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3 years ago

3. A projectile is fired from a vertical cliff edge (height = 100 m) with a horizontal velocity of 20

Physics

1 answer:

Klio2033 [76]3 years ago

6 0

Refer to the diagram shown below.

u = 20 m/s, the horiontal launch velocity
v = 0, initial vertical velocity

Assume that wind resistance may be ignored.

The time, t, of the flight is given by
(1/2)*(9.8 m/s²)*t² = (100 m)
t² = 100/4.9 = 20.4082
t = 4.5175 s

Th horizontal distance traveled is
d = (20 m/s)*(4.5175 s) = 90.35 m

Answer: B
The projectile hits the ground at approximately 90 m from the base of the cliff.

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Rutherford discovered the nucleus of the atom by firing a particles at gold foil. An a particle has a charge of q = +2e and a ma

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Answer: $r_0=3.037\times 10^{-14}m$

Explanation:

Given

charge on alpha particle=+2e

mass of alpha particle= $6.64\times 10^{-27}$ kg

Charge on gold nucleus=+79e

Velocity at r=1m is $1.9\times 10^{7}$

Using Energy conservation

Kinetic energy of particle will be converting to Potential energy as it approaches to nucleus

therefore

$\frac{1}{2}mv^2+U_{r=1m}=U_{closest\ to\ nucleus}$

$\frac{1}{2}\left ( 6.64\times 10^{-27}\right )\left ( 1.9\times 10^{7}^2\right )+\frac{K\left ( 2e\right )\left ( 79e\right )}{1}=\frac{K\left ( 2e\right )\left ( 79e\right )}{r_0}$

$\frac{1}{2}\left ( 6.64\times 10^{-27}\right )\left ( 1.9\times 10^{7}^2\right )=\frac{9\times 10^9\times 158\times \left ( 1.6\times 10^{-19}\right )}{y}\left [\frac{1}{r_0}-\frac{1}{1}\right ]$

on solving we get

$\frac{1}{r_0}=3.292\times 10^{13}$

$r_0=3.037\times 10^{-14}m$

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3 years ago

11. An atom with the same number of protons and electrons has a

balandron [24]

Answer:

the answer is A.

Explanation:

An atom contains equal numbers of protons and electrons . Since protons and electrons have equal and opposite charges , this means that atoms are neutral overall.

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3 years ago

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The hottest stars are _____. -Blue -Orange -White -Red

AnnyKZ [126]

Answer:

yellow

Explanation:

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4 years ago

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A wheel of diameter 78 cm has an axle of diameter 14.8 cm. A force 150 N is exerted along the rim of the wheel.What force should

bezimeni [28]

Answer:

F = 263.51 N

Explanation:

given,

diameter of wheel = 78 cm

diameter of axle = 14.8 cm

Force exerted on the rim of wheel = 150 N

Force applied outside the axle = ?

To prevent rotation wheel from rotating the Force 'F' should be applied outside of the axle.

Net momentum about the center of mass should be zero

now,

Moment of about center due to 150 N = moment about center due to F on axle

$50\times \dfrac{78}{2}=F\times \dfrac{14.8}{2}$

7.4 F = 1950

F = 263.51 N

Hence, Force exerted outside of the axle in order to prevent the wheel from rotating is equal to 263.51 N.

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4 years ago

A student (m = 68 kg) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a

Gnesinka [82]

Answer:

5.7141 m

Explanation:

Here the potential and kinetic energy will balance each other

$mgh=\frac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}$

This is the initial velocity of the system and the final velocity is 0

t = Time taken = 0.04 seconds

F = Force = 18000 N

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

Equation of motion

$v=u+at\\\Rightarrow a=\frac{v-u}{t}$

From Newton's second law

$F=ma\\\Rightarrow F=m\frac{v-u}{t}\\\Rightarrow 18000=68\frac{0-\sqrt{2gh}}{0.04}\\\Rightarrow \frac{18000}{68}\times 0.04=-\sqrt{2\times 9.81\times h}\\\Rightarrow 10.58823=-\sqrt{2\times 9.81\times h}$

Squarring both sides

$112.11061=2\times 9.81\times h\\\Rightarrow h=\frac{112.11061}{2\times 9.81}\\\Rightarrow h=5.7141\ m$

The height from which the student fell is 5.7141 m

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3 years ago