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3 years ago

14

Consider a high-speed bus colliding head-on with an innocent bug. The force of impact splatters the unfortunate bug over the win

dshield. According to Newton's 3rd Law, which is greater, the force on the bug or the force on the bus?

1 answer:

Alex17521 [72]3 years ago

3 0

Answer:

The force on the bug and the force on the bus are the same.

Explanation:

Newton's Third Law of Motion states that, "For every action, there is an equal and opposite reaction" or it can be restated that If a high-speed bus exerts a force on an innocent bug, then the innocent bug must exert a force of equal magnitude and opposite direction back on the high-speed bus.

Note: though the forces are equal but the effect are not, in this case the innocent bug was splattered but the bus was not destroyed.

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Adding resistors in series changes the total resistance of a circuit by

skelet666 [1.2K]

Answer:

Increasing the resistance

Explanation:

6 0

3 years ago

Three particles lie in the xy plane. Particle 1 has mass m1 = 6.7 kg and lies on the x-axis at x1 = 4.2 m, y1 = 0. Particle 2 ha

krek1111 [17]

Answer:

$F=18.58\times 10^{-11}\ N$

$\theta=30.276^{\circ}$

Explanation:

Given:

mass of first particle, $m_1=6.7\ kg$

mass of second particle, $m_2=5.1\ kg$

mass of third particle, $m_3=3.7\ kg$

coordinate position of first particle in meters, $(x_1,y_1)\equiv(4.2,0)$

coordinate position of second particle in meters, $(x_2,y_2)\equiv(0,2.8)$

coordinate position of third particle in meters, $(x_3,y_3)\equiv(0,0)$

<u>Now, gravitational force on particle 3 due to particle 1:</u>

$F_{31}=G\frac{m_1.m_3}{r_{31}^2}$

$F_{31}=6.67\times 10^{-11} \times \frac{6.7\times 3.7}{4.2^2}$

$F_{31}=9.37\times 10^{-11}\ N$

towards positive Y axis.

<u>gravitational force on particle 3 due to particle 2:</u>

$F_{32}=G\frac{m_2.m_3}{r_{21}^2}$

$F_{32}=6.67\times 10^{-11} \times \frac{5.1\times 3.7}{2.8^2}$

$F_{32}=16.05\times 10^{-11}\ N$

towards positive X axis.

<u>Now the net force</u>

$F=\sqrt{F_{31}\ ^2+F_{32}\ ^2}$

$F=\sqrt{(10^{-11})^2(9.37^2+16.05^2)}$

$F=18.58\times 10^{-11}\ N$

<em>For angle in counterclockwise direction from the +x-axis</em>

$tan\theta=\frac{9.37\times 10^{-11}}{16.05\times 10^{-11}}$

$\theta=30.276^{\circ}$

4 0

3 years ago

A turtle takes 3.5 minutes to walk 18 m toward the south along a deserted highway. A truck driver stops and picks up the turtle.

Akimi4 [234]

Answer:

3.59 m/s

Explanation:

The average velocity is defined as:

$\frac{Total displacement}{Total time}$

The turtle first walks 18m south, and then is taken 1,1Km (or 1100m) north. Thus, the total displacement is 1082m north (1100m north - 18m south).

Now we have to calculate the total time, which will be equal to the sum of the time the turtle walked and the time it was taken by truck.

The walking time is 3.5 minutes. Since 1 minute = 60 seconds, then the walking time is 210 seconds.

To calculate the truck time we use the equation:

Time = $\frac{Distance}{Speed}$

Where the distance the truck travelled is 1100m and the speed of the truck is 12m/s.

Thus,

Truck time= $\frac{1100m}{12m/s}$ = 91.67s

The total time is the sum of the walking time and the truck time.

Total time = 210s + 91.67s = 301.67s.

As mencioned previously, the average velocity is equal to total displacement/ total time, thus:

Average velocity = $\frac{1082m}{301.67s}$ = 3.59 m/s North

Since the average velocity is a vector, it has a magnitude and a direction. In this case the magnitude is 3.59 m/s and the direction is north since the turtle's final displacement is north of where it started.

8 0

3 years ago

How can you find the reading of main scale and vernier scale

anygoal [31]

Answer:

this pdf should help you out

Explanation:

7 0

2 years ago

An electric heater containing two heating wires X and Y is connected to a power supply of electromotive force(emf) 9.0V and negl

mina [271]

Answer:

0.4 ohms.

Explanation:

From the circuit,

The voltage reading in the voltmeter = voltage drop across each of the parallel resistance.

1/R' = 1/R1+1/R2

R' = (R1×R2)/(R1+R2)

R' = (2.4×1.2)/(2.4+1.2)

R' = 2.88/3.6

R' = 0.8 ohms.

Hence the current flowing through the circuit is

I = V'/R'................ Equation 1

Where V' = voltmeter reading

I = 6/0.8

I = 7.5 A

This is the same current that flows through the variable resistor.

Voltage drop across the variable resistor = 9-6 = 3 V

Therefore, the resistance of the variable resistor = 3/7.5

Resistance = 0.4 ohms.

7 0

3 years ago