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ZanzabumX [31]
3 years ago
13

A rifleis aimed horizontally to wards the center of target 100m away.If the buller strikes 10 cm below the center

Physics
1 answer:
zmey [24]3 years ago
5 0

Answer:

So, horizontal Velocity is 699.30 m/s

Explanation:

First, we need to determine the time as follow

y = ( U x t ) + 1/2 x ( g x t^{2} )

Where

y = Vertical distance = 10 cm / 1000 = 0.1 m

u = vertical velocity = 0

g = gravity = 9.8

Placing values in the formula

0.1 = ( 0 x t ) + 1/2 x ( 9.8 x t^{2} )

0.1 = 0 + 4.9t^{2}

0.1 = 4.9t^{2}

t^{2} = \sqrt{0.1 / 4.9}

t = 0.143 seconds

Use the following formula to calculate the horizontal velocity

V = x / t

Where

x = Distance = 100 m

t = time = 0.143 seconds

V = 100m / 0.143 seconds

V = 699.30 m/s

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The uniform rods AB and BC weigh 24 ky and kg, respectively,and the small wheel at C is of negligible weight. If the wheel ismov
victus00 [196]

The velocity of pin B after rod AB has rotated through 90* is vb = 3.2549 m/s.

<h3>What is Potential and Kinetic energy?</h3>

Potential energy is the energy that is stored in any item or system as a result of its location or component arrangement. The environment outside of the object or system, such as air or height, has no impact on it. In contrast, kinetic energy refers to the energy of moving particles inside a system or an item.

mass of rod, mab = 2.4kg

mass of rod, mbc = 4kg

conservation of energy

T_{1}  + V_{1} = T_{2}  + V_{2}

h_{ab}  = h_{bc}  = 0.18m

potential energy at position 1,

V1 = m_{ab} gh_{ab}  + m_{bc} gh_{bc}

V1 = 2.5 * 9.81 * 0.18 + 4 * 9.81 * 0.18

V1 = 11.30112

kinetic energy T1 at position 1 is zero

potential energy at position 2 is zero

K.E at position 2,

T_{2} = \frac{1}{2} l_{ab} w^{2}_{ab} +  \frac{1}{2} m_{bc} v^{2}_{G} +  \frac{1}{2} lw^{2}_{bc}

l_{ab} =\frac{m_{ab} l^{2}_{ab}  }{3}

= 1/3 *4 * (0.36)²

=0.10368kg m²

l =\frac{m_{bc} l^{2}_{bc}  }{12}

= 1/12 *4 * (0.6)²

=0.12kg m²

on putting the values in above equation we get,

T₂ = 1.0667vb²

0 + 11.30112 = 1.0667vb² + 0

vb = 3.2549 m/s

to learn more about Kinetic and potential energy go to - brainly.com/question/18963960

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2 years ago
A balloon of mass M is floating motionless in the air. A person of mass less than M is on a rope ladder hanging from the balloon
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Let the mass of the person be m. Total momentum is conserved (because the exterior forces on the system are balanced), especially the component in the vertical direction.

Given that,

Mass of gallon is M

Let man mass be m

Velocity of man is v

Let velocity if ballot be Vb

When the person begin to move we have

Conservation of momentum

mv + MVb=0

MVb=-mv

Vb= -(m/M) v

Given that the mass of man is less than mass of balloon. i.e. m<M

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Therefore, .

Vb= -(m/M) v

Vb< -v

This implies that the velocity of balloon is less than the velocity of man and if is also moving in opposite direction

So the man is moving upward, then the balloon is moving downward and it's velocity is less than the velocity of man,

The answer is C

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3 years ago
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Indeed the answer is c!!
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