Airbags will deploy in a head-on collision, but not in a collision that occurs from an angle.

The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given b

Mnenie [13.5K]

Explanation:

The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given by :

$Q=t^3-2t^2+4t+4$

We need to find the current flowing. We know that the rate of change of electric charge is called electric current. It is given by :

$I=\dfrac{dQ}{dt}\\\\I=\dfrac{d(t^3-2t^2+4t+4)}{dt}\\\\I=3t^2-4t+4$

At t = 1 s,

Current,

$I=3(1)^2-4(1)+4\\\\I=3\ A$

So, the current at t = 1 s is 3 A.

For lowest current,

$\dfrac{dI}{dt}=0\\\\\dfrac{d(3t^2-4t+4)}{dt}=0\\\\6t-4=0\\\\t=0.67\ s$

Hence, this is the required solution.

7 0

3 years ago

A sound is recorded at 19 decibels. What is the intensity of the sound?

sp2606 [1]

$1 \times 10^{-10.1} \mathrm{Wm}^{-2}$ is the intensity of the sound.

Answer: Option B

<u>Explanation:</u>

The range of sound intensity that people can recognize is so large (including 13 magnitude levels). The intensity of the weakest audible noise is called the hearing threshold. (intensity about $1 \times 10^{-12} \mathrm{Wm}^{-2}$ ). Because it is difficult to imagine numbers in such a large range, it is advisable to use a scale from 0 to 100.

This is the goal of the decibel scale (dB). Because logarithm has the property of recording a large number and returning a small number, the dB scale is based on a logarithmic scale. The scale is defined so that the hearing threshold has intensity level of sound as 0.

$\text { Intensity }(d B)=(10 d B) \times \log _{10}\left(\frac{I}{I_{0}}\right)$

Where,

I = Intensity of the sound produced

$I_{0}$ = Standard Intensity of sound of 60 decibels = $1 \times 10^{-12} \mathrm{Wm}^{-2}$

So for 19 decibels, determine I as follows,

$19 d B=(10 d B) \times \log _{10}\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)$

$\log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=\frac{19}{10}$

$\log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=1.9$

When log goes to other side, express in 10 to the power of that side value,

$\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)=10^{1.9}$

$I=1 \times 10^{-12} \mathrm{Wm}^{-2} \times 10^{1.9}=1 \times 10^{-12-1.9}=1 \times 10^{-10.1} \mathrm{Wm}^{-2}$

5 0

3 years ago

How are longitudinal and transverse waves different?

Natalija [7]

Answer:

A longitudinal wave is a wave where the movement of the medium is in the same direction as the wave. On the other hand, a transverse wave is a wave where the movement of the medium is at a right angle to the wave direction.

Explanation:i got this right on a quiz so i know its right

3 0

3 years ago

The first person to map the Milky Way Galaxy using radio waves was

Art [367]

Answer:B

Explanation:Grote reber was the first scientist to map the milky way galaxy using radio waves.

7 0

2 years ago

Read 2 more answers

A water balloon is thrown horizontally from a tower that is 45 m high. It strikes the shoes of an unsuspecting passerby who is 4

LekaFEV [45]

Answer:

14.85 m/s

Explanation:

From the question given above, the following data were obtained:

Height (h) of tower = 45 m

Horizontal distance (s) moved by the balloon = 45 m

Horizontal velocity (u) =?

Next, we shall determine the time taken for the balloon to hit the shoe of the passerby. This is illustrated below:

Height (h) of tower = 45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

45 = ½ × 9.8 × t²

45 = 4.8 × t²

Divide both side by 4.9

t² = 45/4.9

Take the square root of both side

t = √(45/4.9)

t = 3.03 s

Finally, we shall determine the magnitude of the horizontal velocity of the balloon as shown below:

Horizontal distance (s) moved by the balloon = 45 m

Time (t) = 3.03 s

Horizontal velocity (u) =?

s = ut

45 = u × 3.03

Divide both side by 3.03

u = 45/3.03

u = 14.85 m/s

Thus, the magnitude of the horizontal velocity of the balloon was 14.85 m/s

4 0

3 years ago