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3 years ago

Help me please Ima give brainiest

Physics

2 answers:

julsineya [31]3 years ago

4 0

Sand. You can identify tiny sugar crystals.

creativ13 [48]3 years ago

3 0

Answer: vinegar

A heterogeneous mixture is a mixture in which the composition is not uniform throughout the mixture. ... A heterogeneous mixture consists of two or more phases. When oil and water are combined, they do not mix evenly, but instead, form two separate layers. Each of the layers is called a phase.

Hope it helps...

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A 25 kg child plays on a swing having support ropes that are 2.20 m long. A friend pulls her back until the ropes are 42◦ from t

Semmy [17]

Answer:

A) P.E = 138.44 J

B) The velocity of swing at bottom, v = 3.33 m/s

C) The work done, W = -138.44 J

Explanation:

Given,

The mass of the child, m = 25 Kg

The length of the swing rope, L = 2.2 m

The angle of the swing to the vertical position, ∅ = 42°

A) The potential energy at the initial position ∅ = 42° is given by the relation

P.E = mgh joule

Considering h = 0 for the vertical position

The h at ∅ = 42° is h = L (1 - cos∅)

P.E = mgL (1 - cos∅)

Substituting the given values in the above equation

P.E = 25 x 9.8 x 2.2 (1 - cos42°)

= 138.44 J

The potential energy for the child just as she is released, compared to the potential energy at the bottom of the swing is, P.E = 138.44 J

B) The velocity of the swing at the bottom.

At bottom of the swing the P.E is completely transformed into the K.E

∴ K.E = P.E

1/2 mv² = 138.44

1/2 x 25 x v² 138.44

v² = 11.0752

v = 3.33 m/s

The velocity of the swing at the bottom is, v = 3.33 m/s

C) The work done by the tension in the rope from initial position to the bottom

Tension on string, T = Force acting on the swing, F

$W=L\int\limits^0_\phi{F} \, d \phi$

$=L\int\limits^0_\phi{mg.sin \phi} \, d \phi$

= $-Lmg[cos\phi]_{42}^{0}$

= - 2.2 x 25 x 9.8 [cos0 - cos 42°]

= - 138.44 J

The negative sign in the in energy is that the work done is towards the gravitational force of attraction.

The work done by the tension in the ropes as the child swings from the initial position to the bottom of the swing, W = - 138.44 J

3 0

3 years ago

When a parachutist jumps from an airplane, he eventually reaches a constant speed, called the terminal speed. Once he has reache

Masteriza [31]

Answer:

b) True. the force of air drag on him is equal to his weight.

Explanation:

Let us propose the solution of the problem in order to analyze the given statements.

The problem must be solved with Newton's second law.

When he jumps off the plane

fr - w = ma

Where the friction force has some form of type.

fr = G v + H v²

Let's replace

(G v + H v²) - mg = m dv / dt

We can see that the friction force increases as the speed increases

At the equilibrium point

fr - w = 0

fr = mg

(G v + H v2) = mg

For low speeds the quadratic depended is not important, so we can reduce the equation to

G v = mg

v = mg / G

This is the terminal speed.

Now let's analyze the claims

a) False is g between the friction force constant

b) True.

c) False. It is equal to the weight

d) False. In the terminal speed the acceleration is zero

e) False. The friction force is equal to the weight

3 0

3 years ago

A spring scale has a spring with a force constant of 250 N/m and a weighing pan with a mass of 0.075 kg. During one weighing, th

mr Goodwill [35]

elasticity stretches and can also return to it's normal size ..

8 0

2 years ago

An object is dropped from a vertical distance of 25.5 m above the ground, and it takes 2.28 sec to fall that distance. A second

DENIUS [597]

Answer:

The second object takes 2.28 s to fall the 25.5 m.

Explanation:

In this case, both objects take the same time to fall, since no vertical velocity is added to any of them.

You can also confirm this by sepparating the second's object movement into its two directions: in the horizontal one, we have linear uniform motion, and in the vertical one, we have free fall, with exactly the same characteristics as for the first object.

4 0

3 years ago

Enter an expression for the force constant for the floating raft, in terms of L, g, and the density of water, ρ.

andriy [413]

Answer:

K = ρL²g

Explanation:

Consider L as the length of the raft inside the water when the raft is displaced through additional distance y;

Then:

F = upthrust ( restoring force) = weight of the liquid displaced.

$F = V_{\omega} \rho_{\omega} g= A y \rho_{\omega} g$