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Pani-rosa [81]
3 years ago
9

A rock is dropped from a bridge what happens to the magnitude of the acceleration

Physics
1 answer:
White raven [17]3 years ago
3 0

Answer:

9.81 m/s² constant

Explanation:

Any object which is falling has only the acceleration due to gravity acting on it. The value of acceleration due to gravity is 9.81 m/s² which is constant. This is the case if air resistance is not taken into consideration.

The air resistance is a result of the surface area of the object which is falling. This will slow down the object and the velocity reached is called the terminal velocity.

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I’ll literlly give you points <br> and make you brainiest
sesenic [268]

Answer:

a

Explanation:

4 0
3 years ago
Suppose you design a new thermometer called the "x" thermometer. on the x scale, the boiling point of water is 130.0 ox and the
Hoochie [10]

You've told us:

-- 130°x  =  212°F

and

-- 10°x  =  32°F

Thank you.  Those are two points on a graph of °x vs °F .  With those, we can figure out the equation of the graph, and easily convert ANY temperature on one scale to the equivalent temperature on the other scale.

-- If our graph is going to have °x on the horizontal axis and °F on the vertical axis, then the two points we know are  (130, 212)  and  (10, 32) .

-- The slope of the line through these two points is

Slope = (32 - 212) / (10 - 130)

Slope = (-180) / (-120)

Slope = 1.5

So far, the equation of the graph is

F = 1.5 x + (F-intercept)

Plug one of the points into this equation.  I'll use the second point  (10, 32) just because the numbers are smaller:

32 = 1.5 (10) + F-intercept

32 = 15 + (F-intercept)

F-intercept = 17

So the equation of the conversion graph is

F = 1.5 x + 17

There you are !  Now you can plug ANY x temperature in there, and the F temperature jumps out at you.

The question is asking what temperature is the same on both scales. This seems tricky, but it's not too bad.  Whatever that temperature is, since it's the same on both scales, you can take the conversion equation, and write the same variable in BOTH places.

We can write [ x = 1.5x + 17 ], solve it for  x, and the solution will be the same temperature in  F  too.

or

We can write [ F = 1.5F + 17 ], solve it for  F, and the solution will be the same temperature in  x  too.

F = 1.5F + 17

Subtract  F  from each side:  0.5F + 17 = 0

Subtract 17 from each side:   0.5F = -17

Multiply each side by 2 :  F = -34

That should be the temperature that's the same number on both scales.

Let's check it out, using our handy-dandy conversion formula (the equation of our graph):

F = 1.5x + 17

Plug in -34 for  x:  

F = 1.5(-34) + 17

F = -51 + 17

<em>F = -34</em>

It works !  -34 on either scale converts to -34 on the other one too. If the temperature ever gets down to -34, and you take both thermometers outside, they'll both read the same number.

<em>yay !</em>

6 0
3 years ago
The speed of light in a vacuum is 2.99x10^8 m/s. calculate its speed in miles per hour.
34kurt
You'll never get the correct answer without the correct conversion factor. Note carefully that you have no decimal. It should be 
<span>1 km = 0.6214 miles </span>
<span>1000 m = 1 km </span>
<span>60 seconds = 1 minute </span>
<span>60 minutes = 1 hour. </span>
<span>2.998E8 m/s x (1 km/1000m) x (0.6214 miles/km) x (60 sec/min) x (60 min/hr) = ?</span>
6 0
3 years ago
A projectile is shot straight up from the earth's surface at a speed of 10,000 km/hr. how high does it go?
mezya [45]
Naturally we assume that 10000 km/hr is initial velocity (same as being shot from a cannon), and no air resistance. With so high a velocity, the effect of diminishing gravity with increasing radius must be taken into account, so you use an energy solution. M is earth mass, r is earth radius. 
KE/m = (9000000/3600)^2/2 = 3858025 J/kg 
ΔPE/m = (PE(at height) - PE(at surface))/m = -GM/(r+h) + GM/r 
KE/m = ΔPE/m 
KE/m - GM/r = -GM/(r+h)
h = -GM / (KE/m - GM/r) - r = 335665.44 m 
(Using G = 6.673E-11 Nm^2/kg^2, M = 5.9742E24 kg, r = 6378100 m)
7 0
3 years ago
5. A cheetah with a mass of 70 kg was clocked running at 72 mph (32 m's). How many joules of
blsea [12.9K]

Answer:

Kinetic energy = 35840 Joules

Explanation:

Given the following data;

Mass = 70kg

Velocity = 32m/s

To find the kinetic energy;

Kinetic energy can be defined as an energy possessed by an object or body due to its motion.

Mathematically, kinetic energy is given by the formula;

K.E = \frac{1}{2}MV^{2}

Where, K.E represents kinetic energy measured in Joules.

M represents mass measured in kilograms.

V represents velocity measured in metres per seconds square.

K.E = \frac{1}{2}MV^{2}

Substituting into the equation, we have;

K.E = \frac{1}{2}*70*32^{2}

K.E = \frac{1}{2}*70*1024

K.E = 35 * 1024

K.E = 35840 Joules.

Therefore, the kinetic energy possessed by the cheetah is 35840 Joules.

7 0
3 years ago
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