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ycow [4]
3 years ago
8

The car salesman tells you that the car can go from a stop position to 60 mph in six seconds is giving you the car’s rate of

Physics
1 answer:
slava [35]3 years ago
5 0
The salesman is telling you the average magnitude of the car's acceleration.

| Acceleration | = (change in speed) / (time for the change)

| Acceleration | = (60 mi/hr) / (6 sec)

| Acceleration | =  10 miles/hr-sec

That would be 36,000 miles per hour squared,
or 0.0028 mile per second squared.
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The mechanical energy of a bicycle at the top of a hill is 6,000 J. The bicycle stops at the bottom of the hill by applying the
Semmy [17]

Answer:

Thermal energy produce =4,000 J.

Explanation:

Given that

Mechanical  energy at the top of hill = 6,000 J

Mechanical  energy at the bottom of hill = 2,000 J

We know that energy is conserve

Energy at top of hill = energy at bottom of hill + Thermal energy produce

So now by putting the values

Energy at top of hill = energy at bottom of hill + Thermal energy produce

6,000 = 2,000+ Thermal energy produce

Thermal energy produce =6,000-2,000 J

Thermal energy produce =4,000 J.

As we know that thermal energy produce due to friction when one mechanical component slides on the other mechanical component then always heat is generated and this heat is known as thermal energy.

7 0
3 years ago
A toy cart at the end of a string 0.70 m long moves in a circle on a table. The cart has a mass of 2.0 kg and the string has a b
luda_lava [24]
Given that the mass of the toy cart is 2.0 kg and and the acceleration is unknown, the normal formula would be a=f/m where a is acceleration, f is force and m is mass but the string's breaking strength is 40n so I think the formula in this case will be f is greater than m*a
40 is greater than 2a
40 is greater than 2a
40/2 is greater than 2a/2
20m/s² is greater than a 
Therefore the maximum speed the toy cart should have should be less than 20m/s²
8 0
3 years ago
A record player has a velocity of 33.33 RPM. How fast is the record spinning in m/s at a distance of 0.085 m from the center?
anygoal [31]

Answer:

A record player has a velocity of 33.33 RPM. How fast is the record spinning in m/s at a distance of 0.085 m from the center? [0.297 m/s] 6. A merry-go-round a.k.a “the spinny thing” is rotating at 15 RPM, and has a radius of 1.75 m A.

5 0
3 years ago
Read 2 more answers
What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1
notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is  F_{net}=  3.22*10^{-5} \ J

Explanation:

From the question we are told that

    The third charge is  q_3 =  55 nC =  55 *10^{-9} C

    The position of the third charge is  x = -1.220 \ m

     The first charge is q_1 =  -16 nC  =  -16 *10^{-9} \ C

     The position of the first charge is x_1 =  -1.650m

      The second charge is  q_2 =  32 nC  =  32 *10^{-9} C

      The position of the second charge is  x_2 =   0  \ m  

The distance between the first and the third charge is

      d_{1-3} =  -1.650 -(-1.220)

     d_{1-3} = -0.43 \ m

The force exerted on the third charge by the first is  

     F_{1-3} =  \frac{k  q_1 q_3}{d_{1-3}^2}

Where k is the coulomb's constant with a value  9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.

substituting values

      F_{1-3} =  \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}

       F_{1-3} = 4.28 *10^{-5} \ N

 The distance between the second and the third charge is      

  d_{2-3} =  0- (-1.22)

   d_{2-3} =1.220 \ m

The force exerted on the third charge by the first is mathematically evaluated as

       F_{2-3} =  \frac{k  q_2 q_3}{d_{2-3}^2}

substituting values

       F_{2-3} =  \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}

       F_{2-3} =  1.06*10^{-5} N

The net force is

      F_{net} =  F_{1-3} -F_{2-3}

substituting values

    F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}

    F_{net}=  3.22*10^{-5} \ J

6 0
2 years ago
A student walks the hallway for 25 m stops to talk and continues down the hallway another 10 m what is the distance and displace
brilliants [131]

Answer:

Distance = displacement = 35m

Explanation:

The distance of the student is how far he has gone.

Distance = 25m + 10m

Distance = 35m

Displacement is the distance specified in specific direction. Since the student walk in the sane direction, thence the displacement is also 35m

3 0
3 years ago
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