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AfilCa [17]
2 years ago
14

A man standing on top of a 30 m tall building throws a brick downwards with a velocity of 12 m/s. Determine the speed of the bri

ck just before it hits the ground.
Physics
1 answer:
BlackZzzverrR [31]2 years ago
7 0

Answer:

27.1m/s

Explanation:

Given parameters:

Height of the building  = 30m

Initial velocity  = 12m/s

Unknown:

Final velocity  = ?

Solution:

We apply one of the kinematics equation to solve this problem:

         v²  = u²  + 2gh

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity

h is the height

          v²   = 12²  + (2 x 9.8 x 30)

          v  = 27.1m/s

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the force applied when using a simple machine is called the effort force

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A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 58.8m/s2 . The accel
Zanzabum
Split the operation in two parts. Part A) constant acceleration 58.8m/s^2, Part B) free fall.

Part A)
Height reached, y = a*[t^2] / 2 = 58.8 m/s^2 * [7.00 s]^2 / 2 = 1440.6 m

Now you need the final speed to use it as initial speed of the next part.

Vf = Vo + at = 0 + 58.8m/s^2 * 7.00 s = 411.6 m/s

Part B) Free fall

Maximum height, y max ==> Vf = 0

Vf = Vo - gt ==> t = [Vo - Vf]/g = 411.6 m/s / 9.8 m/s^2 = 42 s

ymax = yo + Vo*t - g[t^2] / 2

ymax = 1440.6 m + 411.6m/s * 42 s - 9.8m/s^2 * [42s]^2 /2
ymax = 1440.6 m + 17287.2m - 8643.6m = 10084.2 m

Answer: ymax = 10084.2m
8 0
3 years ago
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3 years ago
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A particle accelerator fires a proton into a region with a magnetic field that points in the +x-direction (a) If the proton is m
pochemuha

Answer:

a) -z direction, b) +z direction, c) F=0  

Explanation:

The magnetic force is given by the expression

        F = q v x B

the bold indicate vectors, this equation can be separated in its module

        F = a v B sin θ

and where θ is the angles between the speed and the magnetic field.

The direction of the force can be found with the right-hand rule. For a positive charge, the thumb goes in the direction of speed, the fingers extended in the direction of the magnetic field and the palm points in the direction of the force, if the charge is negative the force is in the opposite direction.

a) Let's apply this to our case

the proton is positively charged

moves in the direction of + x

The magnetized field goes in the direction of y

therefore applying the right hand rule the force must be in the direction of the negative part of the z-axis (-z)

The right-hand rule is used to find this address.

b)  in this case it indicates that the proton moves in the recode of -y

again we apply the right hand rule and the force is in the direction of + z

c)   The proton moves in the x direction

In this case the force is zero because the angle between the field and the speed is zero and the sine is zero, therefore the force is zero

4 0
3 years ago
Particle-X has a speed of 0.720 c and a momentum of 4.350x1019 kgm/s. What is the mass of the particle? 2.0206 10-27 kg Hints: T
kirill115 [55]

Explanation:

Given that,

Speed of particle = 0.720 c

Momentum = 4.350\times10^{-19}\ kgm/s[/tex]

(I). We need to calculate the mass of the particle

Using formula of momentum

P=mv

m =\dfrac{P}{v}

m=\dfrac{4.350\times10^{-19}}{ 0.720\times3\times10^{8}}

m=2.013\times10^{-27}\ Kg

We need to calculate the rest mass of particle

Using formula of rest mass

m=\dfrac{m_{0}}{\sqrt{1-(\dfrac{v}{c})^2}}

Where, m_{0} = rest mass

Put the value into the formula

m_{0}=2.013\times10^{-27}\times\sqrt{1-(\dfrac{0.720 c}{c})^2}

m_{0}=2.013\times10^{-27}\times\sqrt{1-(0.720)^2}

m_{0}=1.4\times10^{-27}\ kg

(b). We need to calculate the rest energy of the particle

Using formula of energy

E_{0}=m_{0}c^2

Put the value into the formula

E_{0}=1.4\times10^{-27}\times(3\times10^{8})^2

E_{0}=1.26\times10^{-10}\ J

(c).  We need to calculate the kinetic energy of the particle

Using formula of kinetic energy

K.E=mc^2-m_{0}c^2

K.E=(m-m_{0})\timesc^2

K.E=(2.013\times10^{-27}-1.4\times10^{-27})\times3\times10^{8}

K.E=1.84\times10^{-19}\ J

(d). We need to calculate the total energy of the particle

Using formula of energy

E=mc^2

Put the value into the formula

E=2.013\times10^{-27}\times(3\times10^{8})^2

E=1.812\times10^{-10}\ J

Hence, This is the required solution.

8 0
3 years ago
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