A student measured the volume of three rocks. The diagram above shows the

Let T is the time of flight and R is the horizontal distance traveled. As there is no force acting in horizontal direction, so the horizontal velocity remains constant. Let the particle hits the ground with velocity v.

initial horizontal component of velocity, ux = u Cos 30

initial vertical component of velocity, uy = u Sin 30

Time of flight is given by

$T = \frac{2u Sin\theta }{g}$

Final horizontal component of velocity, vx = ux = u Cos 30

Let vy is teh final vertical component of velocity.

Use first equation of motion

vy = uy - gT

$v_{y}=u_{y}- g \times \frac{2u Sin\theta }{g}$

$v_{y}=u Sin 30 - 2u Sin 30$

vy = - u Sin 30

The magnitude of final velocity is given by

$v = \sqrt{v_{x}^{2}+v_{y}^{2}}$

$v = \sqrt{\left (uCos 30 \right )^{2}+\left (uSin 30 \right )^{2}}$

v = u

Thus, the velocity is same as it just reaches the ground.

6 0

3 years ago

An elite tour de france cyclist can maintain an output of 420 watts during a sustained climb. at this output power, how long wou

julia-pushkina [17]

<span>2002 seconds, or 33 minutes, 22 seconds. First, let's calculate how many joules it will take to lift 78 kg against gravity for 1100 meters. So: 78 kg * 9.8 m/s^2 * 1100 m = 840840 kg*m^2/s^2 Now a watt is defined as kg*m^2/s^3, so a division of the required joules should give us a convenient value of seconds. So: 840840 kg*m^2/s^2 / 420 kg*m^2/s^3 = 2002 seconds. And 2002 seconds is the same as 33 minutes, 22 seconds.</span>

7 0

3 years ago

Rotational dynamics about a fixed axis: A person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface

FrozenT [24]

Answer:

I = 2 kgm^2

Explanation:

In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:

$\tau=I\alpha$ (1)

I: moment of inertia of the door

α: angular acceleration of the door = 2.00 rad/s^2

τ: torque exerted on the door

You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:

$\tau=Fd$ (2)

F: force = 5.00 N

d: distance to the hinges = 0.800 m

You replace the equation (2) into the equation (1), and you solve for α:

$Fd=I\alpha\\\\I=\frac{Fd}{\alpha}$

Finally, you replace the values of all parameters in the previous equation for I:

$I=\frac{(5.00N)(0.800m)}{2.00rad/s^2}=2kgm^2$

The moment of inertia of the door around the hinges is 2 kgm^2

3 0

3 years ago