Ask question

Ask question

All categories

3 years ago

11

An object is placed at 0 on a line. It moves 3 units to the rights then 4 to the left and then 6 units to the rigth.the displace

ment of the object is

1 answer:

Maslowich3 years ago

8 0

It would be five, because the object would end up five units away from 0. I just did this assignment lol, good luck!

You might be interested in

According to Newton’s law of universal gravitation, which statements are true?

andreyandreev [35.5K]

Before we solve this, we should know this fact:

According to Newton's Law of Gravitation, the force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The force acts along the line joining the centres of the two objects. It can be shown by this:

$F ∝ \frac{Mm}{ {d}^{2} }$

Now, let us check all the options.

A. As we move to higher altitudes, the force of gravity on us decreases.

<em>This </em><em>statement </em><em>is </em><em>true.</em>

The force of gravity is inversely proportional to the square of distance from the centre of the earth. If, we go up the surface of the earth, the distance from the centre of the earth increases and hence the value of force of gravity decrease. So, force of gravity decreases with altitude.

B. As we move to higher altitudes, the force of gravity on us increases.

<em>This </em><em>statement</em><em> </em><em>is </em><em>false.</em>

We have already got the result in option A. that the force of gravity decreases with altitude. It never increases with altitude.

C. As we gain mass, the force of gravity on us decreases.

<em>This </em><em>statement</em><em> </em><em>is </em><em>false.</em>

The force of gravity is directly proportional to the product of the masses. So, if increase our mass, then the force of gravity will also increase and if we decrease our mass, then the force of gravity decreases.

D. As we gain mass, the force of gravity on us increases.

<em>This </em><em>statement</em><em> is</em><em> </em><em>true.</em>

As mentioned earlier in option C., the force of gravity is directly proportional to the product of the masses of the earth and another object. So, as we gain mass, the force of gravity on us increases.

E. As we move faster, the force of gravity on us increases.

<em>This </em><em>statement</em><em> is</em><em> </em><em>true</em><em>.</em>

Here, we have to consider a different formula. According to Newton's Second Law,

F = ma, where F is the force, m is the mass and a is the acceleration.

In other words,

F ∝ a, i.e., force is directly proportional to acceleration.

We know, acceleration is the rate of change of velocity of an body within a time period.

So, if speed is increased, then acceleration will also be greater, which results in the increase of force. So, as we move faster, the force of gravity on us increases.

<u>Answers:</u>

A: As we move to higher altitudes, the force of gravity on us decreases.

D: As we gain mass, the force of gravity on us increases.

E: As we move faster, the force of gravity on us increases.

Hope you could understand.

If you have any query, feel free to ask.

7 0

2 years ago

What is the net force exerted by these two charges on a third charge q3 = 49.5 nC placed between q1 and q2 at x3 = -1.170 m ? Yo

insens350 [35]

Full Question

Consider two point charges located on the x axis: one charge, Q1 = -12.0 nC , is located at x1 = -1.705m ; the second charge, Q2 = 36.5 nC, is at the origin (x=0.0000).

What is the net force exerted by these two charges on a third charge q3 = 49.5 nC placed between q1 and q2 at x3 = -1.170 m ? Your answer may be positive or negative, depending on the direction of the force.

Answer:

6.79E6 N

Explanation:

Given

Q1 is negative and to the left of Q3 the force will be to the left

Q2 is positive and to the right of Q3 the the force will also be to the left

Net Force is calculated as:

Using Coulomb's law

Coulomb's law: F = kqQ / r²

the constant k = 8.99 x 10^9 N m2 / C2

F = -kQ1*Q3/(r1)² -kQ2*Q3/(r2)²)

F = -kQ3(Q1/(r1)² + Q2/(r2)²)

Where

Q1 = -12nC = -12 * 10^-9C

Q2 = 36.5nC = 36.5 * 10^-9C

Q3 = 49.5nC = 49.5 * 10^-9C

x1 = -1.705m

x2 = x = 0

x3 = -1.170m

r1 = x3 - x1

r1 = -1.170 - -1.705

r1 = -1.170 + 1.705

r1 = 0.535

r1² = 0.286225

r2 = x3

r2 = -1.170

r2² = -1.170²

r2² = 1.3689

So,

F = (-8.99 * 10^9)(49.5 *10^-9) [-12 * 10^-9/0.286225 + 36.5 * 10^-9/1.3689]

F = -445.005 (−4.192505895711E−8 + 2.6663744612462E−8)

F = -445.005 * −1.5261314344648E−8

F = -(8.99 * 10^9) * (49.5 * 10^-9) * [ (-12 * 10^-9) /(-1.770 - -1.705)² + (36.5 * 10^-9)/(-1.170)²]

F = -445.005( −0.000002813572941777538)

F = 0.00000679136118994008324

F = 6.79E6 N

3 0

3 years ago

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km fr

mestny [16]

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030 kg ears is 6.0 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 cm closer.

What is the tension between her ears?

Would the astronaut find it difficult to keep from being torn apart by the gravitational forces?

Answer:

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

Explanation:

Given that:

Orbital radius of the spacecraft (R) = 120 Km = 120 × 10³ m

Mass of the black hole (m) = $5 \ * (M \ _{sun})$

where : $M_{sun} = 1.99*10^{33} \ kg$

Then; we have:

$m = 5*(1.99*10^{30} \ kg ) \\ = 9.95*10^{30} kg$

Schwarzchild radius of the black hole

r - 15.0 km

Mass of each ear $m_{ear} = 0.030 \ kg$

Farther distance between one ear and the black hole (d) = 6.0 cm

= 0.06 m

Closer distance between the other ear and the black home is (d) 6.0 cm

= 0.6 cm

NOW, If we assume that the tension force should be T; then definitely the two ears will posses the same angular velocity .

The net force on the ear closer to the black hole will be:

$\frac{GMm_{ear} }{(R-d)}- T = m_{ear} (R - d) \omega^2$

$\frac{GMm_{ear} }{(R-d)^2}- \frac{T}{(R-d)} = m_{ear} \omega^2 \ ----> \ (1)$

The net force on the ear farther to the black hole is :

$\frac{GMm_{ear} }{(R+d)}- T = m_{ear} (R + d) \omega^2$

$\frac{GMm_{ear} }{(R+d)^2}- \frac{T}{(R+d)} = m_{ear} \omega^2 \ ----> \ (2)$

Equating equation (1) and (2) & therefore making (T) the subject of the formula; we have:

$T = \frac{3GMm_{ear}d}{R^3}$

$T = \frac{3(6.67*10^{-11}N.m^2/kg^2)(1.95*10^{30}kg)(0.03kg)(0.06m)}{(120*10^3m)^3}$

$T = 2073.9 N\\T = 2.07 KN$

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

3 0

4 years ago

Which are the three types of mutations?

WITCHER [35]

Replication, Multiplication, and Substitution.

7 0

4 years ago

What causes tendonitis?

Irina18 [472]

overuse of a muscle Answer:

Explanation:

6 0

3 years ago

Read 2 more answers