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3 years ago

Question 1 (2 points)

Physics

1 answer:

Free_Kalibri [48]3 years ago

8 0

1. A
2. B
3. B
4. A
5. B
6. A
7. C
8. C
9. A
10. D

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Why is a camera lens round but the pictures come out square

sergey [27]

It is round or say spherical to widen the range of photography and it is designed to focus the image as the ray are coming from infinity so !!

I am not an expert of camera but ya it is the main theme !!

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3 years ago

Read 2 more answers

During electrical storms, a bolt of lightning can transfer 10 C of charge in 2.0 µs (the amount of time can vary considerably).

nydimaria [60]

Answer:

Explanation:

charge, q = 10 C

time, t = 2 micro second

Current, i = q / t

i = 10 / (2 x 10^-6) = 5 x 10^6 A

(a)

distance, d = 1 m

the formula for the magnetic field is given by

$B = \frac{\mu_{0}}{4\pi}\frac{2i}{d}$

$B = 10^{-7}\frac{2\times 5\times 10^{6}}{1}$

B = 1 Tesla

Now the distance is d' = 1 km = 1000 m

$B' = \frac{\mu_{0}}{4\pi}\frac{2i}{d}$

$B' = 10^{-7}\frac{2\times 5\times 10^{6}}{1000}$

B' = 0.001 Tesla

(b) The magnetic field of earth is Bo = 3 x 10^-5 tesla

B / Bo = 3.3 x 10^4

B'/Bo = 33.3

4 0

4 years ago

Increasing the concentration of greenhouse gases in Earth's atmosphere decreases the transparency of the atmosphere to infrared

kondaur [170]

Answer:

Decreases the transparency of the atmosphere to infrared light.

Explanation:

When a large amount of green-house gases are present in the atmosphere, the layer of these gases become opaque to infrared radiation and radiation from the sun get trapped into these gases molecules. These excited molecules radiate this energy into our own atmosphere and that why the temperature of Earth is rising due to the Green-House effect.

8 0

3 years ago

A spring is stretched from x=0 to x=d, where x=0 is the equilibrium position of the spring. It is then compressed from x=0 to x=

VARVARA [1.3K]

Answer:

The same amount of energy is required to either stretch or compress the spring.

Explanation:

The amount of energy required to stretch or compress a spring is equal to the elastic potential energy stored by the spring:

$U=\frac{1}{2}k (\Delta x)^2$

where

k is the spring constant

$\Delta x$ is the stretch/compression of the spring

In the first case, the spring is stretched from x=0 to x=d, so

$\Delta x = d-0=d$

and the amount of energy required is

$U=\frac{1}{2}k d^2$

In the second case, the spring is compressed from x=0 to x=-d, so

$\Delta x = -d -0 = -d$

and the amount of energy required is

$U=\frac{1}{2}k (-d)^2= \frac{1}{2}kd^2$

so we see that the amount of energy required is the same.

7 0

4 years ago

A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.90 m/s2

Ahat [919]

Answer:

Approximately $0.608$ (assuming that $g = 9.81\; \rm N\cdot kg^{-1}$ .)

Explanation:

The question provided very little information about this motion. Therefore, replace these quantities with letters. These unknown quantities should not appear in the conclusion if this question is actually solvable.

Let $m$ represent the mass of this car.
Let $r$ represent the radius of the circular track.

This answer will approach this question in two steps:

Step one: determine the centripetal force when the car is about to skid.
Step two: calculate the coefficient of static friction.

For simplicity, let $a_{T}$ represent the tangential acceleration ( $1.90\; \rm m \cdot s^{-2}$ ) of this car.

<h3>Centripetal Force when the car is about to skid</h3>

The question gave no information about the distance that the car has travelled before it skidded. However, information about the angular displacement is indeed available: the car travelled (without skidding) one-quarter of a circle, which corresponds to $90^\circ$ or $\displaystyle \frac{\pi}{2}$ radians.

The angular acceleration of this car can be found as $\displaystyle \alpha = \frac{a_{T}}{r}$ . ( $a_T$ is the tangential acceleration of the car, and $r$ is the radius of this circular track.)

Consider the SUVAT equation that relates initial and final (tangential) velocity ( $u$ and $v$ ) to (tangential) acceleration $a_{T}$ and displacement $x$ :

$v^2 - u^2 = 2\, a_{T}\cdot x$ .

The idea is to solve for the final angular velocity using the angular analogy of that equation:

$\left(\omega(\text{final})\right)^2 - \left(\omega(\text{initial})\right)^2 = 2\, \alpha\, \theta$ .

In this equation, $\theta$ represents angular displacement. For this motion in particular:

$\omega(\text{initial}) = 0$ since the car was initially not moving.
$\theta = \displaystyle \frac{\pi}{2}$ since the car travelled one-quarter of the circle.

Solve this equation for $\omega(\text{final})$ in terms of $a_T$ and $r$ :

$\begin{aligned}\omega(\text{final}) &= \sqrt{2\cdot \frac{a_T}{r} \cdot \frac{\pi}{2}} = \sqrt{\frac{\pi\, a_T}{r}}\end{aligned}$ .

Let $m$ represent the mass of this car. The centripetal force at this moment would be:

$\begin{aligned}F_C &= m\, \omega^2\, r \\ &=m\cdot \left(\frac{\pi\, a_T}{r}\right)\cdot r = \pi\, m\, a_T\end{aligned}$ .

<h3>Coefficient of static friction between the car and the track</h3>

Since the track is flat (not banked,) the only force on the car in the horizontal direction would be the static friction between the tires and the track. Also, the size of the normal force on the car should be equal to its weight, $m\, g$ .

Note that even if the size of the normal force does not change, the size of the static friction between the surfaces can vary. However, when the car is just about to skid, the centripetal force at that very moment should be equal to the maximum static friction between these surfaces. It is the largest-possible static friction that depends on the coefficient of static friction.

Let $\mu_s$ denote the coefficient of static friction. The size of the largest-possible static friction between the car and the track would be:

$F(\text{static, max}) = \mu_s\, N = \mu_s\, m\, g$ .

The size of this force should be equal to that of the centripetal force when the car is about to skid:

$\mu_s\, m\, g = \pi\, m\, a_{T}$ .

Solve this equation for $\mu_s$ :

$\mu_s = \displaystyle \frac{\pi\, a_T}{g}$ .

Indeed, the expression for $\mu_s$ does not include any unknown letter. Let $g = 9.81\; \rm N\cdot kg^{-1}$ . Evaluate this expression for $a_T = 1.90\;\rm m \cdot s^{-2}$ :

$\mu_s = \displaystyle \frac{\pi\, a_T}{g} \approx 0.608$ .

(Three significant figures.)

7 0

3 years ago