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3 years ago

What are the similarities between a car at rest and a car at a constant velocity

1 answer:

8 0

They are both in motion because an object is not at rest, but moving so slow it could be at rest. A car going at the same constant velocity is neither speeding up or slowing down, an object "at rest" is also moving at a constant rate, not speeding up or slowing done.

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The two speakers at S1 and S2 are adjusted so that the observer at O hears an intensity of 6 W/m² when either S1 or S2 is sounde

Zanzabum

Answer:

The minimum frequency is 702.22 Hz

Explanation:

The two speakers are adjusted as attached in the figure. From the given data we know that

$S_1 S_2$ =3m

$S_1 O$ =4m

By Pythagoras theorem

$S_2O=\sqrt{(S_1S_2)^2+(S_1O)^2}\\S_2O=\sqrt{(3)^2+(4)^2}\\S_2O=\sqrt{9+16}\\S_2O=\sqrt{25}\\S_2O=5m$

Now

The intensity at O when both speakers are on is given by

$I=4I_1 cos^2(\pi \frac{\delta}{\lambda})$

Here

I is the intensity at O when both speakers are on which is given as 6 $W/m^2$

I1 is the intensity of one speaker on which is 6 $W/m^2$

δ is the Path difference which is given as

$\delta=S_2O-S_1O\\\delta=5-4\\\delta=1 m$

λ is wavelength which is given as

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

Here

v is the speed of sound which is 320 m/s.

f is the frequency of the sound which is to be calculated.

$16=4\times 6 \times cos^2(\pi \frac{1 \times f}{320})\\16/24= cos^2(\pi \frac{1f}{320})\\0.667= cos^2(\pi \frac{f}{320})\\cos(\pi \frac{f}{320})=\pm0.8165\\\pi \frac{f}{320}=\frac{7 \pi}{36}+2k\pi \\ \frac{f}{320}=\frac{7 }{36}+2k \\\\ {f}=320 \times (\frac{7 }{36}+2k )$

where k=0,1,2

for minimum frequency $f_1$ , k=1

${f}=320 \times (\frac{7 }{36}+2 \times 1 )\\\\{f}=320 \times (\frac{79 }{36} )\\\\ f=702.22 Hz$

So the minimum frequency is 702.22 Hz

5 0

3 years ago

A red train travelling at 72 km/h and a green train travelling at 144 km/h are headed toward each

abruzzese [7]

Answer:

Collision will occur.

Speed of red train when they collide = 0 m/s.

Speed of green train when they collide = 10 m/s.

Explanation:

Speed of red train = 72 km/h = 20 m/s

Speed of green train = 144 km/h = 40 m/s.

Deceleration of trains = 1 m/s²

For red train:-

Equation of motion v = u + at

u = 20 m/s

v = 0 m/s

a = -1 m/s²

Substituting

0 = 20 - 1 x t

t = 20 s.

Equation of motion s = ut + 0.5at²

u = 20 m/s

t = 20 s

a = -1 m/s²

Substituting

s = 20 x 20 - 0.5 x 1 x 20² = 200 m

So red train travel 200 m before coming to stop.

For green train:-

Equation of motion v = u + at

u = 40 m/s

v = 0 m/s

a = -1 m/s²

Substituting

0 = 40 - 1 x t

t = 40 s.

Equation of motion s = ut + 0.5at²

u = 40 m/s

t = 40 s

a = -1 m/s²

Substituting

s = 40 x 40 - 0.5 x 1 x 40² = 800 m

So green train travel 800 m before coming to stop.

Total distance traveled = 800 + 200 = 1000 m>950 m.

So both trains collide.

Distance traveled by green train when red train stops(t=20s)

Equation of motion s = ut + 0.5at²

u = 40 m/s

t = 20 s

a = -1 m/s²

Substituting

s = 40 x 20 - 0.5 x 1 x 20² = 600 m

Total distance after 20 s = 600 + 200 = 800 m< 950m . So they collide after red train stops.

Speed of red train when they collide = 0 m/s.

Distance traveled by green train when they collide = 950 - 200 = 750 m

Equation of motion v² = u² + 2as

u = 40 m/s

s= 750 m

a = -1 m/s²

Substituting

v² = 40² - 2 x 1 x 750 = 100

v = 10 m/s

Speed of green train when they collide = 10 m/s.

6 0

3 years ago

A wave of wavelength 0.3 m travels 900 m in 3.0 s. Calculate its frequency.

zzz [600]

Answer:

1000 Hz

Explanation:

The frequency would be 1000 Hz.

The frequency, wavelength, and speed of a wave are related by the equation:

v = fλ ..................(1)

where v = speed of the wave, f = frequency of the wave, and λ = wavelength of the wave.

Making f the subject of the formula:

f = v/λ.........................(2)

Also, speed (v) = distance/time.

From the question, distance = 900 m, time = 3.0 s

Hence, v = 900/3.0 = 300 m/s

Substitute v = 300 and λ = 0.3 into equation (2):

f = 300/0.3 = 1000 Hz

6 0

3 years ago

Pls answer the 15a answer i cant understand it

Maksim231197 [3]

Materials required for the experiment of limiting force borne by string:-

String balance
weights
light strings
weight hanger
pan for spring balance
Sand

Steps of procedure for for the experiment of limiting force borne by string:-

First we have to tie a light string to the fixed support and then tie the other end with the weight hanger consists of weight.
Add additional weight to the hanger again and again. And continue the same until the string is broken.
Note down the weight (x) where the string is broken.
Suspend spring balance to a support.
Tie the light string at the end of the balance and at the other end suspend the pan for spring balance.
Now place the weights (x-100 grams) in pan.
Observe the reading in the spring balance.
Add a small amount of sand in the pan by observing the readings.
same is to be done till the string is broken.

Learn more about limiting force here:- brainly.com/question/11371672

#SPJ1

6 0

1 year ago

what is the density to the nearest hundredths, of a metal with a volume of 3.00 cm3 and a mass of 8.13g?

I am Lyosha [343]

Since D=M/V, the answer would be 2.7

8 0

3 years ago