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Elena L [17]
3 years ago
5

you want to get more physically fit but often make excuses for why you don't have time to be active what strategy can help with

this problem
Physics
1 answer:
Anna35 [415]3 years ago
6 0
<span>Find an activity you enjoy and make it a priority in your schedule.</span>
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You accidentally drop an eraser out of the window of an apartment 15 m above the ground
docker41 [41]

Answer:

hello, yes or nou sorry jaja

3 0
2 years ago
Two point charges 3q and −8q (with q &gt; 0) are at x = 0 and x = L, respectively, and free to move. A third charge is placed so
riadik2000 [5.3K]

Answer:

Explanation:

The unknown charge can not remain in between the charge given because force on the middle charge will act in the same direction due to both the remaining charges.

So the unknown charge is somewhere on negative side of x axis . Its charge will be negative . Let it be - Q and let it be at distance - x on x axis.

force on it due to rest of the charges will be equal and opposite so

k3q Q / x² =k 8q Q / (L+x)²

8x² = 3 (L+x)²

2√2 x = √3 (L+x)

2√2 x - √3 x = √3 L

x(2√2 - √3 ) = √3 L

x = √3 L / (2√2 - √3 )

Let us consider the balancing force on 3q

force on it due to -Q and -8q will be equal

kQ . 3q / x² = k3q  8q / L²

Q = 8q  (x² / L²)

so charge required = - 8q  (x² / L²)

and its distance from x on negative x side = √3 L / (2√2 - √3 )

3 0
3 years ago
Which statement explains the similarity between momentum and kinetic energy of an object?
Anon25 [30]

Answer:B.

Both increase as the mass and velocity increase.

3 0
3 years ago
Read 2 more answers
A coil of wire with 100. circular turns of radius 8.00 cm is in a uniform magnetic field along the axis of the coil. The resista
andreyandreev [35.5K]
I think Im gonna have to go with C 6.00 T/s but Im not sure
6 0
3 years ago
A yet-to-be-built spacecraft starts from Earth moving at constant speed to the yet-tobe-discovered planet Retah, which is 20 lig
Gre4nikov [31]

Answer:

The  time elapsed at the spacecraft’s frame is less that the time elapsed at earth's  frame

Explanation:

From the question we are told that

The distance between earth and Retah is  d = 20 \ light \ hours =  20 * 3600 *  c =  72000c \ m

Here c is the peed of light with value c =  3.0*10^8 m/s

The time taken to reach Retah from earth is  t =  25 \ hours  =  25 * 3600 =90000 \ sec

The velocity of the spacecraft is mathematically evaluated  as

     v_s =  \frac{d }{t}

substituting values

   v_s =  \frac{72000 * 3.0*10^{8} }{90000}

    v_s =  2.40*10^{8} \ m/s

The time elapsed in the spacecraft’s frame is mathematically evaluated as

      T  =  t *  \sqrt{ 1 -  \frac{v^2}{c^2} }

substituting value

       T  =  90000 *  \sqrt{ 1 -  \frac{[2.4*10^{8}]^2}{[3.0*10^{8}]^2} }

        T = 54000 \ s

=>    T  = 15 \ hours

So  The  time elapsed at the spacecraft’s frame is less that the time elapsed at earth's  frame

       

7 0
3 years ago
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