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pishuonlain [190]
3 years ago
10

Richard is driving home to visit his parents. 135 mi of the trip are on the interstate highway where the speed limit is 65 mph .

Normally Richard drives at the speed limit, but today he is running late and decides to take his chances by driving at 73 mph .
Physics
1 answer:
Elis [28]3 years ago
5 0
<h2>Answer:</h2>

He saves 13.2 minutes

<h2>Explanation:</h2>

Hey! The question is incomplete, but it can be found on the internet. The question is:

How many minutes did he save?

Let's call:

t_{1}:Time \ at \ speed \ 65mph \\ \\ t_{2}:Time \ at \ speed \ 73mph \\ \\ v_{1}=65mph \\ \\ v_{2}=73mph

We know that the 135 miles are on the interstate highway where the speed limit is 65 mph. From this, we can calculate the time it takes to drive on this highway. Assuming Richard maintains constant the speed:

v=\frac{d}{t} \\ \\ d:distance \\ \\ t:time \\ \\ v:velocity \\ \\ t_{1}=\frac{d}{v_{1}} \\ \\ t=\frac{135}{65} \\ \\ t_{1}=2.07 \ hours

Today he is running late and decides to take his chances by driving at 73 mph, so the new time it takes to take the trip is:

t_{2}=\frac{135}{73} \\ \\ t_{2}=1.85 \ hours

So he saves the time t_{s}:

t_{s}=t_{1}-t_{2}=2.07-1.85=0.22 \ hours

In minutes:

t_{s}=0.22h\left(\frac{60min}{1h}\right) \\ \\ \boxed{t_{s}=13.2min}

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Water at room temperature is discharged from a pipe at a rate of 1000 gallons per minute (gpm). Express this flow rate in cubic
marshall27 [118]

Answer

given,

discharge rate from pipe = 1000 gallons/minutes

now,

flow rate in  cubic meters per second

1 gallon = 0.00378541 m³

1 min = 60 s

Q = 1000\times \dfrac{0.00378541\ m^3}{1\ gallon}\times \dfrac{1\ min}{60\ s}

Q = 0.063 m³/s

flow rate in  liters per minute

1 gallon = 3.78541 L

 Q = 1000\times \dfrac{3.78541\ m^3}{1\ gallon}

 Q = 3785.41 m³/min

flow rate in cubic feet per second

 1 gallon = 0.133681 ft³

 1 min = 60 s

Q = 1000\times \dfrac{0.133681\ ft^3}{1\ gallon}\times \dfrac{1\ min}{60\ s}

Q = 2.23 ft³/s

4 0
3 years ago
When monochromatic light illuminates a grating with 7000 lines per centimeter, its second order maximum is at 62.4°. what is the
zhannawk [14.2K]

When red light illuminates a grating with 7000 lines per centimeter, its second maximum is at 62.4°. What is the wavelength of this light?

ans: 633nm

6 0
3 years ago
Read 2 more answers
A charge feels a 7.89 x 10-7 N force when it moves 2090 m/s at a 29.4° angle to a 4.23 x 10-3 T magnetic field. What is the amou
tamaranim1 [39]

We have that the amount of the charge q is

q=1.8*10^{-7}

From the Question we are told that

Force F=7.89 x*10^{-7}

Velocity V=2090m/s

Angle \theta=29.4

Magnetic field B=4.23 * 10^{-3} T

Generally, the equation for Force F is mathematically given by

F=qVBsin\theta\\\\q=\frac{F}{VBsin\theta}

q=\frac{7.89 x*10^{-7}}{4.23 * 10^{-3} T*sin29.4*2090m/s}

q=1.8*10^{-7}

In conclusion

The amount of the charge q is

q=1.8*10^{-7}

For more information on this visit

brainly.com/question/1470439

6 0
3 years ago
Read 2 more answers
1.)Two objects, one of m=20,000 kg, and another of 12,500 kg, are placed at a distance of 5 meters apart. What is the force of g
Delvig [45]

1) 6.67\cdot 10^{-4} N

The force of gravitation between the two objects is given by:

F=G\frac{m_1 m_2}{r^2}

where

G=6.67\cdot 10^{-11} kg^{-1} m^{3} s^{-2} is the gravitational constant

m1 = 20,000 kg is the mass of the first object

m2 = 12,500 kg is the mass of the second object

r = 5 m is the distance between the two objects

Substituting the numbers inside the equation, we find

F=(6.67\cdot 10^{-11})\frac{(20,000 kg)(12,500 kg)}{(5 m)^2}=6.67\cdot 10^{-4} N


2)  2.7\cdot 10^{-3} N

From the formula in exercise 1), we see that the force is inversely proportional to the square of the distance:

F \sim \frac{1}{r^2}

this means that if we cut in a half the distance without changing the masses, the magnitude of the forces changes by a factor

F'\sim \frac{1}{(r/2)^2}=4 \frac{1}{r^2}=4F

So, the gravitational force increases by a factor 4. Therefore, the new force will be

F' = 4 F=4(6.67\cdot 10^{-4} N)=2.7\cdot 10^{-3} N


3)  12.5 Nm

The torque is equal to the product between the magnitude of the perpendicular force and the distance between the point of application of the force and the centre of rotation:

\tau=Fd

Where, in this case:

F = 25 N is the perpendicular force

d = 0.5 m is the distance between the force and the center

By using the equation, we find

\tau=(25 N)(0.5 m)=12.5 Nm


4) 0.049 kg m^2/s

The relationship between angular momentum (L), moment of inertia (I) and angular velocity (\omega) is:

L=I\omega

In this problem, we have

I=0.007875 kgm^2

\omega=6.28 rad/s

So, the angular momentum is

L=I\omega=(0.007875 kgm^2)(6.28 rad/s)=0.049 kg m^2/s

6 0
3 years ago
Hansel pushes a 2 lb. box 5 feet in 20 seconds. How much work has he done?
Licemer1 [7]
Work = force x distance

You can see time doesn’t matter (if we were talking about power, which is the RATE at which work is performed, that would be a different story).

W = 2 x 5 = 10 foot-pounds of work

Foot-pounds are gross units. Better to work in SI units when you can!

8 0
3 years ago
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