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2 years ago

1. You walk along a long straight school corridor for 55

Physics

1 answer:

Sedaia [141]2 years ago

3 0

Explanation:

14 m is the displacement and towards the north

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Which are the three types of mutations?

WITCHER [35]

Replication, Multiplication, and Substitution.

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3 years ago

Find the net electric force that the two charges would exert on an electron placed at point on the xx-axis at xx = 0.200 mm. Exp

UkoKoshka [18]

Answer:

The question has some details missing, here is the complete question ; A -3.0 nC point charge is at the origin, and a second -5.0nC point charge is on the x-axis at x = 0.800 m. Find the net electric force that the two charges would exert on an electron placed at point on the x-axis at x = 0.200 m.

Explanation:

The application of coulonb's law is used to approach the question as shown in the attached file.

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2 years ago

A 5.00 kg crate is on a 21.0 degree hill. Using X-Y axes tilted down the plane, what is the y-component of the normal force?

Rama09 [41]

Answer:

The y-component of the normal force is 45.74 N.

Explanation:

Given that,

Mass of the crate, m = 5 kg

Angle with hill, $\theta=21^{\circ}$

We need to find the y component of the normal force. We know that the y component of the normal force is given by :

$F_y=F\ \cos\theta\\\\F_y=mg\ \cos\theta\\\\F_y=5\times 9.8\ \cos(21)\\\\F_y=45.74\ N$

So, the y-component of the normal force is 45.74 N. Hence, this is the required solution.

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2 years ago

What is the temperature at which a solid melts?

Diano4ka-milaya [45]

Generally speaking, solid turns to a liquid at it's melting point. Ice turns to water at 0 degrees Celcius. Chocolate melts at 25 degrees Celcius-Yum! Ice (solid) thaws when the temperature rises above 32 degrees Fahrenheit, becoming water (liquid). Other solids (oddly) vary. your welcome

3 0

3 years ago

The driver of a car slams on the brakes, causing the car to slow down at a rate of 17ft/s2 as the car skids 285ft to a stop.

ozzi

1) 5.79 s

2) 98.4 ft/s

Explanation:

The motion of the car is a uniformly accelerated motion (it means it travels with constant acceleration), so we can find the time it takes for the car to stop by using the following suvat equation:

$s=vt-\frac{1}{2}at^2$

where

s is the distance travelled

v is the final velocity

t is the time

a is the acceleration of the car

In this problem we have:

s = 285 ft is the distance travelled

$a=-17 ft/s^2$ is the acceleration of the car (negative since the car is slowing down)

v = 0 ft/s is the final velocity of the car, since it comes to a stop

Solving for t, we find:

$t=\sqrt{\frac{-2s}{a}}=\sqrt{\frac{-2(285)}{-17}}=5.79 s$

The initial speed of the car can be found by using another suvat equation, namely:

$v=u+at$

where

v is the final speed

u is the initial speed

a is the acceleration

t is the time

In this problem, we have:

v = 0 is the final speed of the car

$a=-17 ft/s^2$ is the acceleration of the car (negative since the car is slowing down)

t = 5.79 s is the total time of motion (found in part 1)

Therefore, the initial speed of the car is:

$u=v-at=0-(-17)(5.79)=98.4 ft/s$

8 0

3 years ago