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2 years ago

5) A person holds a 1.7 kg bucket and lets it move down at

Physics

1 answer:

Elena L [17]2 years ago

7 0

Answer:

12.5J

Explanation:

Given parameters:

Mass of bucket = 1.7kg

Height = 75cm = 0.75m

Unknown;

Work done on the bucket by the person = ?

Solution:

To solve this problem, we use the work done equation;

Work done = force x distance = mgh

m is the mass

g is the acceleration due to gravity

h is the height

Now, insert parameters and solve ;

Work done = 1.7 x 9.8 x 0.75 = 12.5J

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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

A 26-kg sled is on a snow-covered slope. The coeﬃcients of friction between the sled’s runners and the snow are µs = 0.096 and µ

sweet-ann [11.9K]

Answer:

Explanation:

Given

mass of sled =26 kg

coefficient of static friction $\mu _s=0.096$

coefficient of kinetic friction $\mu _k=0.072$

In order to move sled from rest we need to provide a force greater than static friction which is given by

$f_s=\mu mg=0.096\times 26\times 9.8=24.46 N$

After Moving Sled kinetic friction comes in to play which is less than static friction

$f_k=\mu _kmg=0.072\times 26\times 9.8=18.34 N$

therefore minimum force to keep moving sledge at constant velocity is 18.34 N

3 0

3 years ago

A pilot drops a bomb from a plane flying horizontally with constant velocity. When the bomb hits the ground, the horizontal loca

polet [3.4K]

Answer: The horizontal location of the plane will BE OVER THE BOMB

Explanation:

As soon as the bomb was dropped, the bomb will fall under gravity (free fall) and the location of the plane continues to increase horizontally till the bomb reaches the ground which is a falling distance to be travelled by the bomb at 9.8m/s²

8 0

3 years ago

How does water get to the tops of the tallest trees against the force of gravity

alexira [117]

Water gets to the leaves in the tops of the tallest trees by something called the cohesion-tension theory. Water has two very unique properties called adhesion and cohesion. Cohesion is the tendency of water molecules to stick together with one another. The water sticks together, leaving no room for air, strengthening the "force" of the water going up the tree. The water also sticks to the sides of the xylem inside the tree. In addition to these properties, there are also the factors of negative and positive water potential. For more information, look up more details of the cohesion-tension theory.

4 0

3 years ago

Read 2 more answers

WILL GIVE BRAINLIEST!

Ulleksa [173]

Answer:0kgm/s

Explanation:

Momentum before collision=momentum after collision

Since the momentum of the two blocks have positive sign, it means they are moving in thesame direction

Therefore we use the formula

Momentum (A)+momentum (B)=Momentum (A)+momentum (B)

25+35=60+momentum (B)

60=60+momentum (B)

Momentum (B)=60-60

Momentum (B)=0kgm/s

6 0

3 years ago