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2 years ago

Mr. Smith is designing a race where velocity will be measured. Which course would allow velocity to accurately get a winner?

Physics

2 answers:

liraira [26]2 years ago

6 0

I’m not completely sure but most likely is is the 10 mile bike ride, I hope I can help! (:

Arturiano [62]2 years ago

5 0

Answer:

Option A

Explanation:

(a 5 mile course from the center of town heading north to the outskirts of town)

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In 2014, the Rosetta space probe reached the comet Churyumov Gerasimenko. Although the comet's core is actually far from spheric

Viktor [21]

To solve this problem we will apply the concepts related to gravity according to the Newtonian definitions. From finding this value we will use the linear motion kinematic equations to find the time. Our values are

Comet mass $M = 1.0*10^{13} kg$

Radius $r = 1.6km = 1600 m$

Rock was dropped from a height 'h' from surface = 1m

The relation for acceleration due to gravity of a body of mass 'm' with radius 'r' is

$g = \frac{GM}{R^2}$

Where G means gravitational universal constant and M the mass of the planet

$g = \frac{(6.67408*10^{-11})(1*10^{13})}{1600^2}$

$g = 2.607*10^{-4} m/s^2$

Now calculate the value of the time

$h = \frac{1}{2} gt^2$

$t = \sqrt{\frac{2h}{g}}$

$t = \sqrt{\frac{2(1)}{2.607*10^{-4}}}$

$t = 87.58s$

The time taken for the rock to reach the surface is t = 87.58s

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2 years ago

The position function x(t) of a particle moving along an x axis is x = 4.00 - 6.00t2, with x in meters and t in seconds. (a) at

elena-14-01-66 [18.8K]

The position function x(t) of a particle moving along an x axis is $x=4.00 - 6.00t^2$

a) The point at which particle stop, it's velocity = 0 m/s

So dx/dt = 0

0 = 0- 12t = -12t

So when time t= 0, velocity = 0 m/s

So the particle is starting from rest.

At t = 0 the particle is (momentarily) stop

b) When t = 0

$x=4.00 - 6.00*0^2 = 4m$

SO at x = 4m the particle is (momentarily) stop

c) We have $x=4.00 - 6.00t^2$

At origin x = 0

Substituting

$0 = 4.00 - 6.00t^2\\ \\ t^2 = \frac{2}{3}$

t = 0.816 seconds or t = - 0.816 seconds

So when t = 0.816 seconds and t = - 0.816 seconds, particle pass through the origin.

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2 years ago

The diagram shows a heat engine. In which area of the diagram is unusable thermal energy detected?

Marat540 [252]

Nope, I disagree with the former answer. The answer is definitely Z. <u>W area</u> (boxed with red outline) is represented as the hot reservoir while <u>Z area</u> is the cold reservoir (boxed with blue outline). X area is the heat engine itself and Y area is the work produced from thermal energy from hot reservoir. Typically, all heat engines lose some heat to the environment (based from the second law of thermodynamics) that is symbolically illustrated by the lost energy in the cold reservoir. This lost thermal energy is basically the unusable thermal energy. The higher thermal energy lost, the less efficient your heat engine is.

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2 years ago

Read 2 more answers

Cylinder A is moving downward with a velocity of 3 m/s when the brake is suddenly applied to the drum. Knowing that the cylinder

Xelga [282]

Answer:

Incomplete question

Check attachment for the given diagram

Explanation:

Given that,

Initial Velocity of drum

u=3m/s

Distance travelled before coming to rest is 6m

Since it comes to rest, then, the final velocity is 0m/s

v=3m/s

Using equation of motion to calculate the linear acceleration or tangential acceleration

v²=u²+2as

0²=3²+2×a×6

0=9+12a

12a=-9

Then, a=-9/12

a=-0.75m/s²

The negative sign shows that the cylinder is decelerating.

Then, a=0.75m/s²

So, using the relationship between linear acceleration and angular acceleration.

a=αr

Where

a is linear acceleration

α is angular acceleration

And r is radius

α=a/r

From the diagram r=250mm=0.25m

Then,

α=0.75/0.25

α =3rad/sec²

The angular acceleration is =3rad/s²

b. Time take to come to rest

Using equation of motion

v=u+at

0=3-0.75t

0.75t=3

Then, t=3/0.75

t=4 secs

The time take to come to rest is 4s

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2 years ago

You are in a hot-air balloon that, relative to the ground, has a veloc- ity of 6.0 m/s in a direction due east. You see a hawk m

Ann [662]

Answer:

6.32 m/s 18.43° northeast

Explanation:

We express the velocity of hawk as:

$v_{Hawk}=v_{balloon}+v_{HawkRelativetoBalloon}=6 x+2 y$

We consider positive x towards east and positive y due north. So the magnitude is simply the square root of the square components:

$|v_{hawk}|=\sqrt[]{6^2+2^2}=\sqrt{40}$ ≈ $6.32 m/s$

And the angle with respect to the east should be with:

$arctan(\frac{2}{6} )=18.43 \°$

8 0

2 years ago