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skelet666 [1.2K]
3 years ago
6

What is the speed of a bobsled whose distance-time graph indicates that it traveled 116m in 29s?

Physics
2 answers:
ryzh [129]3 years ago
5 0
Speed = Distance ÷ Time
Speed = 116 ÷ 29
Speed = 4
mars1129 [50]3 years ago
5 0

Speed = (distance covered) / (time to cover the distance)

= (116 meters) / (29 seconds)

=  (116/29) (meters per second)

=  4 meters per second  (4 m/s)
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Lunna [17]

Answer:

a.  v=3.11mls, 29.4^{0}

b.   K.E =-697.8J

Explanation:

To calculate the values in the  question, a deep understanding of perfect inelastic collision is important.

When two bodies undergo inelastic collision, two important parameters must be well understood i.e

Momentum: the momentum is always conserved in perfectly inelastic collision. i.e the total momentum after collision is the sum of the individual momentum before collision

Kinetic energy: Kinetic energy is not conserved due to dissipative force.

a.To calculate the velocity, we first find the total momentum before collision

Momentum of player 1 p_{1} =mv=95kg*5m/s\\p_{1} =475kgm/s\\

Momentum of player 2 p_{2} =mv=90kg*3m/s\\p_{1} =270kgm/s\\

Hence the total momentum p_{12}=p_{1}+p_{2}\\

Note, since the direction of movement before collision is due south and  due north respectively we have to represent the velocity using the rectangular coordinate

Hence  p_{12}=(m_{1}+m_{2})v=p_{1}i+p_{2}j\\

(95+90)v=475i+270j\\

v=2.57i+1.45j\\

solving for the resultant velocity, we have

v=\sqrt{2.75^{2} +1.45^{2}}\\ v=3.11mls

To calculate the direction of movement, we have

\alpha =tan^{-1}=\frac{v_{j} }{v_{i}}\\  \alpha =tan^{-1}=\frac{1.45}{2.57}\\\alpha =29.4^{0}

b. to calculate the decrease in total kinetic energy, before collision, the total kinetic was

K.E_{initial} =\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}.\\K.E_{initial} =((1/2)*95*5^{2})+((1/2)*90*3^{2})\\K.E_{initial} =1187.5+405\\K.E_{initial} =1592.5J\\

And the final kinetic energy after collision is

K.E_{final} =\frac{1}{2}(m_{1}+m_{2} )v^{2}\\  K.E_{final} =\frac{1}{2}(95+90)* 3.11^{2}\\ K.E_{final} =894.7J

The decrease in Kinetic energy is

K.E =K.E_{final}- K.E_{initial}=894.7-1592.5

K.E =-697.8J

The negative sign indicate a decrease in Kinetic energy

4 0
3 years ago
A particle with mass 1.81×10−3 kg and a charge of 1.22×10−8 C has, at a given instant, a velocity v⃗ =(3.00×104m/s)j^. What are
slava [35]

Answer:

The magnitude and direction of the acceleration of the particle is a= 0.3296\ \hat{k}\ m/s^2

Explanation:

Given that,

Mass m = 1.81\times10^{-3}\ kg

Velocity v = (3.00\times10^{4}\ m/s)j

Charge q = 1.22\times10^{-8}\ C

Magnetic field B= (1.63\hat{i}+0.980\hat{j})\ T

We need to calculate the acceleration of the particle

Formula of the acceleration is defined as

F = ma=q(v\times B)

a =\dfrac{q(v\times B)}{m}

We need to calculate the value of v\times B

v\times B=(3.00\times10^{4}\ m/s)j\times(1.63\hat{i}+0.980\hat{j})

v\times B=4.89\times10^{4}

Now, put the all values into the acceleration 's formula

a =\dfrac{1.22\times10^{-8}\times(-4.89\times10^{4}\hat{k})}{1.81\times10^{-3}}

a= -0.3296\ \hat{k}\ m/s^2

Negative sign shows the opposite direction.

Hence, The magnitude and direction of the acceleration of the particle is a= 0.3296\ \hat{k}\ m/s^2

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Answer: 11

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Answer:

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Answer:

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