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Mkey [24]
2 years ago
11

A liquid has a volume of 100 cm and a mass of 85g.

Physics
1 answer:
djyliett [7]2 years ago
6 0

Answer:

B. Its density is lower than that of water

Explanation:

density = mass / volume

density of the liquid = 85 / 100 = 0.85 g/cm^3

now,

density of water is 1 g/cm^3 which is greater than the density of the given liquid ( 0.85 g/cm^3 )

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Of the more than 2,000 known minerals approximately 95% of them belong to what mineral group?
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Your answer is Silicates.
5 0
3 years ago
Write the products for the following neutralization reactions
madam [21]

Answer:

answer 3

Explanation:

5 0
2 years ago
You are traveling in a car toward a hill at a speed of 36.4 mph. The car's horn emits sound waves of frequency 231 Hz, which mov
Marina CMI [18]

Answer:

<em>a. The frequency with which the waves strike the hill is 242.61 Hz</em>

<em>b. The frequency of the reflected sound wave is 254.23 Hz</em>

<em>c. The beat frequency produced by the direct and reflected sound is  </em>

<em>    11.62 Hz</em>

Explanation:

Part A

The car is the source of our sound, and the frequency of the sound wave it emits is given as 231 Hz. The speed of sound given can be used to determine the other frequencies, as expressed below;

f_{1} = f[\frac{v_{s} }{v_{s} -v} ] ..............................1

where f_{1} is the frequency of the wave as it strikes the hill;

f is the frequency of the produced by the horn of the car = 231 Hz;

v_{s} is the speed of sound = 340 m/s;

v is the speed of the car = 36.4 mph

Converting the speed of the car from mph to m/s we have ;

hint (1 mile = 1609 m, 1 hr = 3600 secs)

v = 36.4 mph *\frac{1609 m}{1 mile} *\frac{1 hr}{3600 secs}

v = 16.27 m/s

Substituting into equation 1 we have

f_{1} =  231 Hz (\frac{340 m/s}{340 m/s - 16.27 m/s})

f_{1}  = 242.61 Hz.

Therefore, the frequency which the wave strikes the hill is 242.61 Hz.

Part B

At this point, the hill is the stationary point while the driver is the observer moving towards the hill that is stationary. The frequency of the sound waves reflecting the driver can be obtained using equation 2;

f_{2} = f_{1} [\frac{v_{s}+v }{v_{s} } ]

where f_{2} is the frequency of the reflected sound;

f_{1}  is the frequency which the wave strikes the hill = 242.61 Hz;

v_{s} is the speed of sound = 340 m/s;

v is the speed of the car = 16.27 m/s.

Substituting our values into equation 1 we have;

f_{2} = 242.61 Hz [\frac{340 m/s+16.27 m/s }{340 m/s } ]

f_{2}  = 254.23 Hz.

Therefore, the frequency of the reflected sound is 254.23 Hz.

Part C

The beat frequency is the change in frequency between the frequency of the direct sound  and the reflected sound. This can be obtained as follows;

Δf = f_{2} -  f_{1}  

The parameters as specified in Part A and B;

Δf = 254.23 Hz - 242.61 Hz

Δf  = 11.62 Hz

Therefore the beat frequency produced by the direct and reflected sound is 11.62 Hz

3 0
2 years ago
What is the instantaneous speed of the monkey at time t=5s?
bezimeni [28]
I’m assuming we’re suppose to get some kind of graph but, Instantaneous speed is the speed that is happening right now. Like driving a car at 15k/h. The instantaneous speed of the car 15k/h. On the graph, at 5s. Wherever the line is, will tell you what the speed is.
6 0
3 years ago
A driver's foot presses with a steady force of 20N on a pedal in a car as shown.
azamat

Answer:

160N

Explanation:

Moments must be conserved - so.

20 * 0.4 = F * 0.05

F=\frac{20*0.4}{0.05} = 160

6 0
3 years ago
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