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AfilCa [17]
3 years ago
6

What would you expect to cause a drop in air pressure?

Physics
2 answers:
givi [52]3 years ago
7 0
D. The air temperature staying the same
Murrr4er [49]3 years ago
6 0

Answer:

D

Explanation:

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A small ball with a mass of 0.6 kg and a velocity of 12 m/s hits another ball with the same mass. The firs forward and hits a th
Ket [755]
I think is 30m/s I am not sure
8 0
3 years ago
The ice skaters partner liftes her up a distance of 1 m work done or not work done
SOVA2 [1]

Answer:

Work done.

Explanation:

The skater who lifts has to overcome the partner's weight. When lifted up by 1 meter, her potential energy increases by (mass)x(gravitational acceleration)x(1meter), which is the amount of work done.

(This all assumes lifting vertically and no other forces being part of the picture)

5 0
3 years ago
A student measures the speed of sound by echo destiny classes hands and then measures the time to hear the echo his distance to
777dan777 [17]

Explanation:

∆x=300 m×2

∆t=1.5 s

v=∆x/∆t → v=2×300/1.5 = 400 m/s

6 0
3 years ago
A train traveling at 25 m/s is blowing its whistle at 440 Hz as it crosses a level crossing. You are waiting at the crossing and
ohaa [14]

Answer:

b) 472HZ, 408HZ

Explanation:

To find the frequencies perceived when the bus approaches and the train departs, you use the Doppler's effect formula for both cases:

f_o=f\frac{v_s+v_o}{v_s-v}\\\\f_o=f'\frac{v_s-v_o}{v_s+v}\\\\

fo: frequency of the source = 440Hz

vs: speed of sound = 343m/s

vo: speed of the observer = 0m/s (at rest)

v: sped of the train

f: frequency perceived when the train leaves us.

f': frequency when the train is getTing closer.

Thus, by doing f and f' the subjects of the formulas and replacing the values of v, vo, vs and fo you obtain:

f=f_o\frac{v_s-v}{v_s+v_o}=(440Hz)\frac{340m/s-25m/s}{340m/s}=408Hz\\\\f'=f_o\frac{v_s+v}{v_s-v_o}=(440Hz)\frac{340m/s+25m/s}{340m/s}=472Hz

hence, the frequencies for before and after tha train has past are

b) 472HZ, 408HZ

6 0
3 years ago
During a very quick stop, a car decelerates at 7.6 m/s2. Assume the forward motion of the car corresponds to a positive directio
Mashcka [7]

Answer:

24.57 revolutions

Explanation:

(a) If they do not slip on the pavement, then the angular acceleration is

\alpha = a / r = 7.6 / 0.26 = 29.23 rad/s^2

(b) We can use the following equation of motion to find out the angle traveled by the wheel before coming to rest:

\omega^2 - \omega_0^2 = 2\alpha\Delta \theta

where v = 0 m/s is the final angular velocity of the wheel when it stops, \omega_0 = 95rad/s is the initial angular velocity of the wheel, \alpha = -29.23 rad/s^2 is the deceleration of the wheel, and \Delta \theta is the angle swept in rad, which we care looking for:

0 - 95^2 = 2*29.23\Delta \theta

9025 = 58.46 \Delta \theta

\Delta \theta = 9025 / 58.46 = 154.375 rad

As each revolution equals to 2π, the total revolution it makes before stop is

154.375 / 2π = 24.57 revolutions

8 0
3 years ago
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