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ira [324]
3 years ago
6

What has to be true about mass if a small object (like a metal ball) has a high density? A: The mass has to be high compared to

the volume. B: The mass has to be low compared to the volume. C: The mass is not related to the density. D: The mass would be determined by throwing the ball in water.
Physics
2 answers:
inysia [295]3 years ago
6 0
<span>Since density is equal to mass divided by volume, we can determine the relationship between mass, density, and volume. Therefore, we can automatically know that the answer must either be A or B, since this information is determinable. The prompt lists the object as small. Therefore, we know that density is high, but volume is low. Let's examine the equation (D = M/V), where d = density, m = mass, and v = volume. In order for any number to be divided by a smaller number and have the end result be a higher number, the numerator (i.e. mass) must be bigger than the denominator (i.e. volume). We can also see that this is true by substituting in somewhat random numbers for d, m, and v. D = M / V 5 = 10 / 2 10 > 2 M > V In conclusion, your answer is D.</span>
laiz [17]3 years ago
3 0

Answer:

the mass must be high compared to the volume

Explanation:

i just took a test and got 100%

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A car horn emits a frequency of 400 Hz. A car traveling at 20.0 m/s sounds the horn as it approaches a stationary pedestrian. Wh
Temka [501]

Answer:

The observed frequency by the pedestrian is 424 Hz.

Explanation:

Given;

frequency of the source, Fs = 400 Hz

speed of the car as it approaches the stationary observer, Vs = 20 m/s

Based on Doppler effect, as the car the approaches the stationary observer, the observed frequency will be higher than the transmitted (source) frequency because of decrease in distance between the car and the observer.

The observed frequency is calculated as;

F_s = F_o [\frac{v}{v_s + v} ] \\\\

where;

F₀ is the observed frequency

v is the speed of sound in air = 340 m/s

F_s = F_o [\frac{v}{v_s + v} ] \\\\400 = F_o [\frac{340}{20 + 340} ] \\\\400 = F_o (0.9444) \\\\F_o = \frac{400}{0.9444} \\\\F_o = 423.55 \ Hz \\

F₀ ≅ 424 Hz.

Therefore, the observed frequency by the pedestrian is 424 Hz.

8 0
3 years ago
Express the kinetic energy K in terms of the potential energy U.<br><br><br> K=GMm/2R
max2010maxim [7]

Answer:

K = -½U

Explanation:

From Newton's law of gravitation, the formula for gravitational potential energy is;

U = -GMm/R

Where,

G is gravitational constant

M and m are the two masses exerting the forces

R is the distance between the two objects

Now, in the question, we are given that kinetic energy is;

K = GMm/2R

Re-rranging, we have;

K = ½(GMm/R)

Comparing the equation of kinetic energy to that of potential energy, we can derive that gravitational kinetic energy can be expressed in terms of potential energy as;

K = -½U

7 0
3 years ago
a. A 65-cm-diameter cyclotron uses a 500 V oscillating potential difference between the dees. What is the maximum kinetic energy
Sati [7]

Explanation:

Below is an attachment containing the solution.

8 0
3 years ago
A hydrogen atom has a diameter of about 10 nm. Express this diameter in meters. Express this diameter in millimeters. Express th
Fynjy0 [20]
So, the first question is: how many meters are 10 nm?

1nm =<span>0.000000001 m.

So 10 nanometers are </span><span>0.00000001 m!

Now, how many milimeter are those?
let's start with meters, 1 meter are 1000 milimeters.
so </span>
0.00000001*1000=0.<span><span>00001</span> m!

now, micrometers .1 micrometer are 1000 nanometers.
so 10 nanometers are 0.01 micrometers! (1 nanometer is 0.001 micrometers)
</span>
8 0
3 years ago
A diffraction grating contains 15,000 lines/inch. We pass a laser beam through the grating. The wavelength of the laser is 633 n
MatroZZZ [7]

Answer:

Recall the Diffraction grating formula for constructive interference of a light

y = nDλ/w                                      Eqn 1

Where;

w = width of slit = 1/15000in =6.67x10⁻⁵in =   6.67x10⁻⁵ x 0.0254m = 1.69x10⁻⁶m

D = distance to screen  

λ = wavelength of light  

n = order number  = 1

Given  

y1 = ? from 1st order max to the central  

D = 2.66 m  

λ = 633 x 10-9 m  

and n = 1  

y₁ = 0.994m  

Distance (m) from the central maximum (n = 0) is the first-order maximum (n = 1)                =         0.994m

Q b. How far (m) from the central maximum (m = 0) is the second-order maximum (m = 2) observed?

w = width of slit = 1/15000in =6.67x10⁻⁵in =   6.67x10⁻⁵ x 0.0254m = 1.69x10⁻⁶m

D = distance to screen  

λ = wavelength of light  

n = order number  = 1

Given  

y1 = ? from 1st order max to the central  

D = 2.66 m  

λ = 633 x 10⁻⁹ m  

and n =  2

y₂ = 0.994m  

Distance (m) from the central maximum (n = 0) is the first-order maximum (n = 2) =1.99m

8 0
3 years ago
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