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Novay_Z [31]
2 years ago
9

A ball is thrown vertically up ward with a velocity of 25m/s. a/how fast was it moving after 2sec?​

Physics
1 answer:
Shalnov [3]2 years ago
3 0

Answer:

12.5

Explanation:

given

distance=25

time=2sec

required

=speed

solution

=v=s/t

25/2

=12.5m/s

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In 1995 a research group led by Eric Cornell and Carl Wiemann at the University of Colorado successfully cooled Rubidium atoms t
saveliy_v [14]

Answer:

0.00493 m/s

Explanation:

T = Temperature of the isotope = 85 nK

R = Gas constant = 8.341 J/mol K

M = Molar mass of isotope = 86.91 g/mol

Root Mean Square speed is given by

v_r=\sqrt{\dfrac{3RT}{M}}\\\Rightarrow v_r=\sqrt{\dfrac{3\times 8.314\times 85\times 10^{-9}}{86.91\times 10^{-3}}}\\\Rightarrow v_r=0.00493\ m/s

The Root Mean Square speed is 0.00493 m/s

6 0
3 years ago
The end of a horizontal rope is attatched to a prong of an electricity driven tuning fork that vibrates at 100hz. The other end
Darina [25.2K]

here since string is attached with a mass of 2 kg

so here tension force in the rope is given as

T = mg

here we will have

T = 2(9.8) = 19.6 N

now we will have speed of wave given as

v = \sqrt{\frac{T}{\mu}}

here we will have

v = \sqrt{\frac{19.6}{0.75\times 10^{-2}}}

v = 16.33 m/s

now we know that frequency is given as

F = 100 Hz

now wavelength is given as

\lambda = \frac{v}{F}

\lambda = \frac{16.33}{100} = 0.16 m

so wavelength will be 0.16 m

5 0
3 years ago
Two parallel wires separated by a distance of 0.6 m each carry current in the same direction. One wire is carrying a current of
mart [117]

Answer:

The value is  B =  3.33 *10^{-6} \  T

Explanation:

From the question we are told that

  The distance of separation is  d = 0.6 \  m

  The current on the one wire is I_1 =  9 \  A

  The current on the second wire is I_2 =  4 \ A

Generally the magnitude of the field exerted between the current carrying wire is

        B  =  B_1 - B_2

Here B_1 is the magnetic field due to the first wire which is mathematically represented as

         B_1 = \frac{\mu_o * I_1 }{2 \pi * d_1}

Here d_1 is the distance to the half way point of the separation and the value is  

    d_1 =  0.3 \  m

B_2 is the magnetic field due to the first wire which is mathematically represented as

         B_2  = \frac{\mu_o * I_2 }{2 \pi * d_2}

Here d_2 is the distance to the half way point of the separation and the value is  

    d_2 =  0.3 \  m  

This means that d_1 = d_2 = a =  0.3

So

     B =  \frac{\mu_o * I_1 }{2 \pi * d_1}  -  \frac{\mu_o * I_2 }{2 \pi * d_2}

=>  B =  \frac{\mu_o * (I_1 - I_2)}{2 \pi *0.3 }

=>  B =  \frac{  4\pi * 10^{-7}  * (9- 4)}{2 * 3.142  *0.3 }

=>  B =  3.33 *10^{-6} \  T

5 0
3 years ago
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