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3 years ago

14

A block slides down a frictionless plane having an inclination of 15.0°. The block starts from rest at the top, and the length o

f the incline is 2.00 m.Find its speed when it reaches the bottom of the incline.

1 answer:

nalin [4]3 years ago

8 0

Answer: check the pic

Explanation:

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For elevator questions, does the weight of a person increase when the elevator is going up or if it's going down-

TiliK225 [7]

The weight of a person increase when the elevator is going up.

<h3>Weight of the person in the elevator</h3>

The weight of the person in the elevator is calculated as follows;

<h3>When the person is going up</h3>

F = ma + mg

F = m(a + g)

where;

a is acceleration of the person
g is acceleration due to gravity

<h3>When the person is going down</h3>

F = mg - ma

F = m(g - a)

Thus, the weight of a person increase when the elevator is going up.

Learn more about weight here: brainly.com/question/2337612

#SPJ1

6 0

2 years ago

Will give correct answer brainliest

sineoko [7]

$200 \times 80\%$

<h2><em>calculate</em></h2>

<em> $200 \times \frac{80}{100}$ </em>

<h2><em>reduce </em><em>the </em><em>numbers</em></h2>

<em> $2 \times 80$ </em>

<h2><em>multiply</em></h2>

<em> $= 160$ </em>

<h2><em>there </em><em>for </em><em>we </em><em>have </em><em>a </em><em>solution</em><em> to</em><em> the</em><em> </em><em>equation</em></h2>

<em>hope </em><em>it</em><em> helps</em>

<em>#</em><em>c</em><em>a</em><em>r</em><em>r</em><em>y</em><em> </em><em>on</em><em> learning</em>

<em>mark </em><em>me</em><em> as</em><em> brainlist</em><em> plss</em>

6 0

3 years ago

Read 2 more answers

Two massless bags contain identical bricks, each brick having a mass M. Initially, each bag contains four bricks, and the bags m

stepladder [879]

Answer: $F_{2}=\frac{3}{4}F_{1}$

Explanation:

According to Newton's law of universal gravitation:

$F=G\frac{m_{1}m_{2}}{r^2}$

Where:

$F$ is the module of the force exerted between both bodies

$G$ is the universal gravitation constant.

$m_{1}$ and $m_{2}$ are the masses of both bodies.

$r$ is the distance between both bodies

In this case we have two situations:

1) Two bags with masses $4M$ and $4M$ mutually exerting a gravitational attraction $F_{1}$ on each other:

$F_{1}=G\frac{(4M)(4M)}{r^2}$ (1)

$F_{1}=G\frac{16M^2}{r^2}$ (2)

$F_{1}=16\frac{GM^2}{r^2}$ (3)

2) Two bags with masses $2M$ and $6M$ mutually exerting a gravitational attraction $F_{2}$ on each other (assuming the distance between both bags is the same as situation 1):

$F_{2}=G\frac{(2M)(6M)}{r^2}$ (4)

$F_{2}=G\frac{12M^2}{r^2}$ (5)

$F_{2}=12\frac{GM^2}{r^2}$ (6)

Now, if we isolate $\frac{GM^2}{r^2}$ from (3):

$\frac{F_{1}}{16}=\frac{GM^2}{r^2}$ (7)

Substituting $\frac{GM^2}{r^2}$ found in (7) in (6):

$F_{2}=12(\frac{F_{1}}{16})$ (8)

$F_{2}=\frac{12}{16}F_{1}$ (9)

Simplifying, we finally get the expression for $F_{2}$ in terms of $F_{1}$ :

$F_{2}=\frac{3}{4}F_{1}$

5 0

3 years ago

3. A bunny and a tortoise start a race from rest. The bunny accelerates at a rate ab for a time to until it reaches its

Aleks [24]

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7 0

2 years ago

George walks to a friend's house. He walks 850 meters north before he realizes he walked too far.

anyanavicka [17]

Answer:

<h2>44 m/s</h2>

Explanation:

In this problem we are expected to calculate the velocity of Georges movements.

Given data

Total distance covered by George= 850+250= 1100 meters

Time taken by George to cover the total distance= 25 seconds

We know that velocity is, v= distance/ time

Therefore substituting our data into the expression for velocity we have

v= 1100/ 25= 44 m/s

Hence the velocity in m/s is 44

7 0

3 years ago