Piston 1 in the figure has a diameter of 1.87 cm. Piston 2 has a diameter of 9.46 cm.
In the absence of friction, determine the force F, necessary to support an object with a mass of 991 kg placed on piston 2. (Neglect the height difference between the bottom of the two pistons, and assume that the pistons are massless).

The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

To find the answer, we need to know about the time of flight and range of projectile motion.

U= √{2.20×9.8/sin(73)} = 22.5m/s

= 2.73 s

Thus, we can conclude that the frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

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Answer:

4 C

Explanation:

The strength of an electric field can be defined as: [tex]E = \frac{f}{q}[/tex] where f=force and q=charge, and e=strength

Since we're given the strength and the force we can simply rearrange the equation so that we solve for Q:

Original Equation:

[tex]E=\frac{f}{q}[/tex]

Multiply both sides by q

[tex]E*q = f[/tex]

Divide both sides by E

[tex]q=\frac{f}{e}[/tex]

So now plug the known values into the equation

[tex]q=\frac{1000 N}{250 N/C}[/tex]

Simplify:

[tex]q = 4 c[/tex]

The voltage between the plates is 3.9 × [tex]10^-^3[/tex] V.

The work-energy theorem states that the delta in the equation equals the change in kinetic energy plus the change in potential energy. Here, a charge's potential energy is expressed as qV, where V is the position's electric potential.  The greater the change in voltage per unit distance, the greater the electric field.

The kinetic energy of the electrons = 4.1 × [tex]10^-^1^5[/tex] J

Charge of the electron = 1.602 × [tex]10^-^1^9[/tex] coulomb

Using,

     ΔU = q × ΔV

4.1 × [tex]10^-^1^5[/tex] = 1.602 × [tex]10^-^1^9[/tex] × ΔV

      ΔV  = 3.9 × [tex]10^-^3[/tex] V

Therefore, the voltage between the plates is 3.9 × [tex]10^-^3[/tex] V.

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The image distance is 33.3 cm while the image height is 14.2 cm.

A converging lens will always have a positive focal length hence, we have to find the object distance as follows;

1/f = 1/v + 1/u

1/12 = 1/v + 1/20

1/v = 1/12 - 1/20

1/v = 0.08 - 0.05

v =33.3 cm

Now;

Magnification =  33.3 cm/20.0 cm =1.67

M = Image height/Object height

1.67 =  Image height/8.50 cm

Image height = 1.67  * 8.50 cm

Image height = 14.2 cm

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Answer:

Explanation:

An electric dipole is formed by two point charges +q and −q connected by a vector a. The electric dipole moment is defined as p = qa

a. θ = 41. 4°

b. θ = 60°

c. θ = 75. 5°

d. θ = 90°

From the given information, we would be using the Malus' law

It is given as;

I = I0 cos²θ

Where I0 is the intensity of the polarized light after passing through P

a. To find the angle, compare with the given equation

I = (0.750)I0

I = I0 cos θ

then

cos θ = 0. 750

θ = [tex]cos^-^1(0. 750)[/tex]

θ = 41. 4°

b.  I = (0.500)I0

cos  θ = 0. 500

θ = [tex]cos^-^1(0. 500)[/tex]

θ = 60°

c.  I = (0.250)I0

cos θ = 0. 250

θ = [tex]cos^-^1 (0. 250)[/tex]

θ = 75. 5°

d.  I = 0

cos  θ = 0

θ = [tex]cos^-^1 (0)[/tex]

θ = 90°

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(a) The time the baseball spends in the air is 0.92 s.

(b) The horizontal distance from the roof edge to the point where the baseball lands on the ground is 3.1 m.

h = vt - ¹/₂gt²

-2.1 = (4.05 x sin 34)t  - ¹/₂(9.8)(t²)

-2.1 = 2.26t - 4.9t²

4.9t² - 2.26t - 2.1 = 0

t = 0.92 s

R = Vx(t)

R = (4.05 x cos 34)(0.92)

R = 3.1 m

Thus, the time the baseball spends in the air is 0.92 s.

The horizontal distance from the roof edge to the point where the baseball lands on the ground is 3.1 m.

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A psychologist will need to have good linguistic intelligence in other to be successful.

This is referred to as a professional who specializes in the handling of mental health challenges in individuals.

It is best for such professional to have a good linguistic intelligence as the right words being said to the patient will solve the problem thereby bringing in more success.

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Answer:

To give light yo people and the neutron are to give you health care

The amount of work done to stop a 975- kg car traveling at 105 km/h is 414,808.34J.

The amount of work done by a moving object can be calculated using the following formula:

W (Kinetic energy) = ½ mv²

Where;

According to this question, a car of 975 kg is traveling at 105 km/h. This speed in m/s is 29.17m/s.

K.E = ½ × 975 × 29.17²

K.E = 414,808.34J

Therefore, the amount of work done to stop a 975- kg car traveling at 105 km/h is 414,808.34J.

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The Force on the left hand pole, F' = 0.167N

Force is an agent which produces a change in the motion or state of an object.

Force is a vector quantity.

The general force is calculated as follows:

F = mg/sinθ

m = 17.1 g = 0.0171 kg

g = 9.81 m/s²

θ = 45°

F = 0.0171 * 9.81/sin45

F = 0.237 N

Force on the left hand pole, F' = Fcosθ

F' = 0.237 * cos 45

F' = 0.167N

In conclusion, the force on the left hand pole is the horizontal component of force.

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The two units for measuring the diameter of nucleus atom are femtometre and metre.

How do you measure the size of the nucleus ?

Nucleus size is expressed in fermi, often known as femtometers. between a lighter and a heavier nucleus. Despite its modest size, the nucleus contains the majority of an atom's mass. The weight or mass of the atom's nucleus and neutrons are determined by neutrons.

femtometre (fm), which equals [tex]10^{-15}[/tex] metre.

A nucleus' diameter largely depends as to how many particles it contains, from about 4 fm for a light nucleus like carbon to 15 fm for a heavy nucleus as lead.

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Answer: The mass of the object is 25kg.

The given question deals with Newton's second law of motion and its applications.

Explanation: Given force, F=500N

                                 acceleration, a=20 m/[tex]s^{2}[/tex]

  From Newton's 2nd law of motion , we have

                             F=ma where m=mass of the object

                         ⇒500=m×20

                         ⇒m=500/20=25

                 ∴ Mass of the object is 25 kg .

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The restoring force on the spring is found to have exactly the same magnitude as the stretching force. Option D

The restoring force is the force that seeks to restore the spring to its equilibrium position. It has the same magnitude as the stretching force but acts in opposite direction.

Thus, the restoring force on the spring is found to have exactly the same magnitude as the stretching force.

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(1) The critical angles for the air is 90⁰,

(2) The critical angles for the polystyrene is 39.1 ⁰ and

(3) The critical angles for the flint glass is 37.2 ⁰.

θc = sin⁻¹( 1/η)

where;

Refractive index of air = 1

Refractive index of polystyrene = 1.5865

Refractive index of flint glass = 1.655

θc = sin⁻¹( 1/1)

θc = 90⁰

θc = sin⁻¹( 1/1.5865)

θc =  39.1 ⁰

θc = sin⁻¹( 1/1.655)

θc = 37.2 ⁰

Thus, the critical angles for the air is 90⁰, the critical angles for the polystyrene is 39.1 ⁰ and the critical angles for the flint glass is 37.2 ⁰.

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The maximum speed with which a 1200- kg car can make a turn is 22.02 m/s.

v(max) = √μrg

where;

v(max) = √(0.55 x 90 x 9.8)

v(max) = 22.02 m/s

Thus, the maximum speed with which a 1200- kg car can make a turn is 22.02 m/s.

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Answer:

NE BİLİM

Explanation:

PUAN İÇİN YAZIYORUM SEN BENİ GÖRMEZDEN GEL ANLDIN MI DOSTUM bu arada Suriye >>>>>

Tension in the rope C is 1.24× 10² N.

To find the answer, we need to know about the horizontal component of tension in the rope B.

Horizontal component= tension in rope B × cos80.5°

= 7.52×10² N × cos80.5°

= 1.24×10² N

Thus, we can conclude that the tension in the rope C is 1.24× 10² N.

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The coefficient of friction between the sled and the snow is 0.119.

To find the answer, we need to know about the friction.

                                     [tex]a=0\\\alpha =35^0\\T=75N\\m=57kg[/tex]

Here, acceleration will be equal to zero, because the velocity is given as constant.

                                      [tex]ma=Tcos\alpha -f\\\where\\f=kN\\\N+Tsin\alpha=mg\\Thus,\\N=mg-Tsin\alpha[/tex]

                   [tex]ma= Tcos\alpha -k(mg-Tsin\alpha )[/tex]

                    [tex]0=(75*cos35)-k((57*9.8)-(75sin35))\\\\k((57*9.8)-(75sin35))=(75*cos35)\\\\515.6k=61.44\\\\k=\frac{61.44}{515.6}=0.119[/tex]

Thus, we can conclude that, the coefficient of friction between the sled and the snow is 0.119.

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The sled's coefficient of friction with the snow is 0.119.

We must understand the friction in order to choose the solution.

                               [tex]\alpha =35\\T=75N\\m=57kg\\a=0[/tex]

Because the velocity is specified as constant in this case, the acceleration will be equal to zero.

                             [tex]ma=Tcos\alpha -f\\ where,f=kN\\N+Tsin\alpha =mg\\ thus,\\N=mg-Tsin\alpha[/tex]

                       [tex]ma=Tcos\alpha -k(mg-Tsin\alpha )[/tex]

                                   [tex]k=0.119[/tex]

As a result, we may say that there is 0.119 coefficient of friction between the sled and the snow.

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The distance between two consecutive nodes and the amplitude after 0.56s are m/2 and 1.75×10^(-4) m respectively.

= 2π/(4π÷m)

= m/2

Displacement after 0.56 s = 0.008×cos(50π×0.56s)

=1.75×10^(-4) m

Thus, we can conclude that the distance between consecutive nodes and displacement after 0.56 s are m/2 and 1.75×10^(-4) m respectively.

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: The particle displacement y of air molecules due to a sound wave is given by y=0.008coswtsinkz where k=4π÷m and w=50π rads/s.

Calculate:

I) the distance between 2 consecutive nodes

ii) the amplitude after 0.565s

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From the calculations, the power expended is 43650 W.

Now we can find the acceleration from;

v = u + at

u = 0 m/s

v =  95 km/h or 26.4 m/s

t =  6.8 s

a = ?

Now

v = at

a = v/t

a = 26.4 m/s/ 6.8 s

a = 3.88 m/s^2

Force = ma = 850-kg * 3.88 m/s^2 = 3298 N

The distance covered is obtained from;

v^2 = u^2 + 2as

v^2 = 2as

s = v^2/2a

s = (26.4)^2/2 * 3.88

s = 696.96/7.76

s = 90 m

Now;

Work = Fs

Work =  3298 N * 90 m = 296820 J

Power =  296820 J/ 6.8 s

= 43650 W

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The angle of projection given that the range is 256√3 m and the maximum height reached is 64 m is 30°

R = u²Sine(2θ) / g

256√3 = u²Sine(2θ) / 9.8

Cross multiply

256√3 × 9.8 = u²Sine(2θ)

Divide both sides by Sine(2θ)

u² = 256√3 × 9.8 / Sine(2θ)

H = u²Sine²θ / 2g

64 = [256√3 × 9.8 / Sine(2θ)] × [Sine²θ / 2 × 9.8]

64 = [256√3 / Sine(2θ)] × [Sine²θ / 2]

Recall

Sine²θ = SineθSineθ

Sine2θ = 2SineθCosθ

Thus,

64 = [256√3 / 2SineθCosθ] × [SineθSineθ / 2]

64 = 256√3 × Sineθ / 4Cosθ

Recall

Sineθ / Cosθ = Tanθ

Thus,

64 = 256√3 / 4 × Tanθ

Divide both side by 256√3 / 4

Tanθ = 64 ÷ 256√3 / 4

Tanθ = 0.5774

Take the inverse of Tan

θ = Tan⁻¹ 0.5774

θ = 30°

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The mass of the airplane of area of two wings 395m², and the average speed of lower and upper surface of the wings are, 259 and 288m/s is 386.8×10^3 kg.

To find the answer, we need to know about the Bernoulli's principle.

                   P+ [tex]\frac{1}{2}[/tex] ρv²+ρgh=a constant.

Instead of ρ we take d as density.

                    [tex]A= 395 m^2\\v_l=295 m/s\\v_u=288m/s\\density=1.21kg/m^3[/tex]

                     [tex]P_1+\frac{1}{2}v_1^2d+dgh_1=P_2+ \frac{1}{2}v_2^2d+dgh_2\\here\\h_1=h_2\\thus\\P_1-P_2=\frac{1}{2}d(v_2^2-v_1^2)\\P_1-P_2=\frac{1}{2}*1.21(288^2-259^2)=9597.12 atm\\[/tex]

                       [tex]\frac{F}{A}=9597.12 atm\\\\ F=9597.12*395m^2=37.9*10^5 N\\\\mg=37.9*10^5 N\\\\m=\frac{37.9*10^5 N}{9.8}\\\\ m=386.8*10^3kg[/tex]

Thus, we can conclude that, the mass of the airplane is 386.8×10^3 kg.

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The mass of an airplane with two wings that are 395m2 in size and average wing surface speeds of 259 and 288 m/s is 387x 10^3 kg.

We need to be aware of the Bernoulli principle in order to determine the solution.

                             P+ [tex]\frac{1}{2}[/tex]dv²+dgh = constant.

We substitute d for to represent density.

                           [tex]v_1=259m/s\\v_2=288m/s\\A=395m^2\\d=1.21kg/m^3[/tex]

                            [tex]P_1+\frac{1}{2}dv_1^2+dgh= P_2+\frac{1}{2}dv_2^2+dgh\\\\P_1+\frac{1}{2}dv_1^2=P_2+\frac{1}{2}dv_2^2\\\\P_1-P_2=\frac{1}{2}d(v_2^2-v_1^2)\\\\P_1-P_2=9597 atm[/tex]

                        [tex]F/A= 9597 atm\\mg=9597*395 =38*10^5N\\m=\frac{38*10^5}{9.8} = 387*10^3kg[/tex]

Consequently, we can say that the mass of an airplane with two wings that are 395m2 in size and average wing surface speeds of 259 and 288 m/s is 387 x 10^3 kg.

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The actual weight of the stone is 11 N. It is based on the Archimedes principles.

Actual weight= weight inside water+ weight of water displaced

= 9N + 2N = 11N

Thus, we can conclude that the actual weight of the object is 11N.

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The tension in the cable is 169.43 N and the vertical component of the force exerted by the hi.nge on the beam is 114.77 N.

Apply the principle of moment and calculate the tension in the cable;

Clockwise torque = TL sinθ

Anticlockwise torque = ¹/₂WL

TL sinθ  =  ¹/₂WL

T sinθ  =  ¹/₂W

T = (W)/(2 sinθ)

T = (29 x 9.8)/(2 x sin57)

T = 169.43 N

T + F = W

F = W - T

F = (9.8 x 29) - 169.43

F = 114.77 N

Thus, the tension in the cable is 169.43 N and the vertical component of the force exerted by the hi.nge on the beam is 114.77 N.

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Answer:

B. Weight

Explanation:

Weight is the force of gravity exerted on a body

OR

It's the force exerted on a body by the influence of the earth's gravitational force

The application of total Internal Reflection occurs in a mirage. Option A

1) We know that a charge can be positive or negative. If the charge is negative, we call it an electron. If the charge is positive, we call it a proton. Now we know that the magnitude of charge on each electron is 1.6×10-19 C. Hence, the magnitude of charge on five electrons = 5 * 1.6×10-19 C = 8.0×10-19 C. Option D

2) The details of question 2 are not shown hence the question can not be answered.

3) Looking at the electronic configuration of the element;  1s22s22p63s23p5, it is clear to see that it has the ns2 np5 outermost configuration that is common to halogens thus it;

4) The atom is composed of the protons, neutrons and electrons. Given the nuclide identified as 230/92U we have  92 protons, 138 neutrons and 92 electrons. Option C

5) The application of total Internal Reflection occurs in a mirage. Option A

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Answer:

Every 15 minutes, remove the cover and check the temperatures of the samples using the three thermometers. Wait 30 seconds before recording the thermometer reading. Once the temperatures of the three samples are no more than a degree apart, record the temperatures.

(three 100-milliliter containers (each with a thermometer) in an ice bath inside an uncovered basin, with one container holding 50 grams of sand, one holding 50 grams of cold water, and one holding 100 grams of cold water)

Next, prepare the hot water.

-Fill each of the three 200-milliliter containers with 100 grams of hot tap water. Measure the mass using the mass scale.

(100 grams of hot water in a 200-milliliter container atop a mass scale)

-Prepare a hot-water bath by boiling water in a pot. Use the heat mitts to pour the hot water from the pot into the second tub. Fill the tub to a depth of about 2 inches. Carefully place the three containers of hot water into the bath without submerging them. Cover and wait for five minutes until the temperatures stabilize.

(three 200 mL containers (each holding 100 grams hot water) in a covered tub containing 2 inches of boiling water)

Prepare to mix the cold samples with hot water:

-Have three empty 300-milliliter mixing containers and three thermometers ready. Timing is important. Uncover the cold-water bath, and pour each cold sample into a different mixing container.

(three 300-milliliter containers (each with a thermometer), one holding 50 grams of sand, one holding 50 grams of cold water, and one holding 100 grams of cold water)

-Uncover the hot-water bath, and use a heat mitt to remove the three containers of hot water from the hot-water bath. Pour 100 grams of hot water into each mixing container.

(three 300 mL containers (each with a thermometer), one holding a 150-gram mixture of sand and hot water, one holding a 150-gram mixture of cold water and hot water, and one holding a 200-gram mixture of cold water and hot water)

Explanation:

Some of the techniques used by astronomers to determine the type of galaxy in which we live are:

This refers to the study of heavenly bodies and space and other things that space is made up of.

Hence, we can see that the reason why it is so difficult to determine the shape of our galaxy is that astronomers can only infer its presence from the motions of stars in the galaxy, and a precise shape is difficult to be determined.

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