PHYSICS COLLEGE

In a solution such as salt water, the component that does the dissolving is called the?

Answers

Answer 1
Answer:

Answer:

Solvent

Explanation:

  • A solution is a substance that is made by dissolving one component in another.
  • It is made up of a solvent and a solute.
  • The component that dissolves the other is known as solvent. Examples of solvents include water, ethanol, milk, chloroform, etc.
  • The component that dissolves in the other is known as the solute. Examples of solute include salts, sugar, etc.
  • For example in an aqueous solution of sodium chloride(NaCl); sodium chloride salt is the solute while water is the solvent.

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A rubber ball is shot straight up from the ground with speed v0. Simultaneously, a second rubber ball at height h directly above the first ball is dropped from rest. a) At what height above the ground do the balls collide? Your answer will be a symbolic expression in terms of v0, h, and g .b) What is the maximum value of h for which a collision occurs before the first ball falls back to the ground?
Express your answer in terms of the variable v0 and appropriate constants.
c) For what value of h does the collision occur at the instant when the first ball is at its highest point?
Express your answer in terms of the variable v0 and appropriate constants.

Answers

In this exercise we have to have knowledge about the horizontal launch, so we have to use the known formulas to find that:

a) h*(1 - 1/2 g * h/v_0^2)

b)h = v_0^2/ g

c)h = v_0^2/ g

So we have to remember some famous equations like the position and velocity of an object moving in a constant, like this:

y = y_0 + v_0*t + 1/2 * a * t^2\nv = v_0 + a * t

where:

  • y = height at time t
  • y0 = initial height
  • v0 = initial velocity
  • a = acceleration
  • t = time
  • v = velocity

a) When the balls collide, h1 = h2. Then,

h_1 = h_2\nv_0 * t - 1/2 g * t^2 = h - 1/2 * g * t^2\nv_0 * t = h\nt = h / v_0

Replacing in the equation of the height of the first ball:

h_1 = v_0 * h/v_0 - 1/2g * h^2/v_0^2\nh_1 = h - 1/2 g * h^2/ v_0^2\nh_1 = h*(1 - 1/2 g * h/v_0^2)

b)  that the balls collide at t = h/v0. Then:

h/ v_0 = -v_0/-g\nh = v_0^2/ g

c) The value of h for which the collision occurs at the instant when the first ball is at its highest point is the maximum value,Then:

h = v_0^2/ g

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6. A barber raises his customer's chair by applying a force of 150 N to thehydraulic piston of area 0.01 m2. If the chair is attached to a piston with an
area of 0.1 m², how much force is needed to raise the customer?
STEP 1: List the known
and unknown values F =
A=
A,
STEP 2: Write the
correct equation
STEP 3: Insert the
known values into the
equation to solve for
the unknown value

Answers

Answer:

15N

Explanation:

F¹=150N

A=0.01m2²

F2=?

A2=0.1m²

P=F/A

F1/A2=F2/A1

150/0.1=F2/0.01

Please use Gauss’s law to find the electric field strength E at a distance r from the center of a sphereof radius R with volume charge density ???? = cr 3 and total charge ????. Your answer should NOT contain c. Be sure to consider regions inside and outside the sphere.

Answers

Answer:

See the explaination for the details.

Explanation:

Gauss Law states that the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. The electric flux in an area is defined as the electric field multiplied by the area of the surface projected in a plane and perpendicular to the field.

According to the Gauss law, the total flux linked with a closed surface is 1/ε0 times the charge enclosed by the closed surface.

Please kindly check attachment for the step by step explaination of the answer.

An artificial satellite circling the Earth completes each orbit in 144 minutes. (a) Find the altitude of the satellite.

Answers

Answer:

Explanation:

given,

time taken to complete the each orbit = 144 minutes

                      t = 144 x 60 = 8640 s

mass of the earth = 5.98 x 10²⁴ Kg

radius of earth,R = 6.38 x 10⁶ m

Using Kepler's 3rd law

T^2 = ((4\pi^2)/(GM_e))r^3

r = ((T^2GM_e)/(4\pi^2))^(1/3)

r = ((8640^2* 6.67 * 10^(-11)* 5.98* 10^(24))/(4\pi^2))^(1/3)

r = 9.1 x 10⁶ m

the altitude of the satellite

H = r - R

H =  9.1 x 10⁶  - 6.38 x 10⁶

H = 2.72 x 10⁶ m

Inductance is usually denoted by L and is measured in SI units of henries (also written henrys, and abbreviated H), named after Joseph Henry, a contemporary of Michael Faraday. The EMF E produced in a coil with inductance L is, according to Faraday's law, given byE=−LΔIΔt.
Here ΔI/Δt characterizes the rate at which the current I through the inductor is changing with time t.
Based on the equation given in the introduction, what are the units of inductance L in terms of the units of E, t, and I (respectively volts V, seconds s, and amperes A)?
What EMF is produced if a waffle iron that draws 2.5 amperes and has an inductance of 560 millihenries is suddenly unplugged, so the current drops to essentially zero in 0.015 seconds?

Answers

Answer:

Explanation:

E= −L ΔI / Δt.

L = E Δt / ΔI

Hence the unit of inductance may be V s A⁻¹

or volt s per ampere .

In the given case

change in current ΔI = - 2.5 A

change in time = .015 s

L = .56 H

E = − L ΔI / Δt.

= .56 x 2.5 / .015

= 93.33 V .

A reducing elbow in a horizontal pipe is used to deflect water flow by an angle θ = 45° from the flow direction while accelerating it. The elbow discharges water into the atmosphere. The cross- sectional area of the elbow is 150 cm2 at the inlet and 25 cm2 at the exit. The elevation difference between the centers of the exit and the inlet is 40 cm. The mass of the elbow and the water in it is 50 kg. Determine the anchoring force needed to hold the elbow in place. Take the momentum flux correction factor to be 1.03 at both the inlet and outlet.

Answers

The anchoring force needed to hold the elbow in place is 839.5 N.

How to find anchoring force?

To determine the anchoring force needed to hold the elbow in place, use the following steps:

Apply the conservation of momentum equation to the elbow:

∑F = ˙m(V₂ - V₁)

where:

˙m = mass flow rate

V₁ and V₂ = velocities at the inlet and outlet of the elbow, respectively

F = anchoring force

The mass flow rate is given by:

˙m = ρAV

where:

ρ = density of water (1000 kg/m³)

A = cross-sectional area of the elbow

V = velocity

The velocities at the inlet and outlet of the elbow can be calculated using the following equations:

V₁ = A₁V₁

V₂ = A₂V₂

where:

A₁ and A₂ = cross-sectional areas at the inlet and outlet of the elbow, respectively

Calculate the momentum flux correction factor at the inlet and outlet of the elbow:

β₁ = 1.03

β₂ = 1.03

Substitute all of the above equations into the conservation of momentum equation:

F = ˙m(V₂ - V₁)

F = ρA₁V₁²β₁ - ρA₂V₂²β₂

Calculate the velocity at the inlet of the elbow:

V₁ = A₂V₂/A₁

V₁ = (25 cm²/150 cm²)(V₂)

V₁ = 1/6 V₂

Substitute the above equation into the conservation of momentum equation:

F = ρA₁V₁²β₁ - ρA₂V₂²β₂

F = ρA₁[(1/6 V₂)²](1.03) - ρA₂V₂²(1.03)

F = ρV₂²(1.03)(1/36 A₁ - A₂)

Calculate the anchoring force:

F = (1000 kg/m³)(V₂²)(1.03)(1/36 × 150 cm² - 25 cm²)

F = 839.5 N

Therefore, the anchoring force needed to hold the elbow in place is 839.5 N.

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The anchoring force needed to hold the elbow in place is 932 N.

The anchoring force needed to hold the elbow in place is the sum of the following forces:

The force due to the change in momentum of the water as it flows through the elbow.

The force due to the weight of the elbow and the water in it.

The force due to the buoyancy of the water in the elbow.

The force due to the change in momentum of the water can be calculated using the momentum equation:

F = mΔv

where:

F is the force

m is the mass of the fluid

Δv is the change in velocity of the fluid

In this case, the mass of the fluid is the mass of the water that flows through the elbow per second. This can be calculated using the mass flow rate equation:

m = ρAv

where:

ρ is the density of the fluid

A is the cross-sectional area of the pipe

v is the velocity of the fluid

The velocity of the fluid at the inlet can be calculated using the Bernoulli equation:

P1 + 0.5ρv1^2 = P2 + 0.5ρv2^2

where:

P1 is the pressure at the inlet

v1 is the velocity at the inlet

P2 is the pressure at the outlet

v2 is the velocity at the outlet

In this case, the pressure at the outlet is atmospheric pressure. The velocity at the outlet can be calculated using the continuity equation:

A1v1 = A2v2

where:

A1 is the cross-sectional area at the inlet

A2 is the cross-sectional area at the outlet

The force due to the weight of the elbow and the water in it is simply the weight of the elbow and the water in it. The weight can be calculated using the following equation:

W = mg

where:

W is the weight

m is the mass

g is the acceleration due to gravity

The force due to the buoyancy of the water in the elbow is equal to the weight of the water displaced by the elbow. The weight of the water displaced by the elbow can be calculated using the following equation:

B = ρVg

where:

B is the buoyancy

ρ is the density of the fluid

V is the volume of the fluid displaced

g is the acceleration due to gravity

The volume of the fluid displaced by the elbow is equal to the volume of the elbow.

Now that we have all of the forces, we can calculate the anchoring force needed to hold the elbow in place. The anchoring force is equal to the sum of the forces in the negative x-direction. The negative x-direction is the direction in which the water is flowing.

F_anchor = F_momentum + F_weight - F_buoyancy

where:

F_anchor is the anchoring force

F_momentum is the force due to the change in momentum of the water

F_weight is the force due to the weight of the elbow and the water in it

F_buoyancy is the force due to the buoyancy of the water in the elbow

Plugging in the values for each force, we get:

F_anchor = 1030 N - 490 N + 392 N = 932 N

Therefore, the anchoring force needed to hold the elbow in place is 932 N.

for such more question on momentum

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