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Katena32 [7]
3 years ago
11

a vehicle travels at a constant speed of 65 mph for 4 hours how far has this vehicle travelled in this time

Physics
1 answer:
Nostrana [21]3 years ago
3 0
260 miles.................
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An object traveling at a constant
mel-nik [20]

Answer:

pi / 2  radians / s

Explanation:

One revolution = 2 pi Radians    in 4 seconds

2 pi  /  4   =   pi/2   radians / s

4 0
2 years ago
The position of a ball rolling in a straight line is given by x=2−3.6t+4.4t2, where x is in meters and t in seconds.
kobusy [5.1K]
I'm assuming the question is time it will take for ball to reach ground, if it is then set equation to zero then use the quadratic formula, the possible t value is your answer then
8 0
3 years ago
You are raising up a big bucket of water from a 25.9 m deep well. The combined mass of the water and the bucket is 13.9 kg. The
Mars2501 [29]

The total work done is  5980 Joules and the power expended is 57 Watts.

<h3>What is work done?</h3>

The work done is the work done in the gravitational field as the bucket is raised up Thus work required to remove the bucket Wb;

Wb = 13.9 kg * 25.9 m * 9.8 m/s^2 = 3530 Joules

Height of the center of mass of chain = 25.9 / 2 = 12.95 m  

Work done by the chain Wc;

Wc = 12.95 * 19.3 * 9.8 = 2450 Joules  

Total work = 3530 + 2450 = 5980 Joules

Power expended = W / t = 5980 J / 105 sec = 57 J/s = 57 Watts

Learn more about work done:brainly.com/question/13662169

#SPJ1

3 0
2 years ago
What is the derivative of x^2 - x + 3 at the point x = 5?​
Hunter-Best [27]

Answer:

9

Explanation:

d/dx (x² - x + 3)

= 2x - 1

when x = 5,

2x-1

= 2(5) - 1

= 10 - 1

= 9

5 0
3 years ago
Assume this 1.20-mm-radius copper wire is electrically neutral in the Earth reference frame, in which it is at rest and carrying
agasfer [191]

Answer:

The charge density in the system is 4.25*10^4C/m

Explanation:

To solve this problem it is necessary to keep in mind the concepts related to current and voltage through the density of electrons in a given area, considering their respective charge.

Our data given correspond to:

r=1*10^{-3}m\\v = 5.2*10^{-4}m/s\\e= 1.6*10^{-19}C

We need to asume here the number of free electrons in a copper conductor, at which is generally of 8.5 *10^{28}m^{-3}

The equation to find the current is

I = VenA

Where

I =Current

V=Velocity

A = Cross-Section Area

e= Charge for a electron

n= Number of free electrons

Then replacing,

I = (5.2*10^{-4})(1.6*10^{-19})(88.5 *10^{28})(\pi(1*10^{-3})^2)

I= 22.11a

Now to find the linear charge density, we know that

I = \frac{Q}{t} \rightarrow Q = It

Where:

I: current intensity

Q: total electric charges

t: time in which electrical charges circulate through the conductor

And also that the velocity is given in proportion with length and time,

V_d = \frac{l}{t} \rightarrow l = V_d t

The charge density is defined as

\lambda = \frac{Q}{l}\\\lambda = \frac{It}{V_d t}\\\lambda = \frac{I}{V_d}

Replacing our values

\lambda = \frac{22.11}{5.20*10{-4}}

\lambda= 4.25*10^4C/m

Therefore the charge density in the system is 4.25*10^4C/m

5 0
3 years ago
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