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3 years ago

5

Reflection of the images example

1 answer:

NISA [10]3 years ago

8 0

Answer:

reflection of water ,sound and water waves etc

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In 0.60 seconds, a projectile goes from 0 to 610 m/s. What is the acceleration of the projectile?

IceJOKER [234]

Answer: $a=1016.66 m/s^{2}$

Explanation:

Acceleration $a$ is expressed in the following formula:

$a=\frac{V_{f}-V_{o}}{t}$

Where:

$V_{f}=610 m/s$ is the final velocity of the projectile

$V_{o}=0 m/s$ is the initial velocity of the projectile

$t=0.6 s$ is the time

Solving:

$a=\frac{610 m/s-0 m/s}{0.6 s}$

$a=1016.66 m/s^{2}$ This is the acceleration of the projectile

6 0

3 years ago

Two college teams, A and B, are playing a game of tug of war. Team A applies a force of 103 newtons, but the rope does not move

Kipish [7]

Given that the rope is not moving (acceleration is zero), by the second Law of Newton (F=m*a), the net force acting on the rope is zero.

Then, the force applied by the team B equals the force applied by the tema A: 103 N.

7 0

2 years ago

Read 2 more answers

An aircraft with a mass of 10,000 kg starts from rest at sea level and takes off, then flies to a cruising speed of 620 km/h and

Natasha_Volkova [10]

Answer:

The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

Explanation:

Given that,

Mass of aircraft = 10000 kg

Speed = 620 km/h = 172.22 m/s

Altitude = 10 km = 1000 m

We calculate the change in potential energy

$\Delta P.E=mg(h_{2}-h_{1})$

$\Delta P.E=10000\times9.8\times(10000-0)$

$\Delta P.E=10000\times9.8\times10000$

$\Delta P.E=980000000\ J$

$\Delta P.E=980\ MJ$

For g = 10 m/s²,

The change in potential energy will be 1000 MJ.

We calculate the change in kinetic energy

$\Delta K.E=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)$

$\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2-0^2)$

$\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2)$

$\Delta K.E=148298642\ J$

$\Delta K.E=148.3\ MJ$

For g = 10 m/s²,

The change in kinetic energy will be 150 MJ.

Hence, The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

7 0

2 years ago

Match the object with its characteristic.

Triss [41]

What object do you need to match

8 0

3 years ago

A small branch is wedged under a 200 kg rock and rests on a smaller object. The smaller object is 2.0 m from the large rock and

Alexxandr [17]

Answer:

a

$F =326.7 \ N$

b

$M = 6$

Explanation:

From the question we are told that

The mass of the rock is $m_r = 200 \ kg$

The length of the small object from the rock is $d = 2 \ m$

The length of the small object from the branch $l = 12 \ m$

An image representing this lever set-up is shown on the first uploaded image

Here the small object acts as a fulcrum

The force exerted by the weight of the rock is mathematically evaluated as

$W = m_r * g$

substituting values

$W = 200 * 9.8$

$W = 1960 \ N$

So at equilibrium the sum of the moment about the fulcrum is mathematically represented as

$\sum M_f = F * cos \theta * l - W cos\theta * d = 0$

Here $\theta$ is very small so $cos\theta * l = l$

and $cos\theta * d = d$

Hence

$F * l - W * d = 0$

=> $F = \frac{W * d}{l}$

substituting values

$F = \frac{1960 * 2}{12}$

$F =326.7 \ N$

The mechanical advantage is mathematically evaluated as

$M = \frac{W}{F}$

substituting values

$M = \frac{1960}{326.7}$

$M = 6$

6 0

3 years ago