All categories

3 years ago

What is physics and important of physics

2 answers:

8 0

Physics is the explanation of our reality, both visible and invisible. With it we can understand the world around us, why things are how they are, and why things do what they do. This understanding allows humans to develop objects that improve quality of life.

Arte-miy333 [17]3 years ago

4 0

Answer:

Physics is the answer to all the forces in nature and it is important to know all of that in a day to day life to succeed in life

Explanation:

You might be interested in

How many ohms of resistance are in a 120–volt hair dryer that draws 7.6 amps of current?

miv72 [106K]

From Ohm's law . . . Resistance = (voltage) / (current)

Resistance = (120 volts) / (7.6 Amperes)

<em>Resistance = 15.8 Ω</em>

7 0

3 years ago

The greatest speed with which an athlete can jump vertically is around 5 m/sec. Determine the speed at which Earth would move do

katrin2010 [14]

Answer:

Approximately $2.0 \times 10^{-23}\; \rm m \cdot s^{-1}$ if that athlete jumped up at $1.8\; \rm m \cdot s^{-1}$ . (Assuming that $g = 9.81\; \rm m\cdot s^{-1}$ .)

Explanation:

The momentum $p$ of an object is the product of its mass $m$ and its velocity $v$ . That is: $p = m \cdot v$ .

Before the jump, the speed of the athlete and the earth would be zero (relative to each other.) That is: $v(\text{athlete, before}) = 0$ and $v(\text{earth, before}) = 0$ . Therefore:

$\begin{aligned}& p(\text{athlete, before}) = 0\end{aligned}$ and $p(\text{earth, before}) = 0$ .

Assume that there is no force from outside of the earth (and the athlete) acting on the two. Momentum should be conserved at the instant that the athlete jumped up from the earth.

Before the jump, the sum of the momentum of the athlete and the earth was zero. Because momentum is conserved, the sum of the momentum of the two objects after the jump should also be zero. That is:

$\begin{aligned}& p(\text{athlete, after}) + p(\text{earth, after}) \\ & =p(\text{athlete, before}) + p(\text{earth, before}) \\ &= 0\end{aligned}$ .

Therefore:

$p(\text{athelete, after}) = - p(\text{earth, after})$ .

$\begin{aligned}& m(\text{athlete}) \cdot v(\text{athelete, after}) \\ &= - m(\text{earth}) \cdot v(\text{earth, after})\end{aligned}$ .

Rewrite this equation to find an expression for $v(\text{earth, after})$ , the speed of the earth after the jump:

$\begin{aligned} &v(\text{earth, after}) \\ &= -\frac{m(\text{athlete}) \cdot v(\text{athlete, after})}{m(\text{earth})} \end{aligned}$ .

The mass of the athlete needs to be calculated from the weight of this athlete. Assume that the gravitational field strength is $g = 9.81\; \rm N \cdot kg^{-1}$ .

$\begin{aligned}& m(\text{athlete}) = \frac{664\; \rm N}{9.81\; \rm N \cdot kg^{-1}} \approx 67.686\; \rm N\end{aligned}$ .

Calculate $v(\text{earth, after})$ using $m(\text{earth})$ and $v(\text{athlete, after})$ values from the question:

$\begin{aligned} &v(\text{earth, after}) \\ &= -\frac{m(\text{athlete}) \cdot v(\text{athlete, after})}{m(\text{earth})} \\ &\approx -2.0 \times 10^{-23}\; \rm m \cdot s^{-1}\end{aligned}$ .

The negative sign suggests that the earth would move downwards after the jump. The speed of the motion would be approximately $2.0 \times 10^{-23}\; \rm m \cdot s^{-1}$ .

3 0

3 years ago

A soccer ball of diameter d and mass m rolls up a hill without slipping, reaching a maximum height of h above the base of the hi

stich3 [128]

Answer:

Convert the units to standard units.

r = d / 2 (cm)

r = r in cm [ 1 m / 100 cm ]

m = m in g * [ 1 kg / 1000 g] = m in kg

Write the general equation for this problem.

The KE at the bottom of the hill = the PE at the top.

1/2 I w^2 + 1/2 m v^2 = mgh

I for a hollow sphere = 2/3xmxr^2

1/2 (2/3 x mr^2 x w^2) + 1/2 x m x v^2 = mgh

Now we need to relate w with v

v = w x r

w = v / r put this result in for w

1/2 (2/3 x mxr^2 z v^2 / r^2) + 1/2 m x v^2 = mgh

1/2 (2/3 x m v^2) + 1/2 x m x v^2 = mgh

1/3 m x v^2 + 1/2 mv^2 = mgh

5/6 m x v^2 = mgh Notice that the m's are all the same on both sides. They cancel.

5/6 v^2 = gx h

Solve for v

v^2 = 6/5 x g x h

Find the Rotational Kinetic Energy (KE_r)

PE - KE_translational = KE_r

mgh - 1/2 x m x v^2 = KE_r

Find the rotational rate at the bottom w

KE_r = 1/2 I x w^2

I = 2/3 m R^2

7 0

3 years ago

A .71-kg billiard ball moving at 2.5 m/s in the x-direction strikes a stationary ball of the same mass. After the collision, the

Finger [1]

After the collision, the momentum didn't change, so the total momentum in x and y are the same as the initial.

The x component was calculated by subtracting the initial momentum (total) minus the momentum of the first ball after the collision

In the y component, as at the beginning, the total momentum was 0 in this axis, the sum of both the first and struck ball has to be the same in opposite directions. In other words, both have the same magnitude but in opposite directions

$\begin{gathered} Py=0.71kg\cdot2.17\cdot sin(30) \\ Py=0.77kg\cdot m/s \end{gathered}$

This is for both balls after the collision, but one goes in a positive and the other in a negative direction.

7 0

1 year ago

A steel beam that is 6.50 m long weighs 336 N. It rests on two supports, 3.00 m apart, with equal amounts of the beam extending

FinnZ [79.3K]

Answer:

Explanation:

When beam is balanced and not rotating with Suki standing on it , let reaction force on the supports be R₁ and R₂. Then

R₁ +R₂ = 336 + 590

= 929

Now the moment beam begins to tip , reaction on distant support R₁ = 0

only R₂ will exists on the support near to Suki.

Taking torque about this support of weight of beam acting from the middle point and weight of suki of 590N ,who is x distance from the support towards the other end.

336 x 1.5 = 590 x

x = .85 m

ie , from second support , Suki can not go beyond a distance of .85 m towards the second end.

4 0

3 years ago

What is physics and important of physics​

What is physics and important of physics