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erma4kov [3.2K]
2 years ago
6

Sean, an astronaut stands on the edge of a lunar crater and throws a Frisbee™ horizontally with a velocity of 5.00 m/s. With n

o atmosphere on the moon the Frisbee ™ will not have any force of lift so it will “fly” like a stone. The floor of the crater is 77.0 m below the astronaut. What horizontal distance will the Frisbee™ travel before hitting the floor of the crater? (The acceleration of gravity on the moon is 1/6th that of the Earth).
Physics
1 answer:
motikmotik2 years ago
4 0

Answer:

48.55mm

Explanation:

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9.5 g bullet has a speed of 1.5 km/s what is the kinetic energy of the bullet
Amiraneli [1.4K]

The answer is  0.000824653J

You need to use the formula Mass * Velocity^2 over 2


3 0
3 years ago
Read 2 more answers
If you dribble a basketball with a frequency of 1.77 Hz, how long does it take for you to complete 12 dribbles?
Licemer1 [7]
<h2>It takes 6.78 seconds to complete 12 dribbles.</h2>

Explanation:

Frequency of dribble = 1.77 Hz

That is

         Number of dribbles in 1 second = 1.77

         \texttt{Time taken for 1 dribble = }\frac{1}{1.77}=0.565s

Now we need to find how long does it take for you to complete 12 dribbles.

         Time taken for 12 dribbles = 12 x Time taken for 1 dribble

         Time taken for 12 dribbles = 12 x 0.565

         Time taken for 12 dribbles = 6.78 seconds      

It takes 6.78 seconds to complete 12 dribbles.  

8 0
2 years ago
HURRYYYY pls help due today n i be giving brainliest!!!<br><br><br> n no mfkn links plsss
nordsb [41]

Answer:

1 Frequency

2 Wavelength

3 Amplitude

4 Crest

Hope it helps pls mark brainliest

5 0
2 years ago
What is the intensity of a sound with a sound intensity level (SIL) 67 dB, in units of W/m^2?
vfiekz [6]

Answer:

The intensity of sound (I) = 3.16 x 10⁻⁶ W/m²

Explanation:

We have expression for sound intensity level (SIL),

              L=10log_{10}\left ( \frac{I}{I_0}\right )

Here we need to find the intensity of sound (I).

               L=10log_{10}\left ( \frac{I}{I_0}\right )\\\\log_{10}\left ( \frac{I}{I_0}\right )=0.1L\\\\\frac{I}{I_0}=10^{0.1L}\\\\I=I_010^{0.1L}

Substituting

          L = 67 dB and I₀ = 10⁻¹² W/m² in the equation

          I=I_010^{0.1L}=10^{-12}\times 10^{0.1\times 65}\\\\I_0=10^{-12}\times 10^{6.5}=10^{-5.5}=3.16\times 10^{-6}W/m^2

The intensity of sound (I) = 3.16 x 10⁻⁶ W/m²

8 0
2 years ago
To calibrate the calorimeter electrically, a constant voltage of 3.6 V is applied and a current of 2.6 A flows for a period of 3
hodyreva [135]

Answer : The correct option is, (c) 3.7\times 10^2J/^oC

Explanation :

First we have to calculate the energy or heat.

Formula used :

E=V\times I\times t

where,

E = energy (in joules)

V = voltage (in volt)

I = current (in ampere)

t = time (in seconds)

Now put all the given values in the above formula, we get:

E=(3.6V)\times (2.6A)\times (350s)

E=3276J

Now we have to calculate the heat capacity of the calorimeter.

Formula used :

C=\frac{E}{\Delta T}=\frac{E}{T_{final}-T_{initial}}

where,

C = heat capacity of the calorimeter

T_{initial} = initial temperature = 20.3^oC

T_{final} = final temperature = 29.1^oC

Now put all the given values in this formula, we get:

C=\frac{3276J}{(29.1-20.3)^oC}

C=372.27J/^oC=3.7\times 10^2J/^oC

Therefore, the heat capacity of the calorimeter is, 3.7\times 10^2J/^oC

7 0
2 years ago
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