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2 years ago

I want to take a picture of the question and get the answer

Physics

1 answer:

OleMash [197]2 years ago

5 0

When you take a picture it pulls up people who asked the same question and they’ll have a answer you can use

You might be interested in

usted / el partido de fútbol Usted va al partido de fútbol. Question 1 with 1 blankyo / la biblioteca Question 2 with 1 blanknos

34kurt

Answer:

1- Yo VOY A la biblioteca

2- Nosotros VAMOS A la piscina

3- Mis primas VAN AL gimnasio

4- Tú VAS de excursión

5- Ustedes VAN AL parque municipal

6- Alejandro VA AL museo de ciencias

Explanation:

Las preguntas refieren en su totalidad a la conjugación temporal del verbo "ir". Así, refiere a conjugar dicho verbo en tiempo presente simple, el cual se conjuga de la siguiente manera:

Yo voy, el va, tú vas, nosotros vamos, ellos van, ustedes van

Así, debe interpretarse la persona o personas que realizan la acción, para determinar así como se conjuga dicho verbo.

4 0

2 years ago

An astronaut drops a hammer from 2.0 meters above the surface of the moon. if the acceleration due to gravity on the moon is 1.6

aivan3 [116]

Answer:

1.57 s

Explanation:

Since the motion of the hammer is a uniformly accelerated motion, the distance covered by the hammer in a time t is

$S=\frac{1}{2}at^2$

Where, in this case

S = 2.0 m is the distance covered

a = 1.62 m/s^2 is the acceleration due to gravity

t is the time taken

Re-arranging the equation, we can find the time the hammer takes:

$t=\sqrt{\frac{2S}{a}}=\sqrt{\frac{2(2.0 m)}{1.62 m/s^2}}=1.57 s$

8 0

2 years ago

Read 2 more answers

Even if u dont know the right answer can someone just give me an answer that looks right

Licemer1 [7]

Answer:

The answer is elastic

Explanation:

hope this helps!

3 0

2 years ago

A Tesla Roadster car accelerates from rest at a rate of 7.1m/s for a time of 3.9s

liberstina [14]

The distance covered is 54.0 m

Explanation:

Since the motion of the car is a uniformly accelerated motion, we can use the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where

u is the initial velocity

t is the time

a is the acceleration

s is the distance covered

For the car in this problem, we have

u = 0 (it starts from rest)

$a=7.1 m/s^2$ is the acceleration

t = 3.9 s is the time

Substituting, we find s:

$s=0+\frac{1}{2}(7.1)(3.9)^2=54.0 m$

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

5 0

3 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

3 years ago