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madreJ [45]
2 years ago
15

Relative to some reference points,your nose is in motion when you run. relative to others, it is not in motion. give one example

of each
Physics
1 answer:
Fed [463]2 years ago
8 0
When you run, your nose is moving in a straight line relative to a stone on the ground, and it's also moving back and forth relative to your left foot. But it's at rest relative to your ear.
You might be interested in
A 125-g metal block at a temperature of 93.2 °C was immersed in 100. g of water at 18.3 °C. Given the specific heat of the metal
Nataly_w [17]

Answer:

34.17°C

Explanation:

Given:

mass of metal block = 125 g

initial temperature T_i = 93.2°C

We know

Q = m c \Delta T   ..................(1)

Q= Quantity of heat

m = mass of the substance

c = specific heat capacity

c = 4.19 for H₂O in J/g^{\circ}C

\Delta T = change in temperature

Now

The heat lost by metal = The heat gained by the metal

Heat lost by metal = 125\times 0.9\times (93.2-T_f)

Heat gained by the water = 100\times 4.184\times(T_f -18.3)

thus, we have

125\times 0.9\times (93.2-T_f) = 100\times 4.184\times(T_f -18.3)

10485-112.5T_f = 418.4T_f - 7656.72

⇒ T_f = 34.17^oC

Therefore, the final temperature will be = 34.17°C

6 0
2 years ago
How long does it take for Saturn's equatorial flow, moving at 1500km/h, to encircle the planet?
Ludmilka [50]

Answer:

252.45 hours or 908820 seconds

Explanation:

The equatorial radius of Saturn is 60,268 km

The length of the equator will be circumference

2\pi 60268

Speed of the equatorial flow = 1500 km/h

Time=\dfrac{Distance}{Speed}\\\Rightarrow Time=\dfrac{2\pi 60268}{1500}\\\Rightarrow Time=252.45\ h=252.45\times 60\times 60=908820\ s

It will take 252.45 hours or 908820 seconds for the equatorial flow to encircle the planet.

4 0
1 year ago
Read 2 more answers
Two basketball teams are resting during halftime. While they are resting, a truck driver asks them for help pushing his broken d
viva [34]

Answer:

It is a superordinate goal because both teams could have helped with the task.

Explanation:

If both teams pushed then they could have made it happened

7 0
2 years ago
A golfer hits a golf ball at an angle of 25.0° to the ground. if the golf ball covers a horizontal distance of 301.5 m, what is
kvasek [131]

<u>Answer:</u>

 Maximum height reached = 35.15 meter.

<u>Explanation:</u>

Projectile motion has two types of motion Horizontal and Vertical motion.

Vertical motion:

         We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

         Considering upward vertical motion of projectile.

         In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g m/s^2 and final velocity = 0 m/s.

        0 = u sin θ - gt

         t = u sin θ/g

    Total time for vertical motion is two times time taken for upward vertical motion of projectile.

    So total travel time of projectile = 2u sin θ/g

Horizontal motion:

  We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

  In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 m/s^2 and time taken = 2u sin θ /g

 So range of projectile,  R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}

 Vertical motion (Maximum height reached, H) :

     We have equation of motion, v^2=u^2+2as, where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.

   Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H

   0^2=(usin\theta) ^2-2gH\\ \\ H=\frac{u^2sin^2\theta}{2g}

In the give problem we have R = 301.5 m,  θ = 25° we need to find H.

So  \frac{u^2sin2\theta}{g}=301.5\\ \\ \frac{u^2sin(2*25)}{g}=301.5\\ \\ u^2=393.58g

Now we have H=\frac{u^2sin^2\theta}{2g}=\frac{393.58*g*sin^2 25}{2g}=35.15m

 So maximum height reached = 35.15 meter.

7 0
1 year ago
An organ pipe open at both ends has a radius of 4.0 cm and a length of 6.0 m. what is the frequency (in hz) of the third harmoni
Marysya12 [62]

When air is blown into the open pipe,

L = \frac{nλ}{2}

where nis any integral number 1,2,3,4 etc. and λ is the wavelength of the oscillation

⇒λ=\frac{2L} {n}

Note here that n=1 is for fundamental, n=2 is first harmonic and so on..

⇒ third harmonic will be n=4

Given L=6m, n=4, solving for λ we get:

λ=\frac{(2)*(6)}{4} =3m

Relationship of frequency(f), velocity of sound (c) and wavelength(λ) is:

c=f.λ Or f= \frac{c}{λ}

⇒f=\frac{344}{3}

≈115 Hz

8 0
2 years ago
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