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prohojiy [21]
3 years ago
14

a 5 charge is locataed 1.25 m to the left of a -3 charge. What is the magnitude and direction of the electrostatic force on the

postive charge
Physics
1 answer:
Alika [10]3 years ago
5 0

Answer:

The force is 86.5×10^9 N towards the negative charge (to the right)

Explanation:

The electrostatic force on the charges is given by Coulomb's law;

F= Kq1q2/r^2

This an inverse square law.

F= electrostatic force on the charges

K= constant of Coulomb's law

q1 and q2= magnitude of the charges

Since K= 9.0×10^9Nm^2C^2

F= 9.0×10^9 × 5 × 3/(1.25)^2 = 135×10^9/1.56

F= 86.5×10^9 N

The force is 86.5×10^9 N towards the negative charge.

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3 years ago
a train is moving with an initial velocity of 30 m/s, the brakes are applied so as to produce a uniform acceleration of -1.5 m/s
Pepsi [2]

Answer:

\boxed{\sf Time \ in \ which \ train \ will \ come \ to \ rest = 20 \ sec}

Given:

Initial velocity (u) = 30 m/s

Final speed (v) = 0 m/s

Acceleration (a) = - 1.5 m/,s²

To Find:

Time in which train will come to rest (t).

Explanation:

\sf From \ equation \ of \ motion: \\ \sf \implies \bold{v = u + at} \\ \\ \sf Substituting \ value \ of \ v, \ u \ and \ a:  \\  \sf \implies 0 = 30 + ( - 1.5)(t) \\   \sf  \implies 0 = 30 - 1.5(t) \\  \sf \implies 30 - 1.5(t) = 0 \\  \\  \sf Subtract  \: 30  \: from  \: both  \: sides: \\  \sf \implies (30 -  \boxed{ \sf 30}) - 1.5(t) =  \boxed{ \sf  - 30} \\  \\  \sf 30 - 30 = 0 :  \\  \sf \implies  - 1.5(t) =  - 30 \\  \\  \sf Divide  \: both  \: sides \:  of \:  - 1.5(t) =  - 30 \: by \:  - 1.5 :  \\  \sf \implies  \frac{  - 1.5(t)}{ \boxed{ \sf - 1.5}}  =  \frac{ - 30}{ \boxed{ \sf -1.5 }}  \\  \\  \sf \frac{ \cancel{ \sf 1.5}}{\cancel{ \sf 1.5}}  = 1 :  \\  \sf \implies t =  \frac{ - 30}{ - 1.5}  \\  \\   \sf  \frac{ - 30}{ - 1.5}  =  \frac{\cancel{ \sf 1.5} \times 20}{\cancel{ \sf 1.5}}  = 20 :  \\  \sf  \implies t = 20 \: sec

So,

Time in which train will come to rest = 20 seconds

4 0
3 years ago
A vehicle moving with a uniform a acceleration of 2m/s and has a velocity of 4m/s at a certain time .
Evgesh-ka [11]

Answer:

1 second later the vehicle's velocity will be:

v(1)= 6\,\,\frac{m}{s} \\

5 seconds later the vehicle's velocity will be:

v(5)=14\,\,\frac{m}{s}

Explanation:

Recall the formula for the velocity of an object under constant accelerated motion (with acceleration "a"):

v(t)=v_0+a\,t

Therefore, in this case v_0=4\,\,\frac{m}{s}  and a=2\,\,\frac{m}{s^2}

so we can estimate the velocity of the vehicle at different times just by replacing the requested "t" in the expression:

v(t)=v_0+a\,t\\v(t)=4+2\,\,t\\v(1)=4+2\,(1) = 6\,\,\frac{m}{s} \\v(5)=4+2\,(5)=14\,\,\frac{m}{s}

3 0
3 years ago
Relating to Newton’s Third Law, action and reaction, what is the reaction when a rocket expels gas, smoke and flames from the no
Cloud [144]

Answer:

Propels in the opposite direction

Explanation:

7 0
3 years ago
An electron emitted in the beta decay of bismuth-210 has a mean kinetic energy of 390 keV. (a) Find the de Broglie wavelength of
Sauron [17]

Explanation:

Given that,

The mean kinetic energy of the emitted electron, E=390\ keV=390\times 10^3\ eV

(a) The relation between the kinetic energy and the De Broglie wavelength is given by :

\lambda=\dfrac{h}{\sqrt{2meE}}

\lambda=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 390\times 10^3}}

\lambda=1.96\times 10^{-12}\ m

(b) According to Bragg's law,

n\lambda=2d\ sin\theta

n = 1

For nickel, d=0.092\times 10^{-9}\ m

\theta=sin^{-1}(\dfrac{\lambda}{2d})

\theta=sin^{-1}(\dfrac{1.96\times 10^{-12}}{2\times 0.092\times 10^{-9}})

\theta=0.010^{\circ}

As the angle made is very small, so such an electron is not useful in a Davisson-Germer type scattering experiment.

4 0
3 years ago
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