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iogann1982 [59]
3 years ago
14

Does the speedometer of a car read average speed or instantaneous speed?

Physics
1 answer:
aleksandrvk [35]3 years ago
8 0
Speedometer of car read instantaneous speed of course. Its because when you stop it instantaneous show 0. If speedometer show average speed there should be different value than 0 (much greater) just when you stop.
Maybe there are car with speedometer with average speed but the time from which the average is taken is extremely small. Its simply instantaneous speed.
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The measurement of 50m east is
Nostrana [21]

Answer:idk

Explanation:idk

6 0
3 years ago
In the 1887 experiment by Michelson and Morley, the length of each interferometer arm was 11m. The experimental limit on the mea
Vikentia [17]
For the answer to the question above,
we can get the number of fringes by dividing (delta t) by the period of the light (Which is λ/c). 

fringe = (delta t) / (λ/c) 

We can find (delta t) with the equation: 

delta t = [v^2(L1+L2)]/c^3 

Derivation of this formula can be found in your physics text book. From here we find (delta t): 

600,000^2 x (11+11) / [(3x10^8)^3] = 2.93x10^-13 

2.93x10^-13/ (589x10^-9 / 3x10^8) = 149 fringes 

This answer is correct but may seem large. That is because of your point of reference with the ether which is usually at rest with respect to the sun, making v = 3km/s. 
4 0
3 years ago
Assume: The bullet penetrates into the block and stops due to its friction with the block. The compound system of the block plus
tino4ka555 [31]

Answer:

1)4.7334J

2)225.4m/s

Explanation:

v= the Velocity of both the bullet and the block after collision=?

H= Height of the bullet along circular arc= 10cm=0.1m

g= acceleration due to gravity= 9.81m/s^2

R= Radius of the circular arc= 18cm= 0.18m

m= Mass of the bullet= 30g= 0.03kg

M= Mass of the block = 4.8 kg

Using the law of conservation of energy

Potential energy of the system= Kinectic energy of the system

1/2 mv^2= mgh..............eqn(1)

But we have two mass m and M

We can write eqn(1) as

0.5(m+M)v^2= (m+M)gh ...........eqn(2)

If we make "v" subject of the formula we have

v = √2gh

Then substitute the values we have

= √2 x 9.81 x 0.1 = 1.40m/s

1) We can now calculate the total energy of the system after collision as

KE = 1/2(m+M)v^2

= 1/2 x (0.03+4.8) x (1.40)^2

KE = 4.7334J

Hence, the total energy of the composite system at any time after the collision is 4.7334J

2)to determine the initial velocity of the bullet.

From law of momentum conservation, which can be expressed as

m1u1+m2u2=(m1+m2)v

Where the initial Velocity of the bullet u1= ?

Final velocity of the bullet = 0

the Velocity of both the bullet and the block after collision=v= 1.40m/s

(0.03×u1) +(u×0)= (4.8+0.03)1.4

0.03u1=6.762

U1=225.4m/s

Hence, the initial velocity of the bullet is 225.4m/s

3 0
3 years ago
Two charges are located in the xx–yy plane. If q1=−4.10 nCq1=−4.10 nC and is located at (x=0.00 m,y=1.080 m)(x=0.00 m,y=1.080 m)
Gala2k [10]

Answer:

Explanation:

Due to first charge , electric field at origin will be oriented towards - ve of y axis.

magnitude

Ey = -8.99 x 10⁹ x 4.1 x 10⁻⁹ / 1.08² j

= - 31.6 j N/C

Due to second charge electric field at origin

= 8.99 x 10⁹ x 3.6  x 10⁻⁹ / 1.2²+ .6²

= 8.99 x 10⁹ x 3.6  x 10⁻⁹ / 1.8

= 18 N/C

It is making angle θ where

Tanθ = .6 / 1.2

= 26.55°

this field in vector form

= - 18 cos 26.55 i - 18 sin26.55 j

= - 16.10 i - 8.04 j

Total field

= - 16.10 i - 8.04 j + ( - 31.6 j )

= -16.1 i - 39.64 j .

Ex = - 16.1 i

Ey = - 39.64 j .

8 0
3 years ago
Consider what will happen when a bar magnet is pushed toward the coil. when the coil "feels" the changing magnetic field caused
notka56 [123]
The question just basically explained what happens
4 0
3 years ago
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