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Elina [12.6K]
3 years ago
14

A car travels at 90 km per hour how much distance in metre does the car travel in 25 seconds​

Physics
1 answer:
Volgvan3 years ago
3 0

Answer:

  1. 90000m÷3600s=25m
  2. 25m×25s=625ms
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photoshop1234 [79]

Answer:

It conserves both energy and momentum in the collision at the same time. By design, when the balls collide the strings that hold them up are vertical (assuming balls are only swung from one side).

Explanation:

Hope This Helps!!

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3 years ago
Some sharks can swim at average cruising speeds of three miles per hour. If a shark swam at that average speed for seven hours,
kari74 [83]

Answer:

21 miles

Explanation:

3 miles an hour for 7 hours

Its simply 7m*3m/hr=21 miles

8 0
2 years ago
It is dangerous to stand near the railway leak when train passing by,why?​
4vir4ik [10]

Answer:

Because a person may be pulled in the direction of the moving train. Thereby causing accident

Explanation:

According to Daniel Bernoulli's theorem, he was widely known as a Mathematician. He stated that due to the higher velocity of a moving train, there is higher kinetic energy in terms of volume around it, while the air pressure between the person and the train becomes lower.

As a result, a person near a moving train may be pulled in the direction of the moving train. Thereby causing accidents that may lead to death.

3 0
3 years ago
Which of the following is a composite of a gas and a solid
cricket20 [7]
<span>{c o {a la manzana o
<span>Vapor de agua en el aire en una ducha de agua caliente</span>
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3 0
3 years ago
Two long, parallel wires carry unequal currents in the same direction. The ratio of the currents is 3 to 1. The magnitude of the
astraxan [27]

Answer:

3A is the larger of the two currents.

Explanation:

Let the currents in the two wires be I₁ and I₂

given:

Magnitude of the electric field, B = 4.0μT = 4.0×10⁻⁶T

Distance, R = 10cm = 0.1m

Ratio of the current = I₁ : I₂ = 3 : 1

Now, the magnitude of a magnetic field at a distance 'R' due to the current 'I' is given as

B = \frac{\mu_oI}{2\pi R}

Where \mu_o is the magnitude constant = 4π×10⁻⁷ H/m

Thus, the magnitude of a magnetic field due to I₁ will be

B_1 = \frac{\mu_oI_1}{2\pi R}

B_2 = \frac{\mu_oI_2}{2\pi R}

given,

B = B₁ - B₂ (since both the currents are in the same direction and parallel)

substituting the values of B, B₁ and B₂

we get

4.0×10⁻⁶T =  \frac{\mu_oI_1}{2\pi R} - \frac{\mu_oI_2}{2\pi R}

or

4.0×10⁻⁶T =  \frac{\mu_o}{2\pi R}\times (I_1-I_2 )

also

\frac{I_1}{I_2} = \frac{3}{1}

⇒I_1 = 3\times I_2

substituting the values in the above equation we get

4.0×10⁻⁶T =  \frac{4\pi\times 10^{-7}}{2\pi 0.1}\times (3 I_2-I_2)

⇒I_2 = 1A

also

I_1 = 3\times I_2

⇒I_1 = 3\times 1A

⇒I_1 = 3A

Hence, the larger of the two currents is 3A

3 0
3 years ago
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