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3 years ago

If there are 1.609 km in a mile, convert 135 miles/hour into meters per second. There are 1000 m in a kilometer.

Physics

2 answers:

Effectus [21]3 years ago

4 0

Answer:

97.1037936

Explanation:

lukranit [14]3 years ago

4 0

Answer:

60.3375 m/ per sec

Explanation:

first multiply 1609 (the amount of meters in a mile) by 135( the total number of miles). this brings you to conclude there are a total of 217,215 meters in 135 miles.

So 135 mph = 217,215 meters per hour.

From here I took a simple route so as to not need a calculator.

217,215 meters = 60mins. (divide both sides by 2)

108,607.5 = 30mins. (divide both sides by 2)

54,303.75 = 15mins. (divide both sides by 15)

3620.25 = 1min

3620.25 = 60secs. (divide both sides by 60)

60.3375 = 1sec

60.3375 meters per second = 135miles per hr.

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A girl pulls on a 10-kg wagon with a constant horizontal force of 30 n. if there are no other horizontal forces, what is the wag

olga2289 [7]

Force = mass * acceleration
F = ma

Given m = 10 kg, F = 30 N;

F = ma
30 = 10a

Solving for a:
a = 3 m/s^2

The acceleration is 3 meters per second squared.

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3 years ago

A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the

Margaret [11]

Hi there!

$\boxed{\omega = 0.38 rad/sec}$

We can use the conservation of angular momentum to solve.

$\large\boxed{L_i = L_f}$

Recall the equation for angular momentum:

$L = I\omega$

We can begin by writing out the scenario as a conservation of angular momentum:

$I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)$

$I_m$ = moment of inertia of the merry-go-round (kgm²)

$\omega_m$ = angular velocity of merry go round (rad/sec)

$\omega_f$ = final angular velocity of COMBINED objects (rad/sec)

$I_b$ = moment of inertia of boy (kgm²)

$\omega_b$ = angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

$I = MR^2$

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

$\omega = \frac{v}{r}$

$L_b = MR^2(\frac{v}{R}) = MRv$

Plug in the given values:

$L_b = (20)(3)(5) = 300 kgm^2/s$

Now, we must solve for the boy's moment of inertia:

$I = MR^2\\I = 20(3^2) = 180 kgm^2$

Use the above equation for conservation of momentum:

$600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}$

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2 years ago

What mass of water must evaporate from the skin of a 70.0 kg man to cool his body 1.00 ∘C? The heat of vaporization of water at

romanna [79]

Answer:

100 cc

Explanation:

Heat released in cooling human body by t degree

= mass of the body x specific heat of the body x t

Substituting the data given

Heat released by the body

= 70 x 3480 x 1

= 243600 J

Mass of water to be evaporated

= 243600 / latent heat of vaporization of water

= 243600 / 2420000

= .1 kg

= 100 g

volume of water

= mass / density

= 100 / 1

100 cc

1 / 10 litres.

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3 years ago

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MAVERICK [17]

The ball only accelerates during the brief time that the club is in contact
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adoni [48]

Body heat :)

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