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o-na [289]
3 years ago
15

A train travels a distance of 2000 km at an average speed of 120 km per hour. How long did the trip take? SHOW WORK

Physics
1 answer:
Igoryamba3 years ago
6 0
S = d/t, s = 120, d = 2000, t = ?
Input the values,
120 = 2000/t
Make t the subject of the formula by cross multiplying, Therefore,
120t = 2000
Divide both sides by 120
t = 16.7hrs  to 1 decimal place.
If you're asked to convert it, you can.


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ehidna [41]

Lamina and turbulent flow

Explanation:

mentioning about lamina and turbulent flow we could say that both form in different period of time

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2 years ago
An overhead door is guided by wheels at a and b that roll in horizontal and vertical tracks. when θ = 40°, the velocity of wheel
Karo-lina-s [1.5K]

I think the situation is modeled by the scenario in the attached image. Some specific values seem to be missing (like the height of door d)...

The door forms a right triangles that satisfies

\tan\theta=\dfrac ab\implies\sec^2\theta\dfrac{\mathrm d\theta}{\mathrm dt}=\dfrac{b\frac{\mathrm da}{\mathrm dt}-a\frac{\mathrm db}{\mathrm dt}}{b^2}

We also have

\tan\theta=\dfrac ab\implies\cos\theta=\dfrac bd

so if you happen to know the height of the door, you can solve for b and a.

d is fixed, so

a^2+b^2=d^2\implies2a\dfrac{\mathrm da}{\mathrm dt}+2b\dfrac{\mathrm db}{\mathrm dt}=0\implies\dfrac{\mathrm da}{\mathrm dt}=-\dfrac ba\dfrac{\mathrm db}{\mathrm dt}

We can solve for the angular velocity \dfrac{\mathrm d\theta}{\mathrm dt}:

\dfrac{\mathrm d\theta}{\mathrm dt}=\cos^2\theta\dfrac{b\left(-\frac ba\frac{\mathrm db}{\mathrm dt}\right)-a\frac{\mathrm db}{\mathrm dt}}{b^2}=-\dfrac1a\dfrac{\mathrm db}{\mathrm dt}

At the point when \theta=40^\circ and \dfrac{\mathrm db}{\mathrm dt}=1.8 ft/s, we get

\dfrac{\mathrm d\theta}{\mathrm dt}=-\dfrac{1.8}a\dfrac{\rm deg}{\rm s}=-\dfrac{1.8}{d\sin40^\circ}\dfrac{\rm deg}{\rm s}

6 0
2 years ago
Communications satellites are placed in a circular orbit where they stay directly over a fixed point on the equator as the earth
zalisa [80]
Refer to the diagram shown below.

Given:
R = 6.37 x 10⁶ m, the radius of the earth
h = 3.58 x 10⁷ m, the height of the satellite above the earth's surface.
Therefore
R + h = 4.217 x 10⁷ m

In geosynchronous orbit, the period of rotation is 1 day.
Therefore the period is
T = (24 h)*(60 min/h)*(60 s/min) = 86400 s

The angular velocity is
ω = (2π rad)/(86400 s) = 7.2722 x 10⁻⁵ rad/s

Part (a)
The tangential speed is
v = (R+h)*ω
   = (4.217 x 10⁷ m)*(7.2722 x 10⁻⁵ rad/s) 
   = 3066.7 m/s
   = 3.067 km/s

Part (b)
The centripetal acceleration is
a = v²/(R+h)
   = (3066.7 m/s)²/(4.217 x 10⁷ m)
   = 0.223 m/s²

Answers:
(a) The speed is 3.067 km/s
(b) The acceleration is 0.223 m/s²

7 0
3 years ago
A student conducts an experiment in which a cart is pulled by a variable applied force during a 2 s time interval. In trial 1, t
fredd [130]

Answer:

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The moment is defined by

\to p = m v

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Let's replace the speeds in this equation

\Delta p = m (v_0 + \frac{F}{m t}) - m v_0\\\\\Delta p = m v_0  + F t - m v_0\\\\\Delta p = F t

They see that shift is not directly proportional to the mass of cars since the force and time were the same across all cars.

5 0
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Suppose two carts, one twice as massive as the other, fly apart when the
disa [49]

Answer:

the question this morning

Explanation:

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