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zhannawk [14.2K]
1 year ago
8

Calculate the initial speed of a box which slides on a horizontal floor surface and comes to rest after sliding 2 s. The coeffic

ient of kinetic friction between the box and the surface is 0.25. (g=9.8 ms-2).
Physics
1 answer:
NARA [144]1 year ago
3 0

A box sliding on a horizontal floor surface starts out moving at 4.9 m/s and stops after 2 seconds. The surface and the box have a kinetic friction has friction coefficient of 0.25 (g=9.8 ms-2).

The speed of an object at the start of a measurement, or an initial state, is known as the initial speed. The ratio of distance travelled to travel time, also known as the average speed, is the sum of the initial and final speeds. The difference between initial and ending speeds is the speed change. A force that acts between moving surfaces is referred to as kinetic friction. A force acting in opposition to the direction of a moving body on the surface is felt.

Interval (t) equals 2 seconds

Kinetic friction coefficient is 0.25.

gravity-induced acceleration (g) = 9.8 m/s2.

Coming to a stop at a final velocity  of 0 m/s.

Regular force  equals mg

Kinetic friction force is given by  = k.

N = μkmg

I = (delta P) (delta P)

Vo = (0.25)(9.8)(2) (2)

Vo = 4.9m/s

Learn more about kinetic friction here

brainly.com/question/17237604

#SPJ4

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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
The velocity of an object is the distance it travels per unit time. Suppose the velocity of a gilding bird is measured to be 52.
Elanso [62]

Answer:

d=7.115s

Explanation:

What problem says can be written mathematically as:

v=\frac{d}{t}

Where:

v=Velocity\\t=Time\\d=Distance

The problem itself it's really simple, we only need to replace the data provided in the previous equation, but first, let's convert the units of the velocity from cm/s to m/s because we have to work with the same units and working in meters is the most apropiate action, because is the base unit of length in the International System of Units:

52\frac{cm}{s} *\frac{1m}{100cm} =0.52\frac{m}{s}

Now, we can replace the data in the equation and find the time it will take the bird to travel 3.7 m:

0.52=\frac{3.7}{t}

Solving for t, multiplying by t both sides, and dividing by 0.52 both sides:

t=\frac{3.7}{0.52} =7.115384615s\approx7.115s

5 0
3 years ago
Consider this situation: A baseball player dives head-first
siniylev [52]
Of the forces listed I think the force of him diving and sliding across the infield acted on the player.

I think so because the slowing down was a result of an action, and I don’t think that should count as An action when it is the result of an action. However, the act of diving head-first into second base and sliding across the infield are independent actions and will cause friction, which will act upon the player.
7 0
2 years ago
On the asteroid Cere, the acceleration due to gravity is about 0.27 m/s^2. How long would it take a ball to fall to the ground i
KiRa [710]

Answer:

F. 25.82 s

Explanation:

Given:

Δy = 90 m

v₀ = 0 m/s

a = 0.27 m/s²

Find: t

Δy = v₀ t + ½ at²

90 m = (0 m/s) t + ½ (0.27 m/s²) t²

t = 25.82 s

7 0
3 years ago
at energy can be described by equations. Kinetic energy depends on both mass and velocity, which of following would have the mos
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