Ask question

Ask question

All categories

3 years ago

9

Which title best fits Rahma's note

2 answers:

Alik [6]3 years ago

5 0

Answer:diffusion

Explanation: it right

Semenov [28]3 years ago

4 0

Answer:

diffusion is the answer.

You might be interested in

A merry-go-round on a playground consists of a horizontal solid disk with a weight of 805 N and a radius of 1.58 m. A child appl

lozanna [386]

Answer:

The value is $KE = 259.6 \ J$

Explanation:

From the question we are told that

The weight of the horizontal solid disk is $W = 805 \ N$

The radius of the horizontal solid disk is $r = 1.58 \ m$

The force applied by the child is $F = 49.5 \ N$

The time considered is $t = 2.95 \ s$

Generally the mass of the horizontal solid disk is mathematically represented as

$m_h = \frac{W}{ g}$

=> $m_h = \frac{805}{ 9.8 }$

=> $m_h = 82.14 \ N$

Generally the moment of inertia of the horizontal solid disk is mathematically represented as

$I = \frac{1}{2} * m * r^ 2$

=> $I = \frac{1}{2} * 82.14 * 1.58^ 2$

=> $I = 102.5 \ kg \cdot m^2$

Generally the net torque experienced by the horizontal solid disk is mathematically represented as

$T = I * \alpha = F * r$

=> $\alpha = \frac{ F * r }{ I }$

=> $\alpha = \frac{ 49.5 * 1.58 }{ 102.53 }$

=> $\alpha = 0.7628$

Gnerally from kinematic equation we have that

$w = w_o + \alpha t$

Here $w_o$ is the initial angular velocity velocity of the horizontal solid disk which is $w_o = 0\ rad/s$

So

$w = 0 + 0.7628 * 2.95$

=> $w = 2.2503 \ rad/s$

Generally the kinetic energy is mathematically represented as

$KE = \frac{1}{2} * I * w^2$

=> $KE = \frac{1}{2} * 102.53 * 2.2503 ^2$

=> $KE = 259.6 \ J$

8 0

3 years ago

The increase in resistant strains of insects to chemicals is a result of:

trapecia [35]

Answer:

answer D. all of the above

5 0

3 years ago

Which letter on the map represents Australia?

vovangra [49]

Answer:

C because it literally in Australia but the pfp tho

7 0

3 years ago

A moving rope (parallel to the slope) is used to pull skiers up the mountain. If the slope of the hill is 37" and friction is ne

Charra [1.4K]

Since rope is parallel to the inclined plane so here we can say that net force parallel to the person which is pulling upwards must counterbalance the component of weight of the person.

Now here we will do the components of the weight of the person

given that weight of the person = 500 N

now its components are

$W_x = 500 cos37$

$W_y = 500 sin37$

now here as we can say that one of the component is balanced here by the normal force perpendicular to plane

while the other component of the weight is balanced by the force applied on the rope

So here the force applied on the rope will be given as

$F = W_y = 500 sin37$

$F = 300 N$

so it apply 300 N force along the inclined plane

5 0

4 years ago

Why did you spin counterclockwise rather than clockwise

rodikova [14]

<span>Because of the orbit of the earth and the sun and the moon. </span>

6 0

3 years ago

Read 2 more answers