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2 years ago

A drive find difficulty to drive a car when the ground is cover with ice explain why?

1 answer:

7 0

The ground is less rough, so there is less friction between the ground and the wheels of the car. This makes it unable for the car to move forward

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Calculate the pressure on a man’s foot when a woman who weighs 520 N steps on his foot with her heel which has an area of 0.001

Vinvika [58]

Answer:

520000 or 520000 pa

Force = 520N

Area of contact = 0.001

Pressure: 520000 or 520000

6 0

2 years ago

8 colors according to their alphabetical order

damaskus [11]

Answer:

8 colors in alphabetical order are

Black

blue

brown

Green

Orange

Pink

Purple

Yellow

Explanation:

6 0

2 years ago

How long will it be before 82.0<br> % of an isotope with a half life of 1250<br> y decays?

Solnce55 [7]

Can you give more information

5 0

2 years ago

A point charge of -0.70 μC is fixed to one corner of a square. An identical charge is fixed to the diagonally opposite corner. A

MakcuM [25]

Answer:

The magnitude and algebraic sign of q is $14\sqrt{2}\ \mu C$

Explanation:

Given that,

Point charge = -0.70 μC[/tex]

We need to calculate the force for all charges

The electric force at first corner

$F_{1}=\dfrac{-k0.70\times10^{-6}q}{r^2}$

The electric force at opposite corner

$F_{3}=\dfrac{-k0.70\times10^{-6}q}{r^2}$

The net force is

$F=\sqrt{F_{1}^2+F_{2}^2}$

Put the value into the formula

$F=\sqrt{(\dfrac{-k0.70\times10^{-6}q}{r^2})^2+(\dfrac{-k0.70\times10^{-6}q}{r^2})^2}$

The electric force at second corner

$F_{2}=\dfrac{-kq^2}{2r^2}$

The net force acting on either of the charges is zero.

So, $F=F'$

$\sqrt{(\dfrac{k0.70\times10^{-6}q}{r^2})^2+(\dfrac{-k0.70\times10^{-6}q}{r^2})^2}=\dfrac{kq^2}{2r^2}$

$\sqrt{2}\times\dfrac{0.70\times10^{-6}kq}{r^2}=\dfrac{kq^2}{2r^2}$

$q=14\sqrt{2}\ \mu C$

Hence, The magnitude and algebraic sign of q is $14\sqrt{2}\ \mu C$

8 0

2 years ago

it's nighttime, and you've dropped your goggles into a swimming pool that is 3.2 m deep. If you hold a laser pointer 1.0 m above

Scrat [10]

Answer:

5.2 m

Explanation:

from the question we are given the following

depth of pool (d) = 3.2 m

height of laser above the pool (h) = 1 m

point of entry of laser beam from edge of water (l) = 2.5 m

we first have to calculate the angle at which the laser beam enters the water (∝),

tan ∝ = \frac{1}{2.2}

∝ = 24.44 degrees

from the attached diagram, the angle with the normal (i) = 90 - 24.4 = 65.56 degrees

lets assume it is a red laser which has a refractive index of 1.331 in water, and with this we can find the angle of refraction (r) using the formula below

refractive index = \frac{sin i}{sin r}

1.331 = \frac{sin 65.56}{sin r}

r = 43.16 degrees

we can get the distance (x) from tan r = \frac{x}{3.2}

tan 43.16 = \frac{x}{3.2}

x = 3 m

To get the total distance we need to add the value of x to 2.2 m

total distance = 3 + 2.2 = 5.2 m

3 0

2 years ago