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3 years ago

Help What's the average speed of each of these???

Physics

1 answer:

zalisa [80]3 years ago

5 0

Answer:

1) 1.15666

2)1.07

3)1.36

Explanation:

To find the average take the vaules of the numbers and add then divide by how many numbers are in the eqation and that should give you your average.

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When a can is crushed did it undergo a physical change

brilliants [131]

Because even though the object got crush and misshape it still has the same identity. the identity never change

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3 years ago

A car traveling at 15 m/s starts to decelerate steadily. It comes to a complete stop in 10 seconds. What is it’s acceleration

tresset_1 [31]

So the initial velocity is 15 m/s, the final velocity is 0 since it's at a complete stop and time is 10 seconds. Therefore:

$Acceleration=\frac{v_{final}-v_{initial}}{t_{final}-t_{initial}} =\frac{0m/s-15m/s}{10s-0s} =-1.5\frac{m}{s^2}$

Therefore, the acceleration is -1.5 m/s^2. The reason it's negative is due to the fact that the vector is going against it's original movement since it's decelerating.

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3 years ago

Read 2 more answers

If a car travels 40 km/hr for hours how far has it traveled

Rainbow [258]

It depends on how many hours!
1 hour = 40km
2 hours = 80km

8 0

3 years ago

A silver wire with resistivity 1.59 × 10-8 Ω-m carries a current density of 4.0 A/mm2, What is the magnitude of the electric fie

Nat2105 [25]

Answer:

Electric field, E = 0.064 V/m

Explanation:

It is given that,

Resistivity of silver wire, $\rho=1.59\times 10^{-8}\ \Omega-m$

Current density of the wire, $J=4\ A/mm^2=4\times 10^6\ A/m^2$

We need to find the magnitude of the electric field inside the wire. The relationship between electric field and the current density is given by :

$E=J\times \rho$

$E=4\times 10^6\times 1.59\times 10^{-8}$

E = 0.0636 V/m

E = 0.064 V/m

So, the magnitude of electric field inside the wire is 0.064 V/m. Hence, this is the required solution.

6 0

3 years ago

When the Apollo 11 lunar module Eagle lands on the moon it comes to a stop 10m above the surface of the moon. The last 10m it fr

Scrat [10]

Answer:

i) 3.514 s, ii) 5.692 m/s

Explanation:

i) We can use Newton's second law of motion to find out how long does it take for the Eagle to touch down.

as the equation says for free-falling

h = ut +0.5gt^2

Here, h = 10 m, g = acceleration due to gravity = 1.62 m/s^2( on moon surface)

initial velocity u = 0

10 = 0.5×1.62t^2

t = 3.514 seconds

Therefore, it takes t = 3.514 seconds for the Eagle to touch down.

ii) use Newton's 1st equation of motion to calculate the velocity of the lunar module when it hits the surface of the moon

v = u + gt

v = 0+ 1.62×3.514

v= 5.692 m/s

7 0

3 years ago