All categories

2 years ago

A leaf is floating on the Surface of the water,what will happen to its movement? Explain

Physics

1 answer:

docker41 [41]2 years ago

4 0

Answer:

MRCORRECT has answered the question

Explanation:

surface tension of water helps creatures(mostly of insecta class such as water striders) to walk on water. . it also helps water to move up the xylem tissue ofhigher plants without breaking up

You might be interested in

How does changing the lengthy but not the height of an inclined plane affect the work done to lift a load? PLZ HELP ME NOW'

Serhud [2]

Answer: Work Done would remain same.

Let us assume that the velocity is constant while taking the load up the inclined plane. Then, the kinetic energy would remain the same. This is because kinetic energy is dependent on velocity $(K.E.=\frac{1}{2}mv^2)$ . If that is constant, the kinetic energy would remain same. The potential energy is dependent on the height $(P.E.=mgh)$ . If the height is changed, then potential energy varies. In the question, it is mentioned that without changing the height, the length of the inclined plane is changed. Therefore, the potential energy would be same as before.

We know, work done is equal to potential energy plus kinetic energy. Since there is no change in any of these, the required work done would not change.

4 0

3 years ago

As shown in the diagram below, a 1 kg rock tied to a rope is

Tatiana [17]

Answer:

Explanation:

Angular momentum has a formula of L = mvr. Fillingin:

L = (1.0)(5.0)(1.0)

L = 5.0 kg*m/s

4 0

3 years ago

What is a planet’s period of rotation?

Annette [7]

The answer is the letter "C" ( I have honors science I am good at this type of stuff )

Hope I helped :) ( ask me for help when u need it :)

5 0

3 years ago

Read 2 more answers

Amir pitches a baseball at an initial height of 6 feet with a velocity of 73 feet per second. this can be represented by the fun

Romashka [77]

The values of t are <u>4.643 second</u> for the function $H(t)=-16t^2+73t+6$

What is batter misses?

An out in baseball happens when the umpire declares a batter or baserunner out. A hitter or runner who is out is no longer able to score runs and must go back to the dugout until their subsequent turn at bat. The batting team's turn is over after three outs are recorded in a half-inning.

In order to signal an out, umpires typically make a fist with one hand and then flex that arm, either upward on pop flies or forward on regular plays at first base. To indicate a called strikeout, home plate umpires frequently use a "punch-out" action.When a batter is struck by a pitched ball without making a swing at it, it is referred to as a hit-by-pitch. He consequently gets first base.

We have been given that

s = 6 feet

v = 73 feet per second

Substituting these values in the formula $H(t)=-16t^2+vt+s$

$H(t)=-16t^2+73t+6$

When the ball hits the ground, the height becomes zero. Thus, H(t)=0

$-16t^2+73t+6=0$

We solve the equation using quadratic formula $x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Substituting the values a=-16, b= 73, c=6

$t_{1,2}=\frac{-73 \pm \sqrt{(73)^2-4(-16)(6)}}{2(-16)}\\\Rightarrow t_{1,2}=\frac{-73 \pm \sqrt{5713}}{2(-16)}\\\Rightarrow t_{1,2}=-0.081, 4.643$

Learn more about the batter misses with the help of the given link:

brainly.com/question/19475098

#SPJ4

8 0

1 year ago

Two particles P and Q each moves towards ther along Straight Line (M)(N), 51m long. P starts from M with velocity 5m/s and const

bagirrra123 [75]

The time required by the particles are as follows:

a. t = 1.5 seconds

b. t = 3 seconds

c. t = 0.4 seconds

<h3>What is the time required?</h3>

The time required for the particles to be at several distances apart is calculated using the equation of motion given below:

$S = ut + \frac{1}{2}at^{2}$

a) Time required to be 30 m apart:

Assuming the distance covered by P is S1 and distance covered by Q is S2.

S1 + S2 = 51 - 30

S1 + S2 = 21

Substituting the values of velocity and acceleration in the equation of motion above:

5t + 1/2t^2 + 6t + 3t^2 = 21

2t^2 + 11t - 21 = 0

Solving for time, t by factorization, t = 1.5 seconds

b) Time required to meet:

Assuming the distance covered by P is S1 and distance covered by Q is S2.

S1 + S2 = 51

Substituting the values of velocity and acceleration in the equation of motion above:

5t + 1/2t^2 + 6t + 3t^2 = 51

2t^2 + 11t - 51 = 0

Solving for time, t by factorization, t = 3 seconds

c) Time required for velocity of P is ¾ of the velocity of Q:

Using the equation of motion: V = u + at

Vp = 3/4 Vq

4Vp= 3Vq

Substituting the values:

4(5 + t) = 3(6 + 3t)

5t = 2

t = 0.4 seconds

Learn more about distance, velocity and acceleration at: brainly.com/question/14344386

#SPJ1

7 0

2 years ago