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katrin2010 [14]

2 years ago

6

A truck moves 70 m east, then moves 120 m west, and finally moves east again a distance of 90 m. If east is chosen as the positi

ve direction, what is the truck's resultant displacement

1 answer:

asambeis [7]2 years ago

6 0

Answer:

140m east

Explanation:

If East is positive then lets rephrase the problem into integers

A truck moves +70 m, then moves -120m, and finally moves +90m.

So totally Displacement = +70-120+90= +140m

Since east is positive, the trucks resultant displacement is 140 m east of origin

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Which of the following best describes a plane?

katrin [286]

B, the surface of a flat table.

4 0

3 years ago

Read 2 more answers

A sound has 13 crests and 15 troughs in 3 seconds. When the second crest is produced the first is 2cm away from the source? Calc

Yuki888 [10]

Answer:

The wavelength will be 4 cm, frequency will be 4.66 Hz and wave speed is 18.6 cm/sec

Explanation:

Given:

No. of crest = 13

No. of trough = 15

Time = 3 seconds

Hence, 1 crest or 1 trough = $\frac{1}{2}\lambda$

therefore,

13 C + 15 T = $28(\frac{1}{2}\lambda)$

= $14\lambda$

Time given 3 seconds

= $\frac{3}{14}s$

$\nu= \frac{14}{3}$

$\nu= 4.6 Hz \approx 5 Hz$

2 cm distance is travelled is time period

$\lambda = 4 cm$

Again wave will travel in 1 T = 4 cm

wave speed v = $\lambda\times\nu$

= $4\times\frac{14}{3}$

= 18.6 cm/s

5 0

3 years ago

The distance between the first and fifth minima of a single-slit diffraction pattern is 0.500 mm with the screen 37.0 cm away fr

WARRIOR [948]

In the single-slit experiment, the displacement of the minima of the diffraction pattern on the screen is given by
$y_n= \frac{n \lambda D}{a}$ (1)
where
n is the order of the minimum
y is the displacement of the nth-minimum from the center of the diffraction pattern
$\lambda$ is the light's wavelength
D is the distance of the screen from the slit
a is the width of the slit

In our problem,
$D=37.0 cm=0.37 m$
$\lambda=530 nm=5.3 \cdot 10^{-7} m$
while the distance between the first and the fifth minima is
$y_5-y_1 = 0.500 mm=0.5 \cdot 10^{-3} m$ (2)

If we use the formula to rewrite $y_5, y_1$ , eq.(2) becomes
$\frac{5 \lambda D}{a} - \frac{1 \lambda D}{a} =\frac{4 \lambda D}{a}= 0.5 \cdot 10^{-3} m$
Which we can solve to find a, the width of the slit:
$a= \frac{4 \lambda D}{0.5 \cdot 10^{-3} m}= \frac{4 (5.3 \cdot 10^{-7} m)(0.37 m)}{0.5 \cdot 10^{-3} m}= 1.57 \cdot 10^{-3} m=1.57 mm$

7 0

3 years ago

Help ASAP!<br> Everything on screenshot

BARSIC [14]

Answer:

For the first one, its B) cities B and C

I'm not so sure, but I hope this helps.

7 0

2 years ago

Electrical systems are governed by Ohm’s law, which states that V = IR, where V is the voltage, I is the current, and R is the r

ella [17]

Answer:

$\frac{dR(t)}{dt}=0.06\Omega$

Explanation:

Since $R(t)=\frac{V}{I(t)}$ , we calculate the resistance rate by deriving this formula with respect to time:

$\frac{dR(t)}{dt}=\frac{d}{dt}(\frac{V}{I(t)})=V\frac{d}{dt}(\frac{1}{I(t)})$

Deriving what is left (remember that $(\frac{1}{f(x)})'=-\frac{1}{f(x)^2}f'(x)$ ):

$\frac{d}{dt}(\frac{1}{I(t)})=-\frac{1}{I(t)^2}\frac{dI(t)}{dt}$

So we have:

$\frac{dR(t)}{dt}=-\frac{V}{I(t)^2}\frac{dI(t)}{dt}$

Which for our values is (the rate of <em>I(t)</em> is decreasing so we put a negative sign):

$\frac{dR(t)}{dt}=-\frac{24V}{(56A)^2}(-8A/s)=0.06\Omega$

8 0

3 years ago