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Reptile [31]
3 years ago
15

How does the time at which you see things happen at a baseball game compare to the time when you hear things happen? Explain you

r answer
Physics
1 answer:
hoa [83]3 years ago
7 0

Answer:

Because of the speed of the sound.

Explanation:

The first thing that happens in such cases is to take into account the speed of the sound. First, we see that the player hits the ball with the bat, if we are in the stands far enough we will hear the sound of the batting time later, this time depends on the speed of the sound which is equal to 345 [m/s].

Another visible and practical example is a fireworks display, where people nearby immediately hear the explosion. while those at a great distance will be able to see first the explosion followed by the sound.

With the following equation, we can calculate how long it takes to hear a hit or explosion

t = x / v

where:

x = distance [m]

v = sound velocity = 345 [m/s]

t = time [s]

You might be interested in
At a time t = 2.80 s , a point on the rim of a wheel with a radius of 0.230 m has a tangential speed of 51.0 m/s as the wheel sl
bonufazy [111]

Answer:

Part A) the angular acceleration is α= 44.347 rad/s²

Part B) the angular velocity is 195.13 rad/s

Part C)  the angular velocity is 345.913 rad/s

Part D ) the time is t= 7.652 s

Explanation:

Part A) since angular acceleration is related with angular acceleration through:

α = a/R = 10.2 m/s² / 0.23 m =   44.347 rad/s²

Part B) since angular acceleration is related

since

v = v0 + a*(t-t0) =  51.0 m/s + (-10.2 m/s²)*(3.4 s - 2.8 s) = 44.88 m/s

since

ω = v/R = 44.88 m/s/ 0.230 m = 195.13 rad/s

Part C) at t=0

v = v0 + a*(t-t0) =  51.0 m/s + (-10.2 m/s²)*(0 s - 2.8 s) = 79.56 m/s

ω = v/R = 79.56 m/s/ 0.230 m = 345.913 rad/s

Part D ) since the radial acceleration is related with the velocity through

ar = v² / R → v= √(R * ar) = √(0.23 m  * 9.81 m/s²)= 1.5 m/s

therefore

v = v0 + a*(t-t0) → t =(v -  v0) /a + t0 = ( 1.5 m/s - 51.0 m/s) / (-10.2 m/s²) + 2.8 s = 7.652 s

t= 7.652 s

4 0
2 years ago
……………………………………………………………..?
kotegsom [21]

Answer:

C

Explanation:

First find the electrical wattage

W = I^2 * R

R = 12 ohms

I = 2 amps

Wattage = 2^2 * 12

Wattage = 4* 12

Wattage = 48 watts.

Now you need to use the power formula

Work = Power * Time

Work = ?

Power = 48 watts

Time = 3 minutes = 3 * 60 = 180 seconds.

Work = 48 * 180

Work = 8640 J

That's C

3 0
2 years ago
"Two long parallel wires 24.0 cm apart carry currents of 3.0 A and 8.0 A in the same direction. How far from the wire carrying 3
avanturin [10]

Answer:

6.5454 m

Explanation:

Let the distance from the wire carrying 3 A current is x

Then the distance from the the carrying current 8 A is 24-x

We know that magnetic field due to long wire is given by B=\frac{\mu _0i}{2\pi r}

It is given that magnetic field is zero at some distance so

\frac{\mu _0i_1}{2\pi x}=\frac{\mu _0i_2}{2\pi (24-x)}

Here i_1=3\ A \ and\  i_2=8\ A

So \frac{3}{x}=\frac{8}{24-x}=6.5454\ m

3 0
2 years ago
A book has been lifted 1.5 meters into the air giving it 30J of potential energy. How much force was used to lift it
joja [24]

Answer:

Force on the object is 20 N

Explanation:

As we know that work done to raise the book from initial position to final position is known as potential energy stored in it

So here we know that

U = F.s

here we know that

U = 30 J

s = displacement = 1.5 m

so we have

30 = F(1.5)

F = 20 N

6 0
3 years ago
The force F required to compress a spring a distance x is given by F 2 F0 5 kx where k is the spring constant and F0 is the prel
IrinaVladis [17]

Answer:

a)W=8.333lbf.ft

b)W=0.0107 Btu.

Explanation:

<u>Complete question</u>

The force F required to compress a spring a distance x is given by F– F0 = kx where k is the spring constant and F0 is the preload. Determine the work required to compress a spring whose spring constant is k= 200 lbf/in a distance of one inch starting from its free length where F0 = 0 lbf. Express your answer in both lbf-ft and Btu.

Solution

Preload = F₀=0 lbf

Spring constant k= 200 lbf/in

Initial length of spring x₁=0

Final length of spring x₂= 1 in

At any point, the force during deflection of a spring is given by;

F= F₀× kx  where F₀ initial force, k is spring constant and x is the deflection from original point of the spring.

W=\int\limits^2_1 {} \, Fds \\\\\\W=\int\limits^2_1( {F_0+kx} \,) dx \\\\\\W=\int\limits^a_b {kx} \, dx ; F_0=0\\\\\\W=k\int\limits^2_1 {x} \, dx \\\\\\W=k*\frac{1}{2} (x_2^{2}-x_1^{2}  )\\\\\\W=200*\frac{1}{2} (1^2-0)\\\\\\W=100.lbf.in\\\\

Change to lbf.ft by dividing the value by 12 because 1ft=12 in

100/12 = 8.333 lbf.ft

work required to compress the spring, W=8.333lbf.ft

The work required to compress the spring in Btu will be;

1 Btu= 778 lbf.ft

?= 8.333 lbf.ft----------------cross multiply

(8.333*1)/ 778 =0.0107 Btu.

6 0
3 years ago
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