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2 years ago

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When Jim and Rob ride bicycles, Jim can only accelerate at three-quarters the acceleration of Rob. Both startfrom rest at the bo

ttom of a long, straight road with constant upward slope. If Rob takes 5.0 minutes to reach thetop, how much earlier should Jim start in order to reach the top at the same time as Rob

1 answer:

Natali5045456 [20]2 years ago

7 0

Answer:

46.4 s

Explanation:

5 minutes = 60 * 5 = 300 seconds

Let g = 9.8 m/s2. And $\theta$ be the slope of the road, s be the distance of the road, a be the acceleration generated by Rob, 3a/4 is the acceleration generated by Jim . Both of their motions are subjected to parallel component of the gravitational acceleration $gsin\theta$

Rob equation of motion can be modeled as s = a_Rt_R^2/2 = a300^2/2 = 45000a[/tex]

Jim equation of motion is $s = a_Jt_J^2/2 = (3a/4)t_J^2/2 = 3at_J^2/8$

As both of them cover the same distance

$45000a = 3at_J^2/8$

$t_J^2 = 45000*8/3 = 120000$

$t_J = \sqrt{120000} = 346.4 s$

So Jim should start 346.4 – 300 = 46.4 seconds earlier than Rob in other to reach the end at the same time

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Two astronauts, A and B, both with mass of 60Kg, are moving along a straight line in the same direction in a weightless spaceshi

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Answer:

$V=14m/s$

Explanation:

From the Question we are told that

Mass of A and B is 60kg

Speed of A=2m/s

Speed of B=1m/s

Mass of bag =5kg

Generally the momentum of the astronaut A and bag is mathematically given as

$M_A=(60+5)*2$

$M_A=130kgm/s$

Generally to avoid collision the speed of astronaut be should be less than or equal to that of astronaut A with the bag

Therefore for minimum requirement speed of astronaut A should be given by astronaut B's speed which is equal to 1

Therefore

$130=(60*1)=(5*v)$

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g A projectile is launched with speed v0 from point A. Determine the launch angle ! which results in the maximum range R up the

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Answer:

The range is maximum when the angle of projection is 45 degree.

Explanation:

The formula for the horizontal range of the projectile is given by

$R = \frac{u^{2}Sin2\theta }{g}$

The range should be maximum if the value of Sin2θ is maximum.

The maximum value of Sin2θ is 1.

It means 2θ = 90

θ = 45

Thus, the range is maximum when the angle of projection is 45 degree.

If the angle of projection is 0 degree

R = 0

It means the horizontal distance covered by the projectile is zero, it can move in vertical direction.

If the angle of projection is 30 degree.

$R = \frac{u^{2}Sin60 }{9.8}$

R = 0.088u^2

If the angle of projection is 45 degree.

$R = \frac{u^{2}Sin90 }{g}$

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Xander's friend Corey has a skateboard that he rides at Newton's Skate Park.

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Answer:

0.5 m/s2

Explanation:

Step 1:

Data obtained from the question.

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Net force = 30N

Acceleration =?

Step 2

Determination of the acceleration.

Force = Mass x Acceleration.

With the above equation, we can easily obtain the acceleration as follow:

30 = 60 x Acceleration

Divide both side by 60

Acceleration = 30/60

Acceleration = 0.5 m/s2

Now, we can thus say that the acceleration at that moment is 0.5 m/s2

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Janice's mother often lets her 6-month-old baby sit in front of the television, watching episodes of Sesame Street. What is Jani

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2 years ago

A high school physics instructor catches one of his students chewing gum in class. He decides to discipline the student by askin

KengaRu [80]

a) 219.8 rad/s

b) $20.0 rad/s^2$

c) $2.9 m/s^2$

d) $7005 m/s^2$

e) Towards the axis of rotation

f) $0 m/s^2$

g) 31.9 m/s

Explanation:

a)

The angular velocity of an object in rotation is the rate of change of its angular position, so

$\omega=\frac{\theta}{t}$

where

$\theta$ is the angular displacement

t is the time elapsed

In this problem, we are told that the maximum angular velocity is

$\omega_{max}=35 rev/s$

The angle covered during 1 revolution is

$\theta=2\pi$ rad

Therefore, the maximum angular velocity is:

$\omega_{max}=35 \cdot 2\pi = 219.8 rad/s$

b)

The angular acceleration of an object in rotation is the rate of change of the angular velocity:

$\alpha = \frac{\Delta \omega}{t}$

where

$\Delta \omega$ is the change in angular velocity

t is the time elapsed

Here we have:

$\omega_0 = 0$ is the initial angular velocity

$\omega_{max}=219.8 rad/s$ is the final angular velocity

t = 11 s is the time elapsed

Therefore, the angular acceleration is:

$\alpha = \frac{219.8-0}{11}=20.0 rad/s^2$

c)

For an object in rotation, the acceleration has two components:

- A radial acceleration, called centripetal acceleration, towards the centre of the circle

- A tangential acceleration, tangential to the circle

The tangential acceleration is given by

$a_t = \alpha r$

where

$\alpha$ is the angular acceleration

r is the radius of the circle

Here we have

$\alpha =20.0 rad/s^2$

d = 29 cm is the diameter, so the radius is

r = d/2 = 14.5 cm = 0.145 m

So the tangential acceleration is

$a_t=(20.0)(0.145)=2.9 m/s^2$

d)

The magnitude of the radial (centripetal) acceleration is given by

$a_c = \omega^2 r$

where

$\omega$ is the angular velocity

r is the radius of the circle

Here we have:

$\omega_{max}=219.8 rad/s$ is the angular velocity when the fan is at full speed

r = 0.145 m is the distance of the gum from the centre of the circle

Therefore, the radial acceleration is

$a_c=(219.8)^2(0.145)=7005 m/s^2$

e)

The direction of the centripetal acceleration in a rotational motion is always towards the centre of the axis of rotation.

Therefore also in this case, the direction of the centripetal acceleration is towards the axis of rotation of the fan.

f)

The magnitude of the tangential acceleration of the fan at any moment is given by

The tangential acceleration is given by

$a_t = \alpha r$

where

$\alpha$ is the angular acceleration

r is the radius of the circle

When the fan is rotating at full speed, we have:

$\alpha=0$ , since the fan is no longer accelerating, because the angular velocity is no longer changing

r = 0.145 m

Therefore, the tangential acceleration when the fan is at full speed is

$a_t=(0)(0.145)=0 m/s^2$

g)

The linear speed of an object in rotational motion is related to the angular velocity by the formula:

$v=\omega r$

where

v is the linear speed

$\omega$ is the angular velocity

r is the radius

When the fan is rotating at maximum angular velocity, we have:

$\omega=219.8 rad/s$

r = 0.145 m

Therefore, the linear speed of the gum as it is un-stucked from the fan will be:

$v=(219.8)(0.145)=31.9 m/s$

7 0

2 years ago