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2 years ago

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A brick is released with no initial speed from the roof of a building and strikes the ground in 2.50 s, encountering no apprecia

ble air drag. How tall, in meters, is the building?

1 answer:

Elenna [48]2 years ago

6 0

Answer:A brick is released with no initial speed from the roof of a building and strikes the ground in 2.50 s, encountering no ..

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Medical cyclotrons need efficient sources of protons to inject into their center. In one kind of ion source, hydrogen atoms (i.e

antoniya [11.8K]

Answer:

$r=5.278\times 10^{-4}\ m$

Explanation:

Given that:

magnetic field intensity, $B=0.07\ T$

kinetic energy of electron, $KE=1.2\ eV= 1.2\times 1.6\times 10^{-19}\ J= 1.92\times 10^{-19}\ J$

we have mass of electron, $m=9.1\times 10^{-31}\ kg$

<em>Now, form the mathematical expression of Kinetic Energy:</em>

$KE= \frac{1}{2} m.v^2$

$1.92\times 10^{-19}=0.5\times 9.1\times 10^{-31}\times v^2$

$v^2=4.2198\times 10^{11}$

$v=6.496\times 10^6\ m.s^{-1}$

<u>from the relation of magnetic and centripetal forces we have the radius as:</u>

$r=\frac{m.v}{q.B}$

$r=\frac{9.1\times 10^{-31}\times 6.496\times 10^6 }{1.6\times 10^{-19}\times 0.07}$

$r=5.278\times 10^{-4}\ m$

6 0

3 years ago

A rock is thrown upward from the level ground in such a way that the maximum height of its flight is equal to its horizontal ran

marshall27 [118]

Answer

a) For the rock

$\dfrac{v_t^2sin 2\theta}{g} = \dfrac{v_t^2sin^2\theta}{2g}$

$2sin\thetacos\theta = \dfrac{sin^2\theta}{2}$

$2cos\theta = \dfrac{sin\theta}{2}$

$tan\theta = 4$

$\theta = tan^{-1} 4$

$\theta = 76^0$

b) $\theta = 45^0$ for maximum range

$\dfrac{d_{max}}{d}=\dfrac{(v_tcos 45^0)(2v_tsin 45^0)g}{(v_tcos 76^0)(2v_tsin 76^0)g}$

$\dfrac{d_{max}}{d}=\dfrac{0.707\times 0.707)}{0.97\times 0.242}$

$\dfrac{d_{max}}{d}=2.129$

$d_{max}=2.129 d$

c) The value of θ is the same on every planet as g divides out.

5 0

3 years ago

A uniform electric field with a magnitude of 5750 N/C points in the positive x direction. Find the change in electric potential

castortr0y [4]

Explanation:

Given that,

Electric field = 5750 N/C

Charge $q=+10.5\times10^{-6}\ C$

Distance = 5.50 cm

(a). When the charge is moved in the positive x- direction

We need to calculate the change in electric potential energy

Using formula of electric potential energy

$\Delta U=-W$

$\Delta U=-F\cdot d$

$\Delta U=-q(E\cdot d)$

Put the value into the formula

$\Delta U=-10.5\times10^{-6}\times5750\times5.50\times10^{-2}$

$\Delta U=-3.32\times10^{-3}\ J$

The change in electric potential energy is $-3.32\times10^{-3}\ J$

(b). When the charge is moved in the negative x- direction

We need to calculate the change in electric potential energy

Using formula of electric potential energy

$\Delta U=-W$

$\Delta U=-F\cdot (-d)$

$\Delta U=-q(E\cdot (-d))$

Put the value into the formula

$\Delta U=-10.5\times10^{-6}\times5750\times(-5.50\times10^{-2})$

$\Delta U=3.32\times10^{-3}\ J$

The change in electric potential energy is $3.32\times10^{-3}\ J$

Hence, This is the required solution.

3 0

3 years ago