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inn [45]
3 years ago
11

A 1.2 x10 3 kilogram automobile in motion strikes a 1.0 x 10 -4 kilogram insect as a result the insect is accelerated at a rate

of 1.0 x 10 2 meters per second 2 what is the magnitude of the force the insect exerts on the car
Physics
1 answer:
Mariana [72]3 years ago
6 0
According to Newton's second law, the force applied to an object is equal to the product between the mass of the object and its acceleration:
F=ma
where F is the magnitude of the force, m is the mass of the object and a its acceleration.

In this problem, the object is the insect, with mass m=1.0 \cdot 10^{-4} kg. The acceleration of the insect is a=1.0 \cdot 10^2 m/s^2, therefore we can calculate the force exerted by the car on the insect:
F=ma=(1.0 \cdot 10^{-4} kg)(1.0 \cdot 10^2 m/s^2)=0.01 N

How do we find the force exerted by the insect on the car?
According to Newton's third law (known as action-reaction law), when an object A exerts a force on an object B, object B also exerts a force equal and opposite on object A. Therefore, the force exerted by the insect on the car is equal to the force exerted by the car on the object, so it is 0.01 N.
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Name the regions of a electromagnetic spectrum from the shortest wavelength to the longest wavelength
kaheart [24]

Answer:

From shortest wavelength to longest wavelength:

Gamma Rays

X-Rays

Ultraviolet

Visible Light

Infrared waves

Microwaves

Radio Waves

Explanation:

5 0
3 years ago
The potential energy of a particle as a function of position will be given as U(x) = A x2 + B x + C, where U will be in joules w
satela [25.4K]

Answer:

F = - 2 A x - B

Explanation:

The force and potential energy are related by the expression

      F = - dU / dx i ^ -dU / dy j ^ - dU / dz k ^

Where i ^, j ^, k ^ are the unit vectors on the x and z axis

The potential they give us is

     U (x) = A x² + B x + C

Let's calculate the derivatives

    dU / dx = A 2x + B + 0

The other derivatives are zero because the potential does not depend on these variables.

Let's calculate the strength

      F = - 2 A x - B

3 0
3 years ago
an athlete runs 5.4 laps around a circular track that is 400.0 m long. If this takes 540 s, what is the average velocity of the
Slav-nsk [51]

Answer:

Average velocity is 0.296 m/s.

Average speed is 4.0  m/s.

Explanation:

Given:

Distance of the circular track is, D=400.0\ m

Number of laps ran is, n=5.4

Time taken for the run is, t=540\ s

Now, total distance covered in 5.4 laps = D_T=D\times n=400\times 5.4=2160\ m

Also, since the path is a circle, the final position of the athlete after 5.4 laps will be 0.4 of 400 m ahead of the starting point.

Distance covered in 0.4 laps is, \textrm{Displacemet}=0.4\times 400=160\ m

Therefore, the displacement of the athlete will be 160 m as the athlete is 160 m ahead of the starting point and displacement depends on the initial and final points only.

Now, average velocity is given as:

v_{avg}=\frac{\textrm{Displacemet}}{t}=\frac{160}{540}=0.296\ m/s

Average speed is the ratio of total distance covered to total time taken.

So, average speed = \frac{D_T}{t}=\frac{2160}{540}=4\ m/s

6 0
3 years ago
the skid marks left by the decelerating jet-powered car the spirit of america were 9600 m long before it came to a stop at 198 m
N76 [4]

Answer:

97 s

Explanation:

Given:

Δx = 9600 m

v₀ = 198 m/s

v = 0 m/s

Find: t

Δx = ½ (v + v₀) t

9600 m = ½ (0 m/s + 198 m/s) t

t = 97 s

8 0
3 years ago
Read 2 more answers
A 9 m3 container is filled with 300 kg of r-134a at 10°c. what is the specific enthalpy (kj/kg) of the r-134a in the container?
andreev551 [17]
Given the mass of R-134a m = 300kg; Volume of the container V = 9  cu. meter; Temperature of R-134a T = 10 degrees Celsius; 
Formula of specific volume : v = V / m = 9 / 300 = 0.03 cu. m / kg. 
At T = 10 degrees Celsius from saturated R-134a tables, vf = 0.0007930 cu. m /kg; vg = 0.049403 cu. m/kg. We know v = vf + x (vg - vf), so 0.03 = 0.0007930 + x (0.049403 - 0.0007930), which makes x = 0.601.  
Specific enthalpy of R-134a in the container is h = hf + x*hfg = 65.43 + (0.601 * 190.73). Answer is 180.0587 kJ/kg
8 0
3 years ago
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