Can water boil at 0°?

A child of mass 47 kg sits on the edge of a merry-go-round with radius 1.3 m and moment of inertia 56.3953 kg m2 . The merrygo-r

lakkis [162]

Answer:

$F=156.416\ N$ the child have to exert this much amount of force radially to stay on the wheel.

Explanation:

Given:

mass of the child, $m=47\ kg$

radius of the merry-go round wheel, $r=1.3\ m$

moment of inertia of the wheel, $I=56.3953\ kg.m^2$

angular velocity of the wheel, $\omega=1.6\ rad.s^{-1}$

Since the child is sitting on the edge of the rotating wheel the child will feel and outward throwing force called centrifugal force.

Centrifugal force is mathematically given as:

$F=m.r.\omega^2$

$F=47\times 1.3\times 1.6^2$

$F=156.416\ N$ the child have to exert this much amount of force radially to stay on the wheel.

4 0

2 years ago

In case 1, a force f is pushing perpendicular on an object a distance l/2 from the rotation axis. in case 2 the same force is pu

Serggg [28]

In case 1, the torque is given by the product between the force and the arm:
$\tau_1 = F \cdot \frac{L}{2}$

In case 2, the torque is given by the product between the component of the force perpendicular to the arm and the arm itself, so we have:
$\tau_2 = F \cos 30^{\circ} L=F \frac{ \sqrt{3} }{2} L = \sqrt{3} F \frac{L}{2}= \sqrt{3} \tau_1$

and since $\sqrt{3}$ is larger than 1, than the torque in case 2 is larger than the torque in case 1.

6 0

3 years ago

You toss a rock of mass mm vertically upward. Air resistance can be neglected. The rock reaches a maximum height hh above your h

skad [1K]

Answer:

The speed of the rock when it is at height h/4 is $\dfrac{\sqrt{3gh} }{2}$ .

Explanation:

At maximum height the final velocity of the rock is equal to 0. Let u is the initial velocity of the rock. Using the conservation of energy to find it as :

$u^2=2gh$ .......(1)

We need to find the speed of the rock when it is at height h/4. Let v' is the speed. Using 3rd equation of motion as :

$v'^2=u^2+2as$

here a = -g and s = h/4

$v'^2=u^2-2g\times \dfrac{h}{4}$

Using equation (1) :

$v'^2=(2gh)-2g\times \dfrac{h}{4}\\\\v'^2=\dfrac{3gh}{4}\\\\v'=\dfrac{\sqrt{3gh} }{2}$

So, the speed of the rock when it is at height h/4 is $\dfrac{\sqrt{3gh} }{2}$ . Hence, this is the required solution.

3 0

2 years ago

Read 2 more answers

The gravitational field strength due to its planet is 5N/kg What does it mean?

Dominik [7]

Answer:

The weight of an object is the force on it caused by the gravity due to the planet. The weight of an object and the gravitational field strength are directly proportional. For a given mass, the greater the gravitational field strength of the planet, the greater its weight.

Weight can be calculated using the equation:

weight = mass × gravitational field strength

This is when:

weight (W) is measured in newtons (N)

mass (m) is measured in kilograms (kg)

gravitational field strength (g) is measured in newtons per kilogram (N/kg)

6 0

2 years ago

What affects fuel consumption in automobiles?

maksim [4K]

Answer:

A and C

Explanation:

drag (the area of lower air pressure behind the car when moving) and mostly air resistance (the work to push the air in front of us away to move through - the faster we go, the stronger the air resists to move aside).

4 0

1 year ago