a circle of radius R = 3.0 m. T

Zigmanuir [339]

The block has maximum kinetic energy at the bottom of the curved incline. Since its radius is 3.0 m, this is also the block's starting height. Find the block's potential energy PE :

PE = m g h

PE = (2.0 kg) (9.8 m/s²) (3.0 m)

PE = 58.8 J

Energy is conserved throughout the block's descent, so that PE at the top of the curve is equal to kinetic energy KE at the bottom. Solve for the velocity v :

PE = KE

58.8 J = 1/2 m v ²

117.6 J = (2.0 kg) v ²

v = √((117.6 J) / (2.0 kg))

v ≈ 7.668 m/s ≈ 7.7 m/s

3 0

3 years ago

A 95 kg fullback, running at 8.2 m/s, collides in midair with a 128 kg defensive tackle moving in the opposite direction. Both p

Sindrei [870]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

Below is the answers:

Fullback running

Mo = mass * velocity 
Mo = 95kg * 8.2 m/s =779 kg*m/s (a 

He got stopped Change in Mo = 779 kg*m/s (b 

Both stopped ===> Tackle's mo = - Halfback's Mo = - 779 kg*m/s (c & d 

- 779 = 128 * v 
v= - 6.09 m/s (e

6 0

3 years ago

n outer space, a constant net force with a magnitude of 140 N is exerted on a 32.5 kg probe initially at rest. A) What accelerat

musickatia [10]

Answer:

a) 4.31 m/s²

b) 215.5 m

Explanation:

a) According to Newton's first law of motion

The net force applied to particular mass produced acceleration, a, according to

F = ma

F = 140 N

m = 32.5 kg

a = ?

140 = 32.5 × a

a = 140/32.5 = 4.31 m/s²

b) Using the equations of motion, we can obtain the distance travelled by the object in t = 10 s

u = initial velocity of the probe = 0 m/s (since it was initially at rest)

a = 4.31 m/s²

t = 10 s

s = distance travelled = ?

s = ut + at²/2

s = 0 + (4.31×10²)/2 = 215.5 m

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3 years ago

-Starts to suck on ur neck giving you hickeys n runs my hand down your body biting my lip against your neck

Hoochie [10]

Answer:

I would shout fore help if I was being raped or try to make him or her stop

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3 years ago