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3 years ago

motorcycle starting from rest cover 200 metre distance in 6 second calculate final velocity and acceleration of the motorcycle

1 answer:

6 0

<h2>Answer:</h2><h2>V=200m</h2><h2><em><u>time=6s</u></em></h2><h2><em><u>time=6s initial velocity=0</u></em></h2><h2><em><u>time=6s initial velocity=0 final velocity=200/6</u></em></h2><h2><em><u>time=6s initial velocity=0 final velocity=200/6=100/3m/s </u></em></h2><h2><em><u>time=6s initial velocity=0 final velocity=200/6=100/3m/s v=u+a</u></em></h2><h2><em><u>time=6s initial velocity=0 final velocity=200/6=100/3m/s v=u+aputting value,</u></em></h2><h2><em><u>time=6s initial velocity=0 final velocity=200/6=100/3m/s v=u+aputting value, 200/3=0+ax6</u></em></h2><h2><em><u>time=6s initial velocity=0 final velocity=200/6=100/3m/s v=u+aputting value, 200/3=0+ax6 200/3-ax6 </u></em></h2><h2><em><u>time=6s initial velocity=0 final velocity=200/6=100/3m/s v=u+aputting value, 200/3=0+ax6 200/3-ax6 200/3/6=a </u></em></h2><h2><em><u>time=6s initial velocity=0 final velocity=200/6=100/3m/s v=u+aputting value, 200/3=0+ax6 200/3-ax6 200/3/6=a 200/18-a </u></em></h2><h2><em><u>time=6s initial velocity=0 final velocity=200/6=100/3m/s v=u+aputting value, 200/3=0+ax6 200/3-ax6 200/3/6=a 200/18-a a 100/9m/s^2</u></em></h2>

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(I) A novice skier, starting from rest, slides down an icy frictionless 8.0° incline whose vertical height is 105 m. How fast is

Vlad1618 [11]

Answer:

v = 45.37 m/s

Explanation:

Given,

angle of inclination = 8.0°

Vertical height, H = 105 m

Initial K.E. = 0 J

Initial P.E. = m g H

Final PE = 0 J

Final KE = $\dfrac{1}{2}mv^2$

Using Conservation of energy

$KE_i + PE_i + KE_f + PE_f$

$0 + m g H = \dfrac{1}{2}mv^2 + 0$

$v = \sqrt{2gH}$

$v = \sqrt{2\times 9.8 \times 105}$

v = 45.37 m/s

Hence, speed of the skier at the bottom is equal to v = 45.37 m/s

3 0

3 years ago

The first antiparticle, the positron or antielectron, was discovered in 1932. It had been predicted by Paul Dirac in 1928, thoug

Taya2010 [7]

Answer:

Energy of Photon = 4.091 MeV

Explanation:

From the conservation of energy principle, we know that total energy of the system must remain conserved. So, the energy or particles before collision must be equal to the energy of photons after collision.

K.E OF electron + Rest Energy of electron + K.E of positron + Rest Energy of positron = 2(Energy of Photon)

where,

K.E OF electron = 3.58 MeV

Rest Energy of electron = 0.511 MeV

Rest Energy of positron = 0.511 MeV

K.E OF positron = 3.58 MeV

Energy of Photon = ?

Therefore,

3.58 MeV + 0.511 MeV + 3.58 MeV + 0.511 MeV = 2(Energy of Photon)

Energy of Photon = 8.182 MeV/2

<u>Energy of Photon = 4.091 MeV</u>

8 0

3 years ago

uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is station

Alex787 [66]

Answer:

The magnitude of the tangential velocity is $v= 0.868 m/s$

The magnitude of the resultant acceleration at that point is $a = 4.057 m/s^2$

Explanation:

From the question we are told that

The mass of the uniform disk is $m_d = 40.0kg$

The radius of the uniform disk is $R_d = 0.200m$

The force applied on the disk is $F_d = 30.0N$

Generally the angular speed i mathematically represented as

$w = \sqrt{2 \alpha \theta}$

Where $\theta$ is the angular displacement given from the question as

$\theta = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}$

$=1.257\ rad$

$\alpha$ is the angular acceleration which is mathematically represented as

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

The moment of inertial is mathematically represented as

$I = \frac{1}{2} m_dR^2_d$

Substituting values

$I = 0.5 * 40 * 0.200^2$

$= 0.8kg \cdot m^2$

Considering the equation for angular acceleration

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

Substituting values

$\alph$ $\alpha = \frac{(30.0)(0.200)}{0.8}$

$= 7.5 rad/s^2$

Considering the equation for angular velocity

$w = \sqrt{2 \alpha \theta}$

Substituting values

$w =\sqrt{2 * (7.5) * 1.257}$

$= 4.34 \ rad/s$

The tangential velocity of a given point on the rim is mathematically represented as

$v = R_d w$

Substituting values

$= (0.200)(4.34)$

$v= 0.868 m/s$

The radial acceleration at hat point is mathematically represented as

$\alpha_r = \frac{v^2}{R}$

$= \frac{0.868^2}{0.200^2}$

$= 3.7699 \ m/s^2$

The tangential acceleration at that point is mathematically represented as

$\alpha _t = R \alpha$

Substituting values

$\alpha _t = (0.200) (7.5)$

$= 1.5 m/s^2$

The magnitude of resultant acceleration at that point is

$a = \sqrt{\alpha_r ^2+ \alpha_t^2 }$

Substituting values

$a = \sqrt{(3.7699)^2 + (1.5)^2}$

$a = 4.057 m/s^2$

7 0

3 years ago

The most common atom used in fission is

Eduardwww [97]

Answer:

Uranium

Explanation:

5 0

4 years ago

What changes a ball velocity

Grace [21]

Answer/Explanation: Speed and direction can change with time. When you throw a ball into the air, it leaves your hand at a certain speed. As the ball rises, it slows down. Then, as the ball falls back toward the ground, it speeds up again. When the ball hits the ground, its direction of motion changes and it bounces back up into the air.

5 0

4 years ago

motorcycle starting from rest cover 200 metre distance in 6 second calculate final velocity and acceleration of the motorcycle ​

motorcycle starting from rest cover 200 metre distance in 6 second calculate final velocity and acceleration of the motorcycle