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Ludmilka [50]
3 years ago
9

If a 2 kg object is falling at 3 m/s at what rate is gravity working on the object

Physics
1 answer:
777dan777 [17]3 years ago
4 0

Answer:

+9.8m/s^2

Explanation:

The rate of gravity of the object is constant thriughout the surface of the earth.

For falling object, the rate of gravity is positive since the body is coming down (falling)

The rate of gravity is negative if the body is going up

The constant value for acceleration due to gravity is 9.8m.s^2

Since the object is falling, hence the acceleration due to gravity is positive.

Rate of gravity working on the object will be +9.8m/s^2

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A ball with a mass of 5000 g is floating on the surface of a pool of water.
Vitek1552 [10]

Answer:

d

Explanation:

reply the the question

8 0
3 years ago
The current in the wires of a circuit is 180.0 milliamps. If the resistance of the circuit were doubled ( with no change in volt
Troyanec [42]

Answer:

I = 0.09[amp] or 90 [milliamps]

Explanation:

To solve this problem we must use ohm's law, which tells us that the voltage is equal to the product of the voltage by the current.

V = I*R

where:

V = voltage [V]

I = current [amp]

R = resistance [ohm]

Now, we replace the values of the first current into the equation

V = 180*10^-3 * R

V = 0.18*R (1)

Then we have that the resistance is doubled so we have this new equation:

V = I*(2R) (2)

The voltage remains constant therefore 1 and 2 are equals and we can obtain the current value.

V = V

0.18*R = I*2*R

I = 0.09[amp] or 90 [milliamps]

7 0
3 years ago
A speaker draws 2 amps of current when it is connected to a 16 volt source. What is the resistance of the speaker? Question 2 op
Oksana_A [137]
Did you ever find this answer
5 0
3 years ago
Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 700 nm light.
pogonyaev

Complete Question

Due to blurring caused by atmospheric distortion, the best resolution that can be obtained by a normal, earth-based, visible-light telescope is about 0.3 arcsecond (there are 60 arcminutes in a degree and 60 arcseconds in an arcminute).Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 700 nm light

Answer:

The diameter is  D = 0.59 \  m    

Explanation:

From the question we are told that

      The best resolution is  \theta  =  0.3 \  arcsecond

       The  wavelength is  \lambda  =  700 \  nm =  700 *10^{-9 } \  m

       

Generally the

         1 arcminute  = >  60 arcseconds

=>      x arcminute =>   0.3 arcsecond

So

       x =  \frac{0.3}{60 }

=>    x = 0.005 \  arcminutes

Now

         60 arcminutes  =>  1 degree

          0.005 arcminutes = >  z degrees  

=>       z =  \frac{0.005}{60 }

=>      z =  8.333 *10^{-5}  \ degree

Converting to radian  

           \theta  = z =  8.333 *10^{-5}  * 0.01745 = 1.454 *10^{-6} \  radian

Generally the resolution is mathematically represented as

            \theta  =  \frac{1.22 *  \lambda  }{ D}

=>    D =  \frac{1.22 * \lambda }{\theta }

=>     D =  \frac{1.22 * 700 *10^{-9} }{ 1.454 *10^{-6} }    

=>     D = 0.59 \  m    

4 0
3 years ago
Helppppppppp plzzzzzz
seropon [69]
I think number 1 is incorrect I believe that answer is D. Number 6 I believe would be B. The rest seem to be correct.
4 0
3 years ago
Read 2 more answers
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