Ask question

Ask question

All categories

2 years ago

6

A fast-moving vw beetle traveling at 54 mph hit a mosquito hovering at rest above the road. which bug experienced the largest fo

rce? 1. the insect 2. the vw 3. unable to determine 4. they experienced the same magnitude of force

2 answers:

Ad libitum [116K]2 years ago

8 0

Answer: 4 they both experienced same magnitude of force.

Explanation:

According to Newton's 3rd law of motion,it implies that for every force exerted by an object on another object is the same magnitude directed back.

The force the VW beetle has when it hits the mosquito was the same force exerted on the mosquito on impact.

Therefore the magnitude of their forces are the same.

allsm [11]2 years ago

3 0

1. the insect
because of how fast the car is going and how small they are

You might be interested in

Two men, Joel and Jerry, each pushes an object that are identical on a horizontal frictionless floor starting from rest. Joel an

Nataliya [291]

Answer:

The work done by Joel is greater than the work done by Jerry.

Explanation:

Let suppose that forces are parallel or antiparallel to the direction of motion. Given that Joel and Jerry exert constant forces on the object, the definition of work can be simplified as:

$W = F\cdot \Delta s$

Where:

$W$ - Work, measured in joules.

$F$ - Force exerted on the object, measured in newtons.

$\Delta s$ - Travelled distance by the object, measured in meters.

During the first 10 minutes, the net work exerted on the object is zero. That is:

$W_{net} = W_{Joel} - W_{Jerry}$

$W_{net} = F\cdot \Delta s - F\cdot \Delta s$

$W_{net} = (F-F)\cdot \Delta s$

$W_{net} = 0\cdot \Delta s$

$W_{net} = 0\,J$

In exchange, the net work in the next 5 minutes is the work done by Joel on the object:

$W_{net} = W_{Joel}$

$W_{net} = F\cdot \Delta s$

Hence, the work done by Joel is greater than the work done by Jerry.

7 0

3 years ago

Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,

Marrrta [24]

Answer:

a

Solid Wire $I = 0.01237 \ A$

Stranded Wire $I_2 = 0.00978 \ A$

b

Solid Wire $R = 0.0149 \ \Omega$

Stranded Wire $R_1 = 0.0189 \ \Omega$

Explanation:

Considering the first question

From the question we are told that

The radius of the first wire is $r_1 = 1.53 mm = 0.0015 \ m$

The radius of each strand is $r_0 = 0.306 \ mm = 0.000306 \ m$

The current density in both wires is $J = 1750 \ A/m^2$

Considering the first wire

The cross-sectional area of the first wire is

$A = \pi r^2$

= > $A = 3.142 * (0.0015)^2$

= > $A = 7.0695 *10^{-6} \ m^2$

Generally the current in the first wire is

$I = J*A$

=> $I = 1750*7.0695 *10^{-6}$

=> $I = 0.01237 \ A$

Considering the second wire wire

The cross-sectional area of the second wire is

$A_1 = 19 * \pi r^2$

=> $A_1 = 19 *3.142 * (0.000306)^2$

=> $A_1 = 5.5899 *10^{-6} \ m^2$

Generally the current is

$I_2 = J * A_1$

=> $I_2 = 1750 * 5.5899 *10^{-6}$

=> $I_2 = 0.00978 \ A$

Considering question two

From the question we are told that

Resistivity is $\rho = 1.69* 10^{-8} \Omega \cdot m$

The length of each wire is $l = 6.25 \ m$

Generally the resistance of the first wire is mathematically represented as

$R = \frac{\rho * l }{A}$

=> $R = \frac{ 1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }$

=> $R = 0.0149 \ \Omega$

Generally the resistance of the first wire is mathematically represented as

$R_1 = \frac{\rho * l }{A_1}$

=> $R_1 = \frac{ 1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }$

=> $R_1 = 0.0189 \ \Omega$

3 0

3 years ago

This is an electrical device that changes the voltage of electricity through the use of two set of coils

Fynjy0 [20]

Voltmeter is the right answer

3 0

2 years ago

Read 2 more answers

A car has a mass of 1600 kg. It is stuck in the snow and is being pulled out by a cable that applies a force of 7560 N due north

sergiy2304 [10]

Answer:

Acceleration of the car will be $a=0.1375m/sec^2$

Explanation:

We have given mass of the ball m = 1600 kg

Force in north direction F= 7560 N

Resistance force which opposes the movement of car $F_R=7340N$

So net force on the car $F_{net}=F-F_R=7560-7340=220N$

According to second law of motion we know that F=ma

So $220=1600\times a$

$a=0.1375m/sec^2$

7 0

2 years ago

dangina [55]

Answer:

C 2000v its obviously ans because if o is 1000 2 vo is 2000v

7 0

2 years ago