2

if a moving object travels north for a distance of 105 m in 22 sec, what is it’s speed and velocity ?

artcher [175]

Answer:

Speed: 4.8 m/s

Velocity: 4.8 m/s north

Explanation:

Definitions:

- Speed is a scalar quantity, which is equal to the ratio between distance covered (d) and time taken (t):

$s=\frac{d}{t}$

- Velocity is a vector quantity, whose magnitude is equal to the ratio between the displacement of the object and the time taken:

$v=\frac{disp.}{t}$

And it also has a direction (the same as the displacement).

In this problem:

- The object travels a distance of

d = 105 m

In a time interval of

t = 22 s

So its speed is

$s=\frac{105}{22}=4.8 m/s$

- The displacement of the object is the same as the distance in this case, so still 105 m, covered in a time interval of 22 s; this means that the magnitude of the velocity is the same as the speed:

$v=4.8 m/s$

However, velocity is a vector quantity, so it also has a direction: and since the object has moved north, the direction of the velocity is north as well.

3 0

3 years ago

One end of a 34-m unstretchable rope is tied to a tree; the other end is tied to a car stuck in the mud. The motorist pulls side

anyanavicka [17]

Answer:

Fc = 89.67N

Explanation:

Since the rope is unstretchable, the total length will always be 34m.

From the attached diagram, you can see that we can calculate the new separation distance from the tree and the stucked car H as follows:

L1+L2=34m

$L1^2=L2^2=L^2=2^2+(H/2)^2$ Replacing this value in the previous equation:

$\sqrt{2^2+H^2/4}+ \sqrt{2^2+H^2/4}=34$ Solving for H:

$H=\sqrt{52}$

We can now, calculate the angle between L1 and the 2m segment:

$\alpha = atan(\frac{H/2}{2})=60.98°$

If we make a sum of forces in the midpoint of the rope we get:

$-2*T*cos(\alpha ) + F = 0$ where T is the tension on the rope and F is the exerted force of 87N.

Solving for T, we get the tension on the rope which is equal to the force exerted on the car:

$T=Fc=\frac{F}{2*cos(\alpha) } = 89.67N$

7 0

3 years ago

A point charge A of charge +4micro coloumb and another B of -1 micro coloumb are placed at a distance in air 1m apart then the d

andrew11 [14]

Answer:

Explanation:

Given that,

A point charge is placed between two charges

Q1 = 4 μC

Q2 = -1 μC

Distance between the two charges is 1m

We want to find the point when the electric field will be zero.

Electric field can be calculated using

E = kQ/r²

Let the point charge be at a distance x from the first charge Q1, then, it will be at 1 -x from the second charge.

Then, the magnitude of the electric at point x is zero.

E = kQ1 / r² + kQ2 / r²

0 = kQ1 / x² - kQ2 / (1-x)²

kQ1 / x² = kQ2 / (1-x)²

Divide through by k

Q1 / x² = Q2 / (1-x)²

4μ / x² = 1μ / (1 - x)²

Divide through by μ

4 / x² = 1 / (1-x)²

Cross multiply

4(1-x)² = x²

4(1-2x+x²) = x²

4 - 8x + 4x² = x²

4x² - 8x + 4 - x² = 0

3x² - 8x + 4 = 0

Check attachment for solution of quadratic equation

We found that,

x = 2m or x = ⅔m

So, the electric field will be zero if placed ⅔m from point charge A, OR ⅓m from point charge B.

5 0

3 years ago

I WILL MARK BRAINLIEST!

xxMikexx [17]

Answer:

Explanation:

The answer is C

7 0

3 years ago

In this week's experiment, you send an unpolarized light beam through two polarizers in series. You orient the first polarizer s

g100num [7]

Answer:

Part a)

$\theta = 90 degree$

so the polarization axis of two polarizers must be at 90 degree

Part b)

$\theta = 20 + 14.3 = 34.3 degree$

(vi) set the second polarizer so that its angular scale reads 34.3 degrees.

Explanation:

Part 1)

As per law of Malus we know that the intensity of light coming from the second polarizer and the intensity of the light from first polarizer is related as

$I = I_o cos^2\theta$

now we know that we rotate the second polarizer till the intensity of the light becomes zero

so we will have

$I = 0$

so we will have

$\theta = 90 degree$

so the polarization axis of two polarizers must be at 90 degree

Part b)

when two axis are inclined at 90 degree then scale reads 104.3 degree

so here the scale exceeds the reading by

$\Delta \theta = 104.3 - 90 = 14.3$

so in order to make them inclined at 20 degree we will have

$\theta = 20 + 14.3 = 34.3 degree$

(vi) set the second polarizer so that its angular scale reads 34.3 degrees.

7 0

4 years ago