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crimeas [40]
3 years ago
5

When light moves from air into glass at an angle, is the light refracted toward or away from the normal?

Physics
1 answer:
Whitepunk [10]3 years ago
6 0

Answer: at an angle !!

Explanation:

light refracts at an angle

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A ball is thrown vertically upward from the ground at a speed of 44 m/s. How long in s does it take to return to the ground?
Agata [3.3K]

vf=vi-gt (at peak vf = 0)

vi=gt

44=9.8t

t=4.49 s

time return = time upward = 4.49 s

6 0
3 years ago
Help ill give brainleist
r-ruslan [8.4K]

Transformation of Energy

7 0
3 years ago
Terminal velocity. A rider on a bike with the combined mass of 100kg attains a terminal speed of 15m/s on a 12% slope. Assuming
Firlakuza [10]

Answer:

0.9378

Explanation:

Weight (W) of the rider = 100 kg;

since 1 kg = 9.8067 N

100 kg will be = 980.67 N

W = 980.67 N

At the slope of 12%, the angle θ is calculated as:

tan \ \theta = \dfrac{12}{100} \\ \\  tan \ \theta = 0.12 \\ \\  \theta = tan^{-1}(0.12) \\\\ \theta = 6.84^0

The drag force D = Wsinθ

\dfrac{1}{2}C_v \rho AV^2 = W sin \theta

where;

\rho = 1.23 \ kg/m^3

A = 0.9 m²

V = 15 m/s

∴

Drag coefficient C_D = \dfrac{2 *W*sin \theta}{\rho *A *V^2}

C_D =\dfrac{2 *980.67*sin 6.84}{1.23 *0.9 *15^2}

C_D =0.9378

8 0
3 years ago
A toy car is given a quick push so that it rolls up an inclined ramp. After it is released, it rolls up, reaches its highest poi
kicyunya [14]

Answer: Option (E)

Explanation:

There are only two forces acting on the car: The force of gravity, which point down, and the normal force generated by the interaction of the car and the ramp, which points perpendicular to the ramp.

N=mgcos(\alpha ),

m:mass of the car ; \alpha:angle of the ramp.

g: gravity's acceleration.

N is constante since all parameters are constants.

So, since the force of gravity is also constant, the net force acting on the car is constant. Since velocity decreases over time and the car stars moving down, the force is pointing down the ramp.

5 0
4 years ago
Calculate the amount of heat needed to raise the temperature of 200g of ice from-30°C 50C water. (C = .5 for ice, C 1 for water,
Nitella [24]

<u>Answer:</u> The amount of heat needed is 29000 Cal.

<u>Explanation:</u>

The process involved in this problem are:

(1):H_2O(s)(-30^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(50^oC)

Now, we calculate the amount of heat released or absorbed in all the processes.

  • <u>For process 1:</u>

q_1=mC_{p,s}\times (T_2-T_1)

where,

q_1 = amount of heat absorbed = ?

m = mass of ice = 200 g

T_2 = final temperature = 0^oC

T_1 = initial temperature = -30^oC

Putting all the values in above equation, we get:

q_1=200g\times 0.5Cal/g^oC\times (0-(-30))^oC=3000Cal

  • <u>For process 2:</u>

q_2=m\times L_f

where,

q_1 = amount of heat absorbed = ?

m = mass of water or ice = 200 g

L_f = latent heat of fusion = 80 Cal/g

Putting all the values in above equation, we get:

q_2=200g\times 80Cal/g=16000Cal

  • <u>For process 3:</u>

q_3=m\times C_{p,l}\times (T_{2}-T_{1})

where,

q_3 = amount of heat absorbed = ?

m = mass of water = 200 g

T_2 = final temperature = 50^oC

T_1 = initial temperature = 0^oC

Putting all the values in above equation, we get:

q_3=200g\times 1Cal/g^oC\times (50-0)^oC=10000Cal

Calculating the total heat absorbed, we get:

Q=q_1+q_2+q_3

Q=3000+16000+10000=29000Cal

Hence, the amount of heat needed is 29000 Cal.

7 0
4 years ago
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