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4 years ago

NEED HELP QUICK!

Physics

2 answers:

Olenka [21]4 years ago

6 0

Hi there!
At angle B, the light is refracted at an angle in one direction as it enters the glass. As it exits the glass at angle C, It is refracted at another angle in the opposite direction.
Hope this helps, have a nice day :)

const2013 [10]4 years ago

4 0

the answer is f and a i took the quiz

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See the person on the right side of the front car. Six reference points could be used to show that the person is in / is NOT in

Rasek [7]

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

1. According to person standing on ground ~

The person is in motion

2. According to The car ~

The person is not in motion

3. According to the Seat ~

The person is not in motion

4. According to another person on ride ~

The person is not in motion

5. According to the track ~

The person is in motion

6. According to the Sun ~

The person is in motion

I hope that's what you were looking for, goodluck for your assignment ~

7 0

2 years ago

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

just olya [345]

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

8 0

3 years ago

A copper cylinder is initially at 21.1 ∘C . At what temperature will its volume be 0.163 % larger than it is at 21.1 ∘C?

fomenos

To solve this problem we will apply the concepts related to the final volume of a body after undergoing a thermal expansion. To determine the temperature, we will use the given relationship as well as the theoretical value of the volumetric coefficient of thermal expansion of copper. This is, for example to the initial volume defined as $V_1$ , the relation with the final volume as

$V_2 = V_1 +0.163\% V_1$

$V_2 = V_1 +0.00163V_1$

$V_2 = 1.00163V_1$

Initial temperature = $21.1\°C$

Let T be the temperature after expanding by the formula of volume expansion

we have,

$V_2 = V_1 (1+\gamma \Delta t)$

Where $\gamma$ is the volume coefficient of copper $5.1*10^{-5}/C$

$1.00163V_1 = V_1(1+\gamma(T-21.1\°))$

$1.00163 = 1+5.1*10^{-5}(T-21.1\°)$

$0.00163 = 0.000051T-0.0010761$

$T = 53.0608\°C$

Therefore the temperature is 53.06°C

7 0

4 years ago

Swinging a tennis racket against a ball is an example of a third class lever. Please select the best answer from the choices pro

kari74 [83]

that statement is true

a Third class lever applied when the effort place between the load and the fulcrum.

For example, in a forearm serve
Fulcrum : The elbow
Effort : The effort that putted by the biceps muscle
Load : The arm

7 0

3 years ago

Read 2 more answers

A man walks along a straight path at a speed of 4 ft/s. A searchlight is located on the ground 6 ft from the path and is kept fo

BARSIC [14]

We are given that,

$\frac{dx}{dt} = 4ft/s$

We need to find $\frac{d\theta}{dt}$ when $x=8ft$

The equation that relates x and $\theta$ can be written as,

$\frac{x}{6} tan\theta$

$x = 6tan\theta$

Differentiating each side with respect to t, we get,

$\frac{dx}{dt} = \frac{dx}{d\theta} \cdot \frac{d\theta}{dt}$

$\frac{dx}{dt} = (6sec^2\theta)\cdot \frac{d\theta}{dt}$

$\frac{d\theta}{dt} = \frac{1}{6sec^2\theta} \cdot \frac{dx}{dt}$

Replacing the value of the velocity

$\frac{d\theta}{dt} = \frac{1}{6} cos^2\theta (4)^2$

$\frac{d\theta}{dt} = \frac{8}{3} cos^2\theta$

The value of $cos \theta$ could be found if we know the length of the beam. With this value the equation can be approximated to the relationship between the sides of the triangle that is being formed in order to obtain the numerical value. If this relation is known for the value of x = 6ft, the mathematical relation is obtained. I will add a numerical example (although the answer would end in the previous point) If the length of the beam was 10, then we would have to

$cos\theta = \frac{6}{10}$

$\frac{d\theta}{dt} = \frac{8}{3} (\frac{6}{10})^2$

$\frac{d\theta}{dt} = \frac{24}{25}$

Search light is rotating at a rate of 0.96rad/s

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3 years ago