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3 years ago

Calculate the density of the football. Use the formula D = m/V

Physics

2 answers:

bulgar [2K]3 years ago

4 0

Answer:

0.10

Explanation:

Sindrei [870]3 years ago

3 0

There’s not enough info here i’m sorry

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A car speeds past you sounding its horn at a frequency of 1000Hz. The frequency you hear:

ki77a [65]

Answer:

f' =1064.1025 Hz

Explanation:

We know that from doppler's effect

$f'=\frac{v+v_0}{v-v_s}f$

f'= observed frequency

f= actual frequency =1000 Hz

v= velocity of sound waves = 332 m/s

v_o= velocity of observer =0

v_s= velocity of source. ( let velocity of car be 20 m/s)

therefore,

$f'=\frac{332+0}{332-20}\times1000$

f' =1064.1025 Hz

3 0

3 years ago

A ship sends a 1200 Hz sound wave down in the water, and detect the echo 3 seconds later what would happen if that ship sent out

kiruha [24]

Answer:

D) The sound would travel the same speed, so it would take the same amount to time to return

Explanation:

The speed of sound in water is constant, decreasing the frequency will only serve to increase the wavelenght of the sound wave, it doesn't affect the speed of the Sound wave.

Speed of sound = Freq. X wavelenght

6 0

4 years ago

A 0.500-kg mass suspended from a spring oscillates with a period of 1.50 s. How much mass must be added to the object to change

Daniel [21]

Answer:

389 kg

Explanation:

The computation of mass is shown below:-

$T = 2\pi \sqrt{\frac{m}{k} }$

Where K indicates spring constant

m indicates mass

For the new time period

$T^' = 2\pi \sqrt{\frac{m'}{k} }$

Now, we will take 2 ratios of the time period

$\frac{T}{T'} = \sqrt{\frac{m}{m'} }$

$\frac{1.50}{2.00} = \sqrt{\frac{0.500}{m'} }$

$0.5625 = \sqrt{\frac{0.500}{m'} }$

$m' = \frac{0.500}{0.5625}$

= 0.889 kg

Since mass to be sum that is

= 0.889 - 0.500

0.389 kg

= 389 kg

Therefore for computing the mass we simply applied the above formula.

7 0

3 years ago

Which is an example of making a qualitative observation?

Alisiya [41]

"Describing how copper changes color in chlorine" is the answer I think is right. Hope this helped!

4 0

3 years ago

Read 2 more answers

"An alpha particle (He2+) with a velocity of 2.6x106m/scrosses a uniform magnetic field at an angle of 37.0°to the field lines a

sergey [27]

Answer:

B = 5.59x10⁹ T

Explanation:

The magnetic force (F), on a the alpha particle with charge (q) that is moving at velocity (v) as the cross product of the velocity and magnetic field (B) is:

$F = qvBsin(\theta)$

F = 1.4x10⁻³ N

v = 2.6x10⁶ m/s

θ = 37.0°

q = 2*p = 2*1.6x10⁻¹⁹ C

Hence, the strength of the magnetic field is:

$B = \frac{F}{qvsin(\theta)} = \frac{1.4 \cdot 10^{-3}}{1.6 \cdot 10^{-19} C*2.6 \cdot 10^{6}*sin(37)} = 5.59 \cdot 10^{9} T$

Therefore, the strength of the magnetic field is 5.59x10⁹ T.

I hope it helps you!

7 0

3 years ago