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asambeis [7]
2 years ago
7

Calculate the average speed of a runner who runs for 500 meters in 40 second HELP!

Physics
1 answer:
lina2011 [118]2 years ago
8 0

Answer:

12.5

Explanation:

500 divided by 40 = 12.5

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Place the following into scientific notation...(.000635)<br> *
AleksAgata [21]
6.35x10^-4 OR 6.3x10-4 (if only one decimal number is allowed)
6 0
3 years ago
If you are at latitude 43 degrees north of Earth's equator, what is the angular distance (in degrees) from your zenith to the no
kirill115 [55]

Answer:

Your zenith is 43 N of 90 deg (equator)

Thus, your zenith is 90 - 43 = 47 deg

(At the N pole your zenith would be 0 deg from the N pole)

8 0
2 years ago
A 10 g particle undergoes SHM with an amplitude of 2.0 mm and a maximum acceleration of magnitude 8.0 multiplied by 103 m/s2, an
Nat2105 [25]

Answer:

a)T=0.0031416s

b)v_{max}=6.283\frac{m}{s}

c) E=0.1974J

d)F=80N

e)F=40N

Explanation:

1) Important concepts

Simple harmonic motion is defined as "the motion of a mass on a spring when it is subject to the linear elastic restoring force given by Hooke's Law (F=-Kx). The motion experimented by the particle is sinusoidal in time and demonstrates a single resonant frequency".

2) Part a

The equation that describes the simple armonic motion is given by X=Acos(\omega t +\phi)    (1)

And taking the first and second derivate of the equation (1) we obtain the velocity and acceleration function respectively.

For the velocity:

\frac{dX}{dt}=v(t)=-A\omega sin(\omega t +\phi)   (2)

For the acceleration

\frac{d^2 X}{dt}=a(t)=-A\omega^2 cos(\omega t+\phi)   (3)

As we can see in equation (3) the acceleration would be maximum when the cosine term would be -1 and on this case:

A\omega^2=8x10^{3}\frac{m}{s^2}

Since we know the amplitude A=0.002m  we can solve for \omega like this:

\omega =\sqrt{\frac{8000\frac{m}{s^2}}{0.002m}}=2000\frac{rad}{s}

And we with this value we can find the period with the following formula

T=\frac{2\pi}{\omega}=\frac{2 \pi}{2000\frac{rad}{s}}=0.0031416s

3) Part b

From equation (2) we see that the maximum velocity occurs when the sine function is euqal to -1 and on this case we have that:

v_{max}=A\omega =0.002mx2000\frac{rad}{s}=4\frac{m rad}{s}=4\frac{m}{s}

4) Part c

In order to find the total mechanical energy of the oscillator we can use this formula:

E=\frac{1}{2}mv^2_{max}=\frac{1}{2}(0.01kg)(6.283\frac{m}{s})^2=0.1974J

5) Part d

When we want to find the force from the 2nd Law of Newton we know that F=ma.

At the maximum displacement we know that X=A, and in order to that happens cos(\omega t +\phi)=1, and we also know that the maximum acceleration is given by::

|\frac{d^2X}{dt^2}|=A\omega^2

So then we have that:

F=ma=mA\omega^2

And since we have everything we can find the force

F=ma=0.01Kg(0.002m)(2000\frac{rad}{s})^2 =80N

6) Part e

When the mass it's at the half of it's maximum displacement the term cos(\omega t +\phi)=1/2 and on this case the acceleration would be given by;

|\frac{d^2X}{dt^2}|=A\omega^2 cos(\omega t +\phi)=A\omega^2 \frac{1}{2}

And the force would be given by:

F=ma=\frac{1}{2}mA\omega^2

And replacing we have:

F=\frac{1}{2}(0.01Kg)(0.002m)(2000\frac{rad}{s})^2 =40N

8 0
2 years ago
A sound wave of frequency
Artyom0805 [142]

Answer:

0.283m

Explanation:

Speed (v) = wavelength × Frequency (f)

Wavelength = speed/ frequency

Wavelength = 343/ 1210 = 0.283m

8 0
3 years ago
A thin rod (length = 1.09 m) is oriented vertically, with its bottom end attached to the floor by means of a frictionless hinge.
Murljashka [212]

Answer:

a) w = 4.24 rad / s , b) α  = 8.99 rad / s²

Explanation:

a) For this exercise we use the conservation of kinetic energy,

Initial. Vertical bar

        Emo = U = m g h

Final. Just before touching the floor

       Emf = K = ½ I w2

As there is no friction the mechanical energy is conserved

       Emo = emf

       mgh = ½ m w²

The moment of inertial of a point mass is

        I = m L²

       m g h = ½ (m L²) w²

       w = √ 2gh / L²

The initial height h when the bar is vertical is equal to the length of the bar

         h = L

         w = √ 2g / L

Let's calculate

       w = RA (2 9.8 / 1.09)

       w = 4.24 rad / s

b) Let's use Newton's equation for rotational motion

         τ = I α

         F L = (m L²) α

The force applied is the weight of the object, which is at a distance L from the point of gro

         mg L = m L² α  

          α  = g / L

          α  = 9.8 / 1.09

          α  = 8.99 rad / s²

5 0
3 years ago
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