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3 years ago

If the acceleration of a motorboat is 4.0 m/s2, and the motorboat starts from rest, what is its velocity after 6.0 s?

Physics

1 answer:

umka2103 [35]3 years ago

6 0

Answer:

The velocity of the motorboat after 6s is 24 m/s.

Explanation:

Given;

acceleration of the motorboat, a = 4.0 m/s²

initial velocity of the motorboat, u = 0

time of motion of the motorboat = 6s

Apply the following kinematic equation to determine the velocity of the motorboat after 6 ;

v = u + at

v = 0 + (4 x 6)

v = 24 m/s

Therefore, the velocity of the motorboat after 6s is 24 m/s.

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A starter pistol is fired 150m away from the spectators who hear the gun 0.5s after they see it fired. How fast does sound trave

Natali [406]

Answer:

300m/s

Explanation:

speed = distance/time

150/0.5

8 0

3 years ago

A 70.0 kg base runner moving at a speed of 4.0 m/s begins his slide into second base. The coefficient of friction between his cl

Andre45 [30]

Answer:

B. 560 J

J. 1.2 m

Explanation:

v = Final velocity = 0

u = Initial velocity = 4 m/s

$\mu$ = Coefficient of friction = 0.7

m = Mass of runner = 70 kg

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

Kinetic energy is given by

$K=\dfrac{1}{2}m(v^2-u^2)\\\Rightarrow K=\dfrac{1}{2}\times 70\times (0^2-4^2)\\\Rightarrow K=-560\ \text{J}$

The mechanical energy lost is 560 J

Acceleration is given by

$a=-\mu g\\\Rightarrow a=-0.7\times 9.81\\\Rightarrow a=-6.867\ \text{m/s}^2$

From kinematic equations we get

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-4^2}{2\times -6.867}\\\Rightarrow s=1.165\approx 1.2\ \text{m}$

The runner slides for 1.2 m

4 0

3 years ago

Please answer this question given in the picture

elena-s [515]

A plane mirror forms a virtual image behind the mirror. The image is as far behind the mirror as the object is in front of it. A cannot see his image because the length of the mirror is too short on his side. However, he can see the objects placed at points P and Q, but cannot see the object placed at point R

Hope this helps Buddy!

~ Courtney

6 0

4 years ago

Two guitarists attempt to play the same note of wavelength 6.50 cm at the same time, but one of the instruments is slightly out

PolarNik [594]

Answer:

The two value of the wavelength for the out of tune guitar is

$\lambda _2 = (6.48,6.52) \ cm$

Explanation:

From the question we are told that

The wavelength of the note is $\lambda = 6.50 \ cm = 0.065 \ m$

The difference in beat frequency is $\Delta f = 17.0 \ Hz$

Generally the frequency of the note played by the guitar that is in tune is

$f_1 = \frac{v_s}{\lambda}$

Where $v_s$ is the speed of sound with a constant value $v_s = 343 \ m/s$

$f_1 = \frac{343}{0.0065}$

$f_1 = 5276.9 \ Hz$

The difference in beat is mathematically represented as

$\Delta f = |f_1 - f_2|$

Where $f_2$ is the frequency of the sound from the out of tune guitar

$f_2 =f_1 \pm \Delta f$

substituting values

$f_2 =f_1 + \Delta f$

$f_2 = 5276.9 + 17.0$

$f_2 = 5293.9 \ Hz$

The wavelength for this frequency is

$\lambda_2 = \frac{343 }{5293.9}$

$\lambda_2 = 0.0648 \ m$

$\lambda_2 = 6.48 \ cm$

For the second value of the second frequency

$f_2 = f_1 - \Delta f$

$f_2 = 5276.9 -17$

$f_2 = 5259.9 Hz$

The wavelength for this frequency is

$\lambda _2 = \frac{343}{5259.9}$

$\lambda _2 = 0.0652 \ m$

$\lambda _2 = 6.52 \ cm$

8 0

3 years ago

Suppose that during any period of ¼ second there is one instant at which the crests or troughs of component waves are exactly in

kolezko [41]

<h3>Answer;</h3>

A. 4

<h3>Explanation;</h3>

The period of a wave or periodic time is the time taken for a complete oscillation to occur. For example its is the time taken between two successive crests or troughs.
The beats or oscillation that occur in one second represents the frequency. Frequency is the number of complete oscillations or beats in one second in a wave.
Frequency, measured in Hertz is given by the reciprocal of the periodic time.
Thus; Frequency or beats per second = 1/(1/4) = 4
Hence , 4 beats per second

6 0

4 years ago