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horrorfan [7]
3 years ago
10

The speed on a car speedometer. Is it speed or velocity? Explain Why

Physics
1 answer:
Sliva [168]3 years ago
3 0
Velocity is a speed AND a direction.
When you turn a corner, or go around a curve in the road, your
direction changes, but the reading on the dial doesn't change.
So it can't be showing velocity.  It must be showing only speed.
That's probably a big part of the reason why it's called a speedometer
and not a velocimeter.
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What quantities are conserved in a nuclear transmutation
vredina [299]

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6 0
4 years ago
Your teacher's car can go from rest to 25 m/s (≈55 mph) in 10 seconds. The car's velocity changes at a uniform rate. Below is a
Mice21 [21]

If the speed of the car changes uniformly, as shown in the graph, this means that the acceleration is constant. As the acceleration is the derivative of the speed

a = \frac{\delta v}{\delta t}

Then, this means that the acceleration is the slope of the equation of the line shown in the graph.

Then, we find the acceleration using the following equation:

a = \frac{v_f- v_i}{t_2-t_1}\\\\a = \frac{25-0}{10}\\\\a = 2.5\ m/s ^ 2

Then, the acceleration of the car is 2.5\ m/s^2

5 0
3 years ago
The electric potential, when measured at a point equidistant from two particles that have charges equal in magnitude but of oppo
Reika [66]

Answer:

C) equal to zero

Explanation:

Electric potential is calculated by multiplying constant and charge, then dividing it by distance. The location that we want to measure is equidistant from two particles, mean that the distance from both particles is the same(r2=r1). The charges of the particle have equal strength of magnitude but the opposite sign(q2=-q1). The resultant will be:V = kq/r

ΔV= V1 + V2= kq1/r1 + kq2/r2

ΔV= V1 + V2= kq1/r1 + k(-q1)/(r)1

ΔV= kq1/r1 - kq1/r1

ΔV=0

The electric potential equal to zero

4 0
4 years ago
A body starting from rest and moves with uniform acceleration. The velocities of the body are 40 m/s at the 4th second and 60 m/
Sergeeva-Olga [200]

Answer:

GIVEN

u is 40m/s and v=60m/s

time is 4s

therefore acceleration =60-40/4

=20/4=5m/s^2

b)next 2 sec=70m/s

8 0
3 years ago
Venus has an average distance to the sun of 0.723 AU. In two or more complete sentences, explain how to calculate the orbital pe
Mice21 [21]

As per the question the distance of venus from sun is given as 0.723 AU

We have been asked to calculate the time period of the planet venus.

As per kepler's laws of planetary motion the square of time period of planet is directly proportional to the cube of semi major axis. mathematically

                                        T^{2} \alpha R^{3}

                                         ⇒ T^{2} = KR^{3} where is k is the proportionality  constant

We may solve this problem by comparing with the time period of the earth . We know that time period of earth is 365.5 days

Hence T_{1} =365.5 days

The distance of sun from earth is taken as 1 AU i.e the mean distance of earth from sun

Hence R_{1} =1 AU

The distance of venus from sun is 0.723 AU i.eR_{2} =0.723

From keplers law we know that-\frac{T_{1} ^{2} }{T_{2} ^{2} } =\frac{R_{1} ^{3} }{R_{2} ^{3} }

                            ⇒T_{2} ^{2} =T_{1} ^{2} *\frac{R_{2} ^{3} }{R_{1} ^{3} }

Putting the values mentioned above we get-

                                      T_{2} ^{2} =50,350.132851075

                                         ⇒ T_{2} =\sqrt{50,350.132851075}

                                        ⇒T_{2} = 224.388352752710 days.

Hence the time period of venus is 224.388352752710 days

                                         

                     






                           

7 0
4 years ago
Read 2 more answers
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