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2 years ago

The speed of light of a ray of light traveling through a distance having an absolute index of refraction of 1.1 is?

1 answer:

4 0

<span>The speed of ray of light traveling through a substance whose index of refraction 1.1 is

(c/1.1) = (299,792,458 m/s) / 1.1 = </span>272,538,598 m/s

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What is true of an object pulled inward in an electric field?

slava [35]

Answer:

option b

Explanation:

There is an object pulled inward in an electric field.

We have to find out of the four options given which is true.

a) The object has a neutral charge is false since when electric field pulls the object inward, there is a charge inside.

b) The object has a charge opposite that of the field, this option is correct since there will be an equal and opposite charge created by the object

c) The object has a negative charge will be correct only if the original charge was positive hence wrong

d) The object has a charge the same as that of the field is incorrect since this would be opposite the charge

So only option b is right

5 0

2 years ago

An amusement park ride spins you around in a circle of radius 2.5 m with a speed of 8.5 m/s. If your mass is 75 kg, what is the

zhannawk [14.2K]

B is the answer to this truly did the math

3 0

2 years ago

Read 2 more answers

7. Imagine you are pushing a 15 kg cart full of 25 kg of bottled water up a 10o ramp. If the coefficient of friction is 0.02, wh

pentagon [3]

Answer:

The frictional force needed to overcome the cart is 4.83N

Explanation:

The frictional force can be obtained using the following formula:

$F= \mu R$

where $\mu$ is the coefficient of friction = 0.02

R = Normal reaction of the load = $mgcos\theta$ = $25 \times 9.81 \times cos 10$ = $241.52N$

Now that we have the necessary parameters that we can place into the equation, we can now go ahead and make our substitutions, to get the value of F.

$F=0.02 \times 241.52N$

F = 4.83 N

Hence, the frictional force needed to overcome the cart is 4.83N

4 0

2 years ago

A sample of nitrogen gas exerts a pressure of 9.80 atm at 32 C. What would its temperature be (in C) when its pressure is increa

Harlamova29_29 [7]

Answer:

T₂ = 111.57 °C

Explanation:

Given that

Initial pressure P₁ = 9.8 atm

T₁ = 32°C = 273 + 32 =305 K

The final pressure P₂ = 11.2 atm

Lets take the final temperature = T₂

We know that ,the ideal gas equation

If the volume of the gas is constant ,then we can say that

$\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}$

$T_2=\dfrac{P_2}{P_1}\times T_1$

Now by putting the values in the above equation ,we get

$T_2=\dfrac{11.2}{9.8}\times 305\ K$

$T_2=348.57\ K$

T₂ = 384.57 - 273 °C

T₂ = 111.57 °C

3 0

2 years ago

A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turn

Cerrena [4.2K]

Answer:

e. Not enough information to determine

Explanation:

This question is incomplete. Here is the complete question with my solution afterwards;

A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turntable (initially at rest) begins to rotate with its rate of rotation constantly increasing.

What is the first event that will occur?(Assume non-zero frictional force and the same coefficients of friction for both bugs.)

a. The ladybug begins to slide

b. The gentleman bug begins to slide

c. Both bugs begin to slide at the same time

d. Nothing ever happens

e. Not enough information to determine

The centripetal force acting on a rotating body or bugs can be written as,

$F=mrw^2$

$m=$ mass of the corresponding bugs

$r=$ corresponding radial distance of each bug

$w=$ angular speed of the turntable

The centripetal force tries to slide the bugs in an outward direction and it is directly proportional to the products of its mass and radial distance from the axis of rotation of the turntable

$F$ ∝ $mr$

Since the radial distance from the axis of rotation of the turntable for each bug is given, but the mass is not given, the given information is therefore not enough to determine which bugs will slide first.

Option "e" is correct.

7 0

2 years ago

Read 2 more answers