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wlad13 [49]
2 years ago
7

The ____________________ of your voice refers to its intensity or loudness.

Physics
1 answer:
JulijaS [17]2 years ago
7 0
That's the amplitude.
You might be interested in
After landing on an unexplored Klingon planet, Spock tests for the direction of the magnetic field by firing a beam of electrons
BARSIC [14]

Answer:

Magnetic field is in south west direction .

Explanation:

Let us represent various direction by  i , j, k . i representing east , j representing north and k representing vertically upward direction .

magnetic field is represented vectorially as follows

B = B₀ ( - i - j )

In the first case velocity of electron

v = v k

Force = q ( v x B )

= -e [ vk x B₀ ( - i - j ) ]

= evB₀ ( j -i )

Direction of force is north -west .

In the second case velocity of electron

v = vj

Force = -e [ vj x B₀ ( - i - j ) ]

= - evB₀ k

force is downward

In the third case, velocity of electron

v = v( -j +i )

Force = -e [ v( -j +i ) x B₀ ( - i - j ) ]

= 2 evB₀ k

Force is upward.

3 0
3 years ago
A tree falls in a forest. How many years must pass before the 14C activity in 1.03 g of the tree's carbon drops to 1.02 decay pe
Illusion [34]

Answer:

t = 5.59x10⁴ y

Explanation:

To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:

A_{t} = A_{0}\cdot e^{- \lambda t}    (1)

<em>where A_{t}: is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.</em>

To find A₀ we can use the following equation:  

A_{0} = N_{0} \lambda   (2)

<em>where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon </em>

From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:        

N_{0} = \frac{m_{T} \cdot N_{A} \cdot abundance}{m_{^{12}C}}

<em>where m_{T}: is the tree's carbon mass, N_{A}: is the Avogadro's number and m_{^{12}C}: is the ¹²C mass.  </em>

N_{0} = \frac{1.03g \cdot 6.022\cdot 10^{23} \cdot 1.3\cdot 10^{-12}}{12} = 6.72 \cdot 10^{10} atoms ^{14}C    

Similarly, from equation (2) λ is:

\lambda = \frac{Ln(2)}{t_{1/2}}

<em>where t 1/2: is the half-life of ¹⁴C= 5700 years </em>

\lambda = \frac{Ln(2)}{5700y} = 1.22 \cdot 10^{-4} y^{-1}

So, the initial activity A₀ is:  

A_{0} = 6.72 \cdot 10^{10} \cdot 1.22 \cdot 10^{-4} = 8.20 \cdot 10^{6} decays/y    

Finally, we can calculate the time from equation (1):

t = - \frac{Ln(A_{t}/A_{0})}{\lambda} = - \frac {Ln(\frac{1.02decays \cdot 24h \cdot 365d}{1h\cdot 1d \cdot 1y \cdot 8.20 \cdot 10^{6} decays/y})}{1.22 \cdot 10^{-4} y^{-1}} = 5.59 \cdot 10^{4} y              

I hope it helps you!

4 0
3 years ago
Assume that the speed of light in a vacuum has the hypothetical value of 18.0 m/s. A car is moving at a constant speed of 14.0 m
denis23 [38]

Answer:

4.245s

Explanation:

Given that,

Hypothetical value of speed of light in a vacuum is 18 m/s

Speed of the car, 14 m/s

Time given is 6.76 s, and we're asked to find the observed time, T

The relationship between the two times can be given as

T = t / √[1 - (v²/c²)]

The missing variable were looking for is t, and we can find it if we rearrange the formula and make t the subject

t = T / √[1 - (v²/c²)]

And now, we substitute the values and insert into the equation

t = 6.76 * √[1 - (14²/18²)]

t = 6.76 * √[1 - (196/324)]

t = 6.76 * √(1 - 0.605)

t = 6.76 * √0.395

t = 6.76 * 0.628

t = 4.245 s

Therefore, the time the driver measures for the trip is 4.245s

8 0
3 years ago
A car is moving at 19 m/s along a curve on a horizontal plane with radius of curvature 49m.
JulsSmile [24]

Answer:

\mu =0.75

Explanation:

<u>Frictional Force </u>

When the car is moving along the curve, it receives a force that tries to take it from the road. It's called centripetal force and the formula to compute it is:

F_c=m.a_c

The centripetal acceleration a_c is computed as

\displaystyle a_c=\frac{v^2}{r}

Where v is the tangent speed of the car and r is the radius of curvature. Replacing the formula into the first one

F_c=m.\frac{v^2}{r}

For the car to keep on the track, the friction must have the exact same value of the centripetal force and balance the forces. The friction force is computed as

F_r=\mu N

The normal force N is equal to the weight of the car, thus

F_r=\mu .m.g

Equating both forces

\displaystyle \mu .m.g=m.\frac{v^2}{r}

Simplifying

\displaystyle \mu =\frac{v^2}{rg}

Substituting the values

\displaystyle \mu =\frac{19^2}{(49)(9.8)}

\boxed{\mu =0.75}

7 0
3 years ago
A car traveling at 60 mph has how much more energy than a car going at 15 mph?
VikaD [51]

Answer:

No, if a car is going faster. The RPM is obviously higher. If that is higher, you can burn through gas and energy much faster. A car going at 15mph would be cruising and wouldn't have to worry too much about burning our your vehicle.

Explanation:

7 0
3 years ago
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