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Nadusha1986 [10]
3 years ago
11

Why echo sound cannot be heard in small room

Physics
1 answer:
Art [367]3 years ago
7 0
They can't hear an echo in small room because in it the sound can't be reflected back. For an echo of a sound to be heard,the minimum distance between the source of sound and the walls of the roomshould be 17.2 m. Obviously,in asmall room echoes cannot beheard.
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A strip of copper 150 µm thick and 4.50 mm wide is placed in a uniform magnetic field of magnitude B = 0.74 T, that is perpendic
Sophie [7]

Answer:

V = 6.55*10^{-6} v

Explanation:

The number density can be determined by using below formula:

n = \frac{Bi}{Vle}

where,

B  is uniform magnetic field 0.74

i is current 18 A

V is hall potential difference

l is thickness 150 MICRO METER

e is electron charge 1.6 *10^{-19} C

therefore V can be determined as

V = \frac{iB}{nle}

V = \frac{18*0.74}{8.47*10^{28}*150*10^{-6}*1.6 *10^{-19}}

V = 6.55*10^{-6} v

8 0
3 years ago
Sonic boom from a plane is an example of a shockwave. Give another example where a shockwave is produced
julia-pushkina [17]
Sonic boom can also be created from an atomic bomb going off.
7 0
4 years ago
A lightbulb has a power of 100 W and is used for 4 hours. A microwave has a power of 1200 W and is used for 5 minutes. How much
ladessa [460]

Answer:

a. E=1440KJ

b. E=360KJ

c. E=1.8 J

Explanation:

I have the power (Watts) is expressed as Energy (Joules) / Time (seconds), also I have to1hour * \frac{60min}{1hour}*\frac{60s}{1min}\\1h=3600s

so, for the lightbulb

100W= \frac{Energy}{3600s*4}\\Energy=100W*14400s\\ E=1440000J =1440KJ

Analogously, for the microwave

1200W= \frac{Energy}{60s*5}\\Energy=1200W*300s\\ E=360000J =360KJ

Now, I have to express the efficiency as Heat energy / power * 100 so1.8= \frac{Heat Energy}{100W}*100\\Heat Energy=1.8J

Done

4 0
4 years ago
A car drives over a hilltop that has a radius of curvature 0.120 km at the top of the hill. At what speed would the car be trave
Mashutka [201]

Answer:

34.3 m/s

Explanation:

Newton's Second Law states that the resultant of the forces acting on the car is equal to the product between the mass of the car, m, and the centripetal acceleration a_c (because the car is moving of circular motion). So at the top of the hill the equation of the forces is:

mg-R = m a_c = m\frac{v^2}{r}

where

(mg) is the weight of the car (downward), with m being the car's mass and g=9.8 m/s^2 is the acceleration due to gravity

R is the normal reaction exerted by the road on the car (upward, so with negative sign)

v is the speed of the car

r = 0.120 km = 120 m is the radius of the curve

The problem is asking for the speed that the car would have when it tires just barely lose contact with the road: this means requiring that the normal reaction is zero, R=0. Substituting into the equation and solving for v, we find:

v=\sqrt{gr}=\sqrt{(9.8 m/s^2)(120 m)}=34.3 m/s

6 0
3 years ago
I need help with this question ASAP!!!!
Bezzdna [24]
The answer is  <span>A) Velocity describes speed with direction.

Speed can be in any direction </span>
5 0
4 years ago
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