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2 years ago

A baseball player runs 27.4 meters from the

Physics

2 answers:

swat322 years ago

8 0

Answer:

Option (3) 3.0 m longer

Explanation:

This is a question of displacement here. Let's look at the problem again and analyse.

The first distance run by the base = 3 m

The second distance to the base = 3 m

The total distance run = 6 m

The first base lags by 3 m, so the distance = 6 - 3

= 3 m.

sleet_krkn [62]2 years ago

6 0

6.0 m longer because the player ran 3 and came back 3 at the very end, which looks like he went nowhere but in reality he ran 6.

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A charge Q is distributed uniformly along the x axis from x1 to x2. What would be the magnitude of the electric field at x0 on t

Lena [83]

Answer:

E = k Q 1 / (x₀-x₂) (x₀-x₁)

Explanation:

The electric field is given by

dE = k dq / r²

In this case as we have a continuous load distribution we can use the concept of linear density

λ= Q / x = dq / dx

dq = λ dx

We substitute in the equation

∫ dE = k ∫ λ dx / x²

We integrate

E = k λ (-1 / x)

We evaluate between the lower limits x = x₀- x₂ and higher x = x₀-x₁

E = k λ (-1 / x₀-x₁ + 1 / x₀-x₂)

E = k λ (x₂ -x₁) / (x₀-x₂) (x₀-x₁)

We replace the density

E = k (Q / (x₂-x₁)) [(x₂-x₁) / (x₀-x₂) (x₀-x₁)]

E = k Q 1 / (x₀-x₂) (x₀-x₁)

8 0

2 years ago

When the Glen Canyon hydroelectric power plant in Arizona is running at capacity, 690 m3 of water flows through the dam each sec

bixtya [17]

Answer:

1340.2MW

Explanation:

Hi!

To solve this problem follow the steps below!

1 finds the maximum maximum power, using the hydraulic power equation which is the product of the flow rate by height by the specific weight of fluid

W=αhQ

α=specific weight for water =9.81KN/m^3

h=height=220m

Q=flow=690m^3/s

W=(690)(220)(9.81)=1489158Kw=1489.16MW

2. Taking into account that the generator has a 90% efficiency, Find the real power by multiplying the ideal power by the efficiency of the electric generator

Wr=(0.9)(1489.16MW)=1340.2MW

the maximum possible electric power output is 1340.2MW

3 0

2 years ago

A floating leaf oscillates up and down two complete cycles in one second as a water wave passes by. The wave's wavelength is 10

postnew [5]

Answer:

C) 20 m/s

Explanation:

Wave: A wave is a disturbance that travels through a medium and transfers energy from one point to another, without causing any permanent displacement of the medium itself. Examples of wave are, water wave, sound wave, light rays, radio waves. etc.

The velocity of a moving wave is

v = λf ............................ Equation 1

Where v = speed of the wave, λ = wave length, f = frequency of the wave.

Given: f = 2 Hz (two complete cycles in one seconds), λ = 10 meters

Substituting these values into equation 1

v = 2×10

v = 20 m/s.

Thus the speed of the wave = 20 m/s

The right option is C) 20 m/s

7 0

2 years ago

A.Whale communication. Blue whales apparently communicate with each other using sound of frequency 17.0 Hz, which can be heard n

Y_Kistochka [10]

A. 90.1 m

The wavelength of a wave is given by:

$\lambda=\frac{v}{f}$

where

v is the speed of the wave

f is its frequency

For the sound emitted by the whale, v = 1531 m/s and f = 17.0 Hz, so the wavelength is

$\lambda=\frac{1531 m/s}{17.0 Hz}=90.1 m$

B. 102 kHz

We can re-arrange the same equation used previously to solve for the frequency, f:

$f=\frac{v}{\lambda}$

where for the dolphin:

v = 1531 m/s is the wave speed

$\lambda=1.50 cm=0.015 m$ is the wavelength

Substituting into the equation,

$f=\frac{1531 m/s}{0.015 m}=1.02 \cdot 10^5 Hz=102 kHz$

C. 13.6 m

Again, the wavelength is given by:

$\lambda=\frac{v}{f}$

where

v = 340 m/s is the speed of sound in air

f = 25.0 Hz is the frequency of the whistle

Substituting into the equation,

$\lambda=\frac{340 m/s}{25.0 Hz}=13.6 m$

D. 4.4-8.7 m

Using again the same formula, and using again the speed of sound in air (v=340 m/s), we have:

- Wavelength corresponding to the minimum frequency (f=39.0 Hz):

$\lambda=\frac{340 m/s}{39.0 Hz}=8.7 m$

- Wavelength corresponding to the maximum frequency (f=78.0 Hz):

$\lambda=\frac{340 m/s}{78.0 Hz}=4.4 m$

So the range of wavelength is 4.4-8.7 m.

E. 6.2 MHz

In order to have a sharp image, the wavelength of the ultrasound must be 1/4 of the size of the tumor, so

$\lambda=\frac{1}{4}(1.00 mm)=0.25 mm=2.5\cdot 10^{-4} m$

And since the speed of the sound wave is

v = 1550 m/s

The frequency will be

$f=\frac{v}{\lambda}=\frac{1550 m/s}{2.5\cdot 10^{-4} m}=6.2\cdot 10^6 Hz=6.2 MHz$

3 0

2 years ago

A bar 4.0m long weights 400 N. Its center of gravity is 1.5m from on end. A weight of 300N is attached at the light end. What ar

Debora [2.8K]

Answer:

The resulting, needed force for equilibrium is a reaction from a support, located at 2.57 meters from the heavy end. It is vertical, possitive (upwards) and 700 N.

Explanation:

This is a horizontal bar.

For transitional equilibrium, we just need a force opposed to its weight, thus vertical and possitive (ascendent). Its magnitude is the sum of the two weights, 400+300 = 700 N, since weight, as gravity is vertical and negative.

Now, the tricky part is the point of application, which involves rotational equilibrium. But this is quite simple if we write down an equation for dynamic momentum with respect to the heavy end (not the light end where the additional weight is placed). The condition is that the sum of momenta with respect to this (any) point of the solid bar is zero:

$0=\Sigma_{i}M=400\cdot1.5+300\cdot4-d\cdot700$

Where momenta from weights are possitive and the opposed force creates an oppossed momentum, then a negative term. Solving our unknown d:

$d=\frac{400\cdot1.5+300\cdot4}{700} =2.57 m$

So, the resulting force is a reaction from a support, located at 2.57 meters from the heavy end (the one opposed to the added weight end).

8 0

2 years ago