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2 years ago

Please help its for science

Physics

2 answers:

elena-14-01-66 [18.8K]2 years ago

7 0

Answer:

reflected by clouds so a is correct

vampirchik [111]2 years ago

3 0

Answer:

This answers idReflected by clouds

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Some help please ,.......

serious [3.7K]

It's c bro, .2x5 equals 1, which accounts for the 1m/s accelerations.

4 0

3 years ago

Read 2 more answers

Physics help please

zhuklara [117]

Answer: 37.981 m/s

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the ball has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$x=52 m$ is the point where the ball strikes ground horizontally

$V_{o}$ is the ball's initial speed

$\theta=0$ because we are told the ball is thrown horizontally

$t$ is the time since the ball is thrown until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=120m$ is the initial height of the ball

$y=0$ is the final height of the ball (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Knowing this, let's start by finding $t$ from (2):

$0=y_{o}+V_{o}sin(0\°) t+\frac{gt^{2}}{2}$ (3)

$0=y_{o}+\frac{gt^{2}}{2}$

$t=\sqrt{\frac{-2 y_{o}}{g}}$ (4)

$t=\sqrt{\frac{-2 (120 m)}{-9.8m/s^{2}}}$ (5)

$t=4.948 s$ (6)

Then, we have to substitute (6) in (1):

$x=V_{o}cos(0\°) t$ (7)

And find $V_{o}$ :

$V_{o}=\frac{x}{t}$ (8)

$V_{o}=\frac{52 m}{4.948 s}$ (9)

$V_{o}=10.509 m/s$ (10)

On the other hand, since we are dealing with constant acceleration (due gravity) we can use the following equation to find the value of the ball's final velocity $V$ :

$V=V_{o} + gt$ (11)

$V=10.509 m/s + (-9.8 m/s^{2})(4.948 s)$ (12)

$V=-37.981 m/s$ (13) This is the ball's final velocity, and the negative sign indicates its direction is downwards.

However, we were asked to find the ball's final speed, which is the module of the ball's final vleocity vector. This module is always positive, hence the speed of the ball just before it strikes the ground is 37.981 m/s (positive).

5 0

3 years ago

What is the net charge of a metal ball if there are 21,749 extra electrons in it?

pickupchik [31]

Answer:

$Q=3.47\times 10^{-15}\ C$

Explanation:

Given that,

Number of extra electrons, n = 21749

We need to find the net charge on the metal ball. Let Q is the net charge.

We know that the charge on an electron is $q=1.6\times 10^{-19}\ C$

To find the net charge if there are n number of extra electrons is :

Q = n × q

$Q=21749\times 1.6\times 10^{-19}\ C$

$Q=3.47\times 10^{-15}\ C$

So, the net charge on the metal ball is $3.47\times 10^{-15}\ C$ . Hence, this is the required solution.

6 0

3 years ago

If you put a object in front of the mirror why does it appear the opposite way

UNO [17]

It is because that is how mirrors work, they reflect light, and since we see objects because we are seeing the light these objects reflect, what is reflected back by the mirror is what we see.

4 0

3 years ago

A skateboarder has an acceleration of −1.9 m/s2. If her initial speed is 6 m/s, how long does it take her to stop?

SSSSS [86.1K]

You said that she's losing 1.9 m/s of her speed every second.

So it'll take

(6 m/s) / (1.9 m/s²) = 3.158 seconds (rounded)

to lose all of her initial speed, and stop.

6 0

3 years ago