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3 years ago

10

The chart shows data for an object moving at a constant acceleration. Which values best complete the chart? Time (s) Velocity (m

/s) X: 0 Y: 0 Z: 0 0 1 2. 3 0 X Y Z O X: 2 Y: 4 Z: 6 OX: 3 Y: 3 Z: 3 OX: 1 Y: 5 Z: 8

2 answers:

irina1246 [14]3 years ago

8 0

Answer:the answer is b

Explanation:

dangina [55]3 years ago

3 0

Answer:

A.

x: 0

y: 0

z: 0

Explanation:

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A car travels along a straight line at a constant speed of 53.0 mi/h for a distance d and then another distance d in the same di

Shalnov [3]

A distance of d is covered with 53 mile/hr initially. Time taken to cover this distance t1 = d/53 hour Next distance of d is covered with x mile hours. Time taken to cover this distance t2 = d/x hours. We have average speed = 26.5 mile / hour

= Total distance traveled/ total time taken

= $\frac{2d}{\frac{d}{53}+\frac{d}{x}} = \frac{2}{\frac{1}{53}+\frac{1}{x} } = \frac{106x}{x+53}$

$26.5 = \frac{106x}{x+53} \\ \\ 79.5 x = 1404.5\\ \\ x = 17.67 miles/hour$

5 0

3 years ago

Two cars are traveling along a straight road. Car A maintains a constant speed of 95 km/h and car B maintains a constant speed o

natulia [17]

Answer

given,

Speed of car A = 95 Km/h

= 95 x 0.278 = 26.41 m/s

Speed of Car B = 121 Km/h

= 121 x 0.278 = 33.64 m/s

Distance between Car A and B at t=0 = 41 Km

a) Distance travel by car B

d = 26.41 t + 41000

speed of the car A = 33.64 m/s

distance = s x t

26.41 t + 41000 = 33.64 x t

7.23 t = 41000

t = 5670.82 s

time taken by Car B to cross Car A is equal to t = 5670.82 s

distance traveled by car A

D = s x t = 26.41 x 5670.82 = 149766.25 m = 149.76 Km

b) distance travel by the car B in 30 s after overtaking car A

D' = s x t = 33.64 x 30 = 1009.2 m = 1 Km

8 0

3 years ago

In a coiled spring, the particles of the medium vibrate to and fro about their

Kitty [74]

Answer:

In a coiled spring, the particles of the medium vibrate to and fro about their mean positions at an angle of

A. 0° to the direction of propagation of wave

Explanation:

The waveform of a coiled spring is a longitudinal wave, which is made up of vibrations of the spring which are in the same direction as the direction of the wave's advancement

As the coiled spring experiences a compression force and is then released, it experiences a sequential movement of the wave of the compression that extends the length of the coiled spring which is then followed by a stretched section of the coiled spring in a repeatedly such that the direction of vibration of particles of the coiled is parallel to direction of motion of the wave

From which we have that the angle between the direction of vibration of the particles of the coiled spring and the direction of propagation of the wave is 0°.

8 0

2 years ago

Two identical loudspeakers are placed on a wall 1.00 m apart. A listener stands 4.00 m from the wall directly in front of one of

natita [175]

Answer:

The phase difference is 0.659 rad.

Explanation:

Given that,

Distance between two identical loudspeakers d= 1.00 m

Distance between speakers and listener r= 4.00 m

Frequency = 300 Hz

Suppose we need to find the phase difference in radian between the waves from the speakers when they reach the observer

We need to calculate the r'

Using Pythagorean theorem

$r'=\sqrt{d^2+r^2}$

Where, d = distance between two identical loudspeakers

r = distance between speakers and listener

Put the value into the formula

$r'=\sqrt{(1.00)^2+(4.00)^2}$

$r'=\sqrt{1.00+16.00}$

$r'=4.12\ m$

We need to calculate the path difference

Using formula of path difference

$|r'-r|=4.12-4.00$

$|r'-r|=0.12\ m$

We need to calculate the wavelength

Using formula of wavelength

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

Where, v = speed of sound

f = frequency

Put the value into the formula

$\lambda=\dfrac{343}{300}$

$\lambda=1.143\ m$

We need to calculate the phase difference

Using formula of phase difference

$\phi=\dfrac{2\pi\times|r'-r| }{\lambda}$

$\phi=\dfrac{2\pi\times0.12}{1.143}$

$\phi=0.659\ rad$

Hence, The phase difference is 0.659 rad.

7 0

3 years ago

A cat leaps into the air to catch a bird with an initial speed of 2.74 m/s at an angle of 60.0° above the ground. What is the hi

Volgvan

Answer: D. 0.29 m

Explanation:

We will use the following equations to describe the leap of the cat:

$y=V_{o}sin\theta t-\frac{gt^{2}}{2}$ (1)

$V_{y}=V_{oy}-gt$ (2)

Where:

$y$ is the height of the cat

$V_{oy}=V_{o}sin\theta$ is the cat's initial velocity

$\theta=60\°$

$g=9.8m/s^{2}$ is the acceleration due gravity

$t$ is the time

$V_{y}$ is the y-component of the velocity

Now the cat will have its maximum height $y_{max}$ when $V_{y}=0$ . So equation (2) is rewritten as:

$0=V_{oy}-gt$ (3)

Finding $t$ :

$t=\frac{V_{oy}}{g}=\frac{V_{o}sin\theta}{g}$ (4)

$t=\frac{2.74 m/s sin(60\°)}{9.8m/s^{2}}$ (5)

$t=0.24 s$ (6)

Substituting (6) in (1):

$y_{max}=(2.74 m/s)sin(60\°) (0.24 s)-\frac{(9.8m/s^{2})(0.24 s)^{2}}{2}$ (7)

Finally:

$y_{max}=0.287 m \approx 0.29 m$ (8)

3 0

3 years ago