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Tatiana [17]
4 years ago
15

Bernie drove home from a vacation in 1.5 hours. He traveled a total distance of 60 miles. What was his average speed in miles pe

r hour
Physics
2 answers:
Lorico [155]4 years ago
8 0

Speed = (distance / (time)

Speed = (60 miles) / (1.5 hours)

Speed = (60 / 1.5) (mile/hour)

<em>Speed = 40 miles/hour</em>

mafiozo [28]4 years ago
6 0

Answer: 40

Explanation:

I believe this is correct. I did 60/1.5 to get 40/mph

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Which part of the light waves does color correspond to?
MariettaO [177]

Answer:

The colour of visible light depends on its wavelength. These wavelengths range from 700 nm at the red end of the spectrum to 400 nm at the violet end. Visible light waves are the only electromagnetic waves we can see. We see these waves as the colors of the rainbow.

Explanation:

7 0
4 years ago
Read 2 more answers
At the nose of a missile in flight, the pressure and temperature are 5.6 atm and 850°R, respectively. Calculate the density and
Contact [7]

To solve this problem we will apply the definition of the ideal gas equation, where we will clear the density variable. In turn, the specific volume is the inverse of the density, so once the first term has been completed, we will simply proceed to divide it by 1. According to the definition of 1 atmosphere, this is equivalent in the English system to

1atm = 2116lb/ft^2

The ideal gas equation said us that,

PV = nRT

Here,

P = pressure

V = Volume

R = Gas ideal constant

T = Temperature

n = Amount of substance (at this case the mass)

Then

\frac{n}{V} = \frac{P}{RT}

The amount of substance per volume is the density, then

\rho = \frac{P}{RT}

Replacing with our values,

\rho = \frac{5.6*2116}{1716*850}

\rho = 0.00812slug/ft^3

Finally the specific volume would be

v = \frac{1}{\rho}

v = 123ft^3/slug

6 0
3 years ago
An air-filled pipe is found to have successive harmonics at 945 Hz , 1215 Hz , and 1485 Hz . It is unknown whether harmonics bel
Aleonysh [2.5K]

Answer:

L = 0.635m

Explanation:

This problem involves the concept of stationary waves in pipes. For pipes closed at one end,

The frequency f = nv/4L for n = 1,3,5....n

For pipes open at both ends

f = nv/2L for n = 1,2,3,4...n

Assuming the pipe is closed at one end and that velocity of sound is 343m/s in air. If we are right we will obtain a whole number for n.

The film solution can be found in the attachment below.

8 0
3 years ago
A ball is launched up a semicircular chute in such a way that at the top of the chute, just before it goes into free fall, the b
Mariulka [41]

We have centripetal acceleration = v^2/r = 2g

So, v = \sqrt{2gr}

Now by equation of motion we have S= ut +0.5at^2

S =displacement, u = initial velocity, a= acceleration and t = time

Here S =  2r and a = g , u = 0

2r = 0+\frac{1}{2} *g*t^2

t = \sqrt{\frac{4r}{g} }

Distance traveled in horizontal direction = \sqrt{2gr}*\sqrt{\frac{4r}{g} }= \sqrt{8r^2} =2r\sqrt{2}

8 0
4 years ago
A cheerleader lifts his 59.6 kg partner straight up off the ground a distance of 0.749 m before
Airida [17]

m = mass of the partner which the cheerleader lifts = 59.6 kg

h = height to which the partner is lifted by the cheerleader = 0.749 m

g = acceleration due to gravity = 9.8 m/s²

work done by the cheerleader in lifting the partner is same as the potential energy gained by the partner.

W = work done by the cheerleader in lifting the partner

PE = potential energy gained

so  W = PE

potential energy is given as

PE = mgh

hence

W = mgh

inserting the values in the above formula

W = 59.6 x 9.8 x 0.749

W = 437.5 J

this is the work done in lifting the partner once.

the cheerleader does this 30 times , hence the total work done is given as

W' = 30 W

W' = 30 x 437.5

W' = 13125 J


5 0
4 years ago
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