If you observe an object in the universe that is roughly 8 megaparsecs in size, what is the object most likely to be?

A 60 kg adult and a 30kg child are passengers on a rotor ride at an amusement park as shown in the diagram above. When the rotat

Ulleksa [173]

Answer:

fr ’= ½ F

Explanation:

For this exercise we use the translational equilibrium equation, on the axis parallel to the wall

fr - W = 0

fr = W

for the adult man they indicate that the friction force is equal to F

F = M g

we write the equilibrium equation for the child

fr ’= w’

fr ’= m g

in the statement they tell us that the mass of the adult is 2 times the mass of the child

M = 2m

we substitute

fr ’= M / 2 g

fr ’= ½ Mg

we substitute

fr ’= ½ F

therefore the force of friction in the child is half of the friction in the adult

8 0

3 years ago

Arrange the core steps of the scientific method in sequential order.

andriy [413]

<span>the picture bellow gives you all the answers, this picture was given to my class by my science teacher a few years ago so it is accurate

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4 0

3 years ago

Read 2 more answers

vA child is in danger of drowning in the Merimac river. The Merimac river has a current of 3.1 km/hr to the east. The child is 0

ikadub [295]

Answer:

$d = 2.26 km$

Explanation:

Let the child is moving with speed same as the speed of water flow

So here the position of child with respect to flow must be zero

And if the boat start at an angle with the vertical

so its relative speed with flow of water is given as

$v_x = 24.8 sin\theta$

$v_y = 24.8 cos\theta$

now the time to reach the child is given as

$\frac{0.6}{24.8 cos\theta} = \frac{2.5}{24.8 sin\theta }$

so now we have

$\theta = 76.5 degree$

So the time to catch the child is given as

$t = \frac{0.6}{24.8 cos78.2}$

$t = 0.104 h$

So distance moved by it in 0.104 h

distance moved by the boat in upstream direction given as

$x = (24.8 sin 74.8 - 3.1)(0.104)$

$x = 2.18 km$

In y direction the displacement of boat is

$y = 0.6 km$

net displacement of the ball is given as

$d = \sqrt{x^2 + y^2}$

$d = \sqrt{2.18^2 + 0.6^2}$

$d = 2.26 km$

4 0

3 years ago

A long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire. The wire has a charge

Alexxx [7]

Explanation:

It is given that, a long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire.

The charge per unit length of the wire is $\lambda$ and the net charge per unit length is $2 \lambda$ .

We know that there exist zero electric field inside the metal cylinder.

(a) Using Gauss's law to find the charge per unit length on the inner and outer surfaces of the cylinder. Let $\lambda_i\ and\ \lambda_o$ are the charge per unit length on the inner and outer surfaces of the cylinder.

For inner surface,

$\phi=\dfrac{q_{enclosed}}{\epsilon_o}$

$E.A=\dfrac{q_{enclosed}}{\epsilon_o}$

$0=\dfrac{\lambda_i+\lambda}{\epsilon_o}$

$\lambda_i=-\lambda$

For outer surface,

$\lambda_i+\lambda_o=2\lambda$

$-\lambda+\lambda_o=2\lambda$

$\lambda_o=3\lambda$

(b) Let E is the electric field outside the cylinder, a distance r from the axis. It is given by :

$E_o=\dfrac{\lambda_o}{2\pi \epsilon_o r}$

$E_o=\dfrac{3\lambda}{2\pi \epsilon_o r}$

Hence, this is the required solution.

6 0

3 years ago

Jill applies a force of 250 N to a machine. The machine applies a force of 25 N to an object. What is the mechanical advantage o

Vinvika [58]

Mechanical advantage is defined as the ratio of output load to the input load. The mechanical advantage of the machine will be 0.1.

<h3>What is mechanical advantage?</h3>

Mechanical advantage is a measure of the ratio of output force to input force in a system,

It is used to obtain the efficiency of forces in levers and pulleys. It is an effective way of amplifying the force in simple machines like levers.

The theoretical mechanical advantage is defined as the ratio of the force responsible for the useful work in the system to the applied force.

Given

applied force = 250 N

Output force = 25

Mechanical advantage = work output / work input

$\rm{Mechanical advantage}=\frac{F_O}{F_I}$