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2 years ago

If two people standing on a scooter board push off of each other what happens ( Newton’s 3rd law )

Physics

1 answer:

Keith_Richards [23]2 years ago

7 0

Answer:They will go off in opposite directions with the same force.

Explanation:

Newton's Third Law states that for every action, there is an equal but opposite reaction. If two people are standing on a scooter and push off each other (following the Law), it should come to the conclusion that they could go off in opposite with the same force.

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A 0.5-kg ball moving at 5 m/s strikes a wall and rebounds in the opposite direction with a speed of 2 m/s. If the impulse occurs

mart [117]

Given that:

mass of the ball (m) = 0.5 Kg ,

ball strikes the wall (v₁) = 5 m/s ,

rebounds in opposite direction (v₂) = 2 m/s,

time duration (t) = 0.01 s,

Determine the force (F) = ?

We know that from Newton's II law,

F = m. a Newtons

(velocity acting in opposite direction, so a = ( (v₁ + v₂)/t

= m × (v₁ + v₂)/t

= 0.5 × (5 + 2)/0.01

= 350 N

The force acting up on the ball is 350 N

6 0

3 years ago

A point charge (–5.0 µC) is placed on the x axis at x = 4.0 cm, and a second charge (+5.0 µC) is placed on the x axis at x = –4.

AveGali [126]

Answer:

The magnitude of electric force is $7.2\times10^{-3} N$

Explanation:

Coulomb's Law:

The force of attraction or repletion is

directly proportional to the products of charges i.e $F\propto q_1q_2$

inversely proportional to the square of distance i.e $F\propto \frac{1}{r^2}$

$\therefore F\propto \frac{q_1q_2}{r^2}$

$\Rightarrow F=k \frac{q_1q_2}{r^2}$ [ k is proportional constant=9×10⁹N m²/C²]

There are two types of force applied on Q=+2.5 μC=2.5×10⁻⁶ C

Let F₁ force be applied on Q =+2.5 μC by q₁= -5.0 μC = - 5.0×10⁻⁶ C

and F₂ force be applied on Q=+2.5 μC by q₂= 5.0 μC= 5.0×10⁻⁶ C

Since the magnitude of F₁ and F₂ are same. Therefore their y component cancel.

If we draw a line from q₁ to Q .

The it forms a triangle whose base = 4.0 cm and altitude =3.0 cm.

Let hypotenuse = r

Therefore, $r=\sqrt{altitude^2+base^2} =\sqrt{3^2+4^2} =5$

we know,

$cos \theta = \frac{base }{hypotenuse}$

$\Rightarrow cos \theta = \frac{4 }{r}$

Total force $F_Q = 2.F_1 cos\theta \hat{i}$

$=2k\frac{Qq_1}{r^2} cos\theta \hat i$

$=2\ \frac{9\times1 0^9\times2.5 \times 5\times 10^{-12}}{r^2} \frac{4}{r} \hat i$

$=8\ \frac{9\times10^9\times2.5 \times 5\times 10^{-12}}{5^3} \hat i$ [ r=5]

$=7.2\times10^{-3}\hat i$ N

The magnitude of electric force is $7.2\times10^{-3} N$

3 0

3 years ago

What may result when the energy that builds up at plate boundaries is released because the plates suddenly overcome the force of

Arlecino [84]

It has to be Earthquake

4 0

2 years ago

Read 2 more answers

You normally drive a 12-h trip at an average speed of 100 km/h . Today you are in a hurry. During the first two-thirds of the di

kherson [118]

Answer:

78 km/h

Explanation:

If I normally drive a 12 hour trip at an average speed of 100 km/h, my destination has a total distance of:

100 km/h · 12 h = 1,200 km

Today, I drive the first 2/3 of the distance at 116 km/h. Let's first calculate what 2/3 of the normal distance is.

1,200 * 2/3 = 800 km

I've driven 800 km already. I need to drive 400 km more to reach my final destination. I need to figure out my average speed during this last 1/3 of the distance.

To do this, I first need to calculate how much time I spent driving 116 km/h for the past 800 km.

116 km/1 h = 800 km/? h
800 = 116 · ?
? = 800/116
? = 6.89655172

I spent 6.89655172 hours driving during the first 2/3 of the distance.

Now, I need to subtract this value from 12 hours to find the remaining time I have left.

12 h - 6.89655172 h = 5.10344828 h

Using this remaining time and my remaining distance, I can calculate my average speed.

? km/1 hr = 400 km/5.10344828 h
5.10344828 · ? = 400
? = 400/5.10344828
? = 78.3783783148

My average speed during the last third of the distance is around 78 km/h.

8 0

2 years ago

A quantity of a gas is an absolute pressure of 400 K PA in the absolute temperature of 110° Kelvin when the temperature of the g

aliina [53]

Given:

P1 = 400 kPa
T1 = 110 K

T2 = 235K

Required:

Solution:

Apply Gay-Lussac’s law where P/T = constant

P1/T1 = P2/T2

P2 = T2P1/T1

P2 = (235K)(400kPa) / (110K)

P2 = 855 kPa

4 0

3 years ago