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2 years ago

You are driving 12 meters per second. What is your speed in miles per hour? (1.6km=1 mike) EXPLAIN

Physics

1 answer:

Blizzard [7]2 years ago

8 0

1.6km = 1 mike? Wow that guy is tall.

I think you meant 1.6 km = 1 mile? Okay then if we're going 12 meters per second how much would you travel in one hour? First we need to figure out how many seconds are in an hour. There are 60 seconds in a minute and 60 minutes in an hour so:

60×60=3600 seconds in an hour

Now we will multiply that by 12 meters per second and we get:

$12\frac{m}{s} \times 3600 s=43200m$

And 43200 meters is 43.2 km (1000 meters in 1 kilometer) meaning 43.2 kilometers an hour. Since there are 1.6 km in one mile we must divide 43.2km to 1.6.

$\frac{43.2}{1.6}=27mph$

And so your speed is 27 miles per hour.

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The amount of heat energy required to raise the temperature of a unit mass of a material one degree is

charle [14.2K]

The amount of heat energy required to raise the temperature of a unit mass of a material to one degree is called D. its heat capacity.

The relationship of the heat when applied to the object and the change in temperature of the object when heat is being applied is directly proportional to each other. This means that when heat is applied to the object, the temperature of the object increases and when heat is not applied to the object, the temperature of the object decreases.

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2 years ago

The train passes point A with a speed of 30 m/s and begins to decrease its speed at a constant rate of at = - 0.25 m/s^2. Determ

prisoha [69]

Explanation:

At point B, the velocity speed of the train is as follows.

$\nu^{2}_{B} = \nu^{2}_{A} + 2a_{t} (s_{B} - s_{A})$

= $(30)^{2} + 2(-0.25(412 - 0))$

= 26.34 m/s

Now, we will calculate the first derivative of the equation of train.

y = $200 e^{\frac{x}{1000}}$

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

Now, second derivative of the train is calculated as follows.

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

$\frac{d^{2}y}{dx^{2}} = 0.2 (10^{-3}) e^{\frac{x}{1000}}$

Radius of curvature of the train is as follows.

$\rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}$

= $\frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}$

= 3808.96 m

Now, we will calculate the normal component of the train as follows.

$a_{n} = \frac{\nu^{2}_{B}}{\rho}$

= $\frac{(26.34)^{2}}{3808.96}$

= 0.1822 $m/s^{2}$

The magnitude of acceleration of train is calculated as follows.

a = $\sqrt{(a_{t})^{2} + (a_{n})^{2}}$

= $\sqrt{(-0.25)^{2} + (0.1822)^{2}}$

= $0.309 m/s^{2}$

Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is $0.309 m/s^{2}$ .

6 0

2 years ago

A 36,287 kg truck has a momentum of 907,175 kg • . What is the truck’s velocity?

Snowcat [4.5K]

By definition,
Momentum = Mass * Velocity

Let v = the velocity of the truck, m/s
The mass of the truck is 36,287 kg.
The momentum is 907,175 (kg-m)/s.

Therefore
907,175 (kg-m)/s = (36287 kg)*(v m/s)
v = 907175/36287 = 25 m/s

Answer: 25 m/s

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2 years ago

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yaroslaw [1]

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2 years ago