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1 year ago

A person on a bicycle travels a distance of 12.5 km in 2.1 hrs, what was the speed of the person on the bicycle?

Physics

1 answer:

Leona [35]1 year ago

8 0

Answer:

5.9 km/hour

Explanation:

divide 12.5 by 2.1 and the answer is 5.9

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Kelly Clarkson is running between the Patronas towers in Kuala Lumpur on a tightrope at a speed of 15 m/s. Kelly currently weigh

Tomtit [17]

KE = 8887.5 J

<h3>Further explanation</h3>

Given

speed = 15 m/s

weight = 79 kg

Required

KE-Kinetic energy

Solution

Energy is the ability to do work

Energy because its motion is expressed as Kinetic energy (KE) which can be formulated as:

$\large {\boxed {\bold {KE = \frac {1} {2} mv ^ 2}}}$

Input the value :

KE = 1/2 x 79 kg x 15² m/s²

KE = 8887.5 J

8 0

3 years ago

Read 2 more answers

He is best known for publishing an almanac based on my astronomical calculations. Who is it?

slega [8]

Benjamin Banneker did this in 1792. Hope this helps

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3 years ago

A crane lifts a 425 kg steel beam vertically upward a distance of 66 m. How much work does the crane do on the beam if the beam

iogann1982 [59]

Answer:

W = 311074.5 [J]

Explanation:

In order to solve this problem we must analyze two parts, in the first part by means of Newton's second law we can determine the acceleration of the beam, remembering that the sum of the forces is equal to the product of mass by acceleration.

∑F = m*a

F = forces acting on the beam [N]

m = mass = 425 [kg]

a = acceleration = 1.8 [m/s²]

The forces acting on the beam are the force of the crane up (positive) and the weight of the beam down (negative)

$F_{crane}-(425*9.81)= 425*1.8\\F_{crane}=4713.25 [N]$

Now in the second part, we use the definition of work, which is equal to the product of the force applied in the direction of displacement, that is, the product of force by distance.

$W=F*d$

where:

W = work [J]

F = force = 4713.25 [N]

d = distance = 66 [m]

$W=4713.25*66\\W=311074.5[J]$

5 0

3 years ago

A projectile is launched at an angle of 30 and lands 20 s later at the same height as it was launched. (a) What is the initial s

Pavlova-9 [17]

Answer:

(a) 196 m/s

(b) 490 m

(c) 3394.82 m

(d) 2572.5 m

Explanation:

First of all, let us know one thing. When an object is thrown in the air, it experiences two forces acting in two different directions, one in the horizontal direction called air resistance and the second in the vertically downward direction due to its weight. In most of the cases, while solving numerical problems, air resistance is neglected unless stated in the numerical problem. This means we can assume zero acceleration along the horizontal direction.

Now, while solving our numerical problem, we will discuss motion along two axes according to our convenience in the course of solving this problem.

<u>Given:</u>

Time of flight = t = 20 s
Angle of the initial velocity of projectile with the horizontal = $\theta = 30^\circ$

<u>Assume:</u>

Initial velocity of the projectile = u
R = Range of the projectile during the time of flight
H = maximum height of the projectile
D = displacement of the projectile from the initial position at t = 15 s

Let us assume that the position from where the projectile was projected lies at origin.

Initial horizontal velocity of the projectile = $u\cos \theta$
Initial horizontal velocity of the projectile = $u\sin \theta$

Part (a):

During the time of flight the displacement of the projectile along the vertical is zero as it comes to the same vertical height from where it was projected.

$\therefore u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow u\sin \theta t=\dfrac{1}{2}(g)t^2\\\Rightarrow u=\dfrac{gt^2}{2\sin \theta t}\\\Rightarrow u=\dfrac{9.8\times 20^2}{2\sin 30^\circ \times 20}\\\Rightarrow u=196\ m/s$

Hence, the initial speed of the projectile is 196 m/s.

Part (b):

For a projectile, the time take by it to reach its maximum height is equal to return from the maximum height to its initial height is the same.

So, time taken to reach its maximum height will be equal to 10 s.

And during the upward motion of this time interval, the distance travel along the vertical will give us maximum height.

$\therefore H = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow H = 196\times \sin 30^\circ \times 10 + \dfrac{1}{2}\times(-9.8)\times 10^2\\ \Rightarrow H =490\ m$

Hence, the maximum altitude is 490 m.

Part (c):

Range is the horizontal displacement of the projectile from the initial position. As acceleration is zero along the horizontal, the projectile is in uniform motion along the horizontal direction.

So, the range is given by:

$R = u\cos \theta t\\\Rightarrow R = 196\times \cos 30^\circ \times 20\\\Rightarrow R =3394.82\ m$

Hence, the range of the projectile is 3394.82 m.

Part (d):

In order to calculate the displacement of the projectile from its initial position, we first will have to find out the height of the projectile and its range during 15 s.

$\therefore h = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow h = 196\times \sin 30^\circ \times 15 + \dfrac{1}{2}\times(-9.8)\times 15^2\\ \Rightarrow h =367.5\ m\\r = u\cos \theta t\\\Rightarrow r = 196\times \cos 30^\circ \times 15\\\Rightarrow r =2546.11\ m\\\therefore D = \sqrt{r^2+h^2}\\\Rightarrow D = \sqrt{2546.11^2+367.5^2}\\\Rightarrow D =2572.5\ m$

Hence, the displacement from the point of launch to the position on its trajectory at 15 s is 2572.5 m.

6 0

2 years ago

What does a black asphalt road become hotter than a white cement sidewalk in the same amount of sunlight?

victus00 [196]

The black means that it is a great emitter/absorber of the electromagnetic spectrum. The electromagnetic radiation is reflected of the white and absorbed nurture black meaning that the temperature of the black tarmac increases to that greater the the white

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3 years ago