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Mkey [24]
3 years ago
10

With what initial velocity must an object be thrown upward from a height of 2 meters to reach a maximum hieght of 200 meters

Physics
1 answer:
larisa86 [58]3 years ago
6 0
The equations of motion to use for a constant acceleration a = g = -9.81:
v = gt + v_0
y = \frac{1}{2} g t^{2} + v_0t + y_0

at the highest point the velocity must be zero:
0 = gt + v_0 \\ v_0 = -gt
combine both equations:
y =  \frac{1}{2}  \frac{v_0^{2} }{g}  -   \frac{v_0^{2} }{g}  + y_0=  -\frac{1}{2}  \frac{v_0^{2} }{g} + y_0
Solve for v₀:
v_0 =   \sqrt{2g(y_0 - y)}
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An electron travels west to east with a kinetic energy of 10 keV. The Earth's magnetic field in Pittsburgh is 19,911.5 nT in the
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Explanation:

It is given that,

Kinetic energy of the electron, E_k=10\ keV=10^4\ eV=1.6\times 10^{-15}\ J

Let the east direction is +x direction, north direction is +y direction and vertical direction is +z direction.    

The magnetic field in north direction, B_y=19911.5\ nT

The magnetic field in west direction, B_x=-3257.1\ nT

The magnetic field in vertical direction, B_z=48381.8 \ nT

Magnetic field, B=(-3257.1i+19911.5j+48381.8k)\ nT

Firstly calculating the velocity of the electron using the kinetic energy formulas as :

E_k=\dfrac{1}{2}mv^2

v=\sqrt{\dfrac{2E_k}{m}}

v=\sqrt{\dfrac{2\times 1.6\times 10^{-15}}{9.1\times 10^{-31}}}

v=5.92\times 10^7 i\ m/s (as it is moving from west to east)

The force acting on the charged particle in the magnetic field is given by :

F=q(v\times B)

F=1.6\times 10^{-19}\times (5.92\times 10^7 i\times (-3257.1i+19911.5j+48381.8k)\times 10^{-9})

Since, i\times j=k\ \\j\times k=i\\k\times i=j

And, i\times i=j\times j=k\times k=0

F=1.6\times 10^{-19}\times [1178 k-2864.20j]

|F|=1.6\times 10^{-19}\times \sqrt{1178^2+2864.20^2}

F=4.95\times 10^{-16}\ N

(b) Let a is the acceleration of the electron. It can be calculated as :

a=\dfrac{F}{m}

a=\dfrac{4.95\times 10^{-16}}{9.1\times 10^{-31}}

a=5.43\times 10^{14}\ m/s^2

Hence, this is the required solution.

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