Plzz answer correctly

Thus, the distance traveled by automobile is [tex]=1.31m[/tex]

The mass of the person is [tex]m=65kg[/tex]

The deceleration of the automobile is [tex]a=30g[/tex]

The acceleration due to gravity is [tex]g=9.81m/s.[/tex]

The initial velocity of the automobile is [tex]v_{i}=100 km/h*\frac{1m/s}{3.6km/h} =27.78m/s[/tex]

Determination of the force exerted on the person:

The force applied to the person can be calculated using Newton's second law as follows:

[tex]F=ma[/tex]

[tex]30g[/tex]Here, m denotes the person's mass, while a denotes the car's acceleration. Because the driver is moving at 30g, the car's acceleration is assumed to be positive.

Substitute the values as 70 kg for [tex]m,(30g)[/tex] for a, and[tex]9.81m/s^{2}[/tex]

for g in the above equation.

[tex]F=70kg*30g[/tex]

   [tex]=70kg*30g*9.81m/s^{2} (\frac{1N}{1kg*m/s^{2} })[/tex]

   [tex]=20,601N[/tex]

Thus, the force acting on the person is [tex]=20,601N[/tex]

Determination of the distance traveled by automobile:

The final velocity of the car can be calculated using Newton's third law and represented as follows:

[tex]v_{f} ^{2}=v_{i} ^{2}+2ad[/tex]

[tex]d=\frac{v_{f} ^{2}=v_{i} ^{2}}{2a}[/tex]

Here [tex]v_{f}[/tex] is the final velocity of the automobile. The final velocity of the automobile becomes 0 m/s as the automobile comes to rest. The acceleration is taken to be negative because the automobile is decelerating at [tex]30g[/tex]

Substitute the values as [tex]0 m/s[/tex] for[tex]v_{f} =27.78m/s[/tex], for[tex]v_{i} ,(-30g)[/tex],  for a, and  [tex]9.81m/s^{2}[/tex]for g in the above equation.

[tex]d=\frac{(0 m/s)^{2} - (27.78m/s)^{2} }{2*(-30)*9.81m/s^{2} }[/tex]

[tex]=1.31m[/tex]

Thus, the distance traveled by automobile is[tex]=1.31m[/tex]

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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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The period of M2, in years. is mathematically given as

T2= 134.3968years

M1 = 4.00 kg

M2 = 1.00 kg

T1 = 34.0 years.

Generally, the equation for is  mathematically given as

T2 = T1 (a2/a1)3/2

T2=  34.0  * (5/2)^{3/2}

T2= 134.3968years

In conclusion, the period of M2, in years

T2= 134.3968years

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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:

The density of the planet is determined as 1,974.26 kg/m³.

√(⁴/₃πGρ) = 2π/(2.35 x 3600)

where;

√(⁴/₃πGρ) = 2π/(2.35 x 3600)

√(⁴/₃πGρ) = 2π/8460

(⁴/₃πGρ)  = (2π/8460)²

⁴/₃πGρ = 4π²/(8460)²

ρ = 12π/(8460² x 4G)

ρ = (12π) / (8460² x 4 x 6.67 x 10⁻¹¹)

ρ = 1,974.26 kg/m³

Thus, the density of the planet is determined as 1,974.26 kg/m³.

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The resultant displacement of the car can be determined using Pythagorean theorem. The magnitude of car's displacement here is 47.3 Km.

Displacement is a vector quantity measuring the change in position of an object. It can be used in short  term distances. Displacement have both magnitude and direction.

Given , distance to travelled to east = 15 km

to the north = 45 km

then,  the two points if connected will form a triangle where we can use the Pythagorean theorem to find the resultant vector.

The resultant displacement = √(45² + 15 ²) = 47.3 km

Therefore, the displacement of the car is 47.3 km.

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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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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 required horsepower engine is 32 horsepower.

The force of the 30 skiers is calculated as follows:

Mass of the skiers = 30 * 67kg = 2010 kg

Net force acting on the skiers along the x-axis

where f is the frictional force

The kinetic frictional force, f = μN

where

μ = The coefficient of the kinetic friction

N = normal reaction

Net force acting on the skiers along y axis, the

Fy = ma

N = mg cos θ

Substituting for N above

f = μk mg cos θ

Substituting for f in  (1)

F = mg sin θ + μk mg cos θ

F = mg(sinθ + μk cos θ)

Work done by the engine in pilling up the skiers, W = Fx

W = mg ( sinθ + μk cos θ)x

x = 300 m

W = (2010 kg) (9.81 m/s²) (sin 23° + (0.10) cos 23°) (300 m)

Work done, W = 2.86 * 10⁶ J

Time taken, t = 2.0 * 60sec = 120 s

1 horsepower = 746 W

Power = 2.86 * 10⁶/ 120 * 1/746

Power = 31.9 horsepower

In conclusion, the power of the engine is the ratio of the work done and time taken.

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(a) The work done by the applied force is 26.65 J.

(b) The work done by the normal force exerted by the table is 0.

(c) The work done by the force of gravity is 0.

(d) The work done by the net force on the block is 26.65 J.

W = Fdcosθ

W = 14 x 2.1 x cos25

W = 26.65 J

W = Fₙd

W = mg cosθ x d

W = (2.5 x 9.8) x cos(90) x 2.1

W = 0 J

The work done by force of gravity is also zero, since the weight is at 90⁰ to the displacement.

∑W = 0 + 26.65 J = 26.65 J

Thus, the work done by the applied force is 26.65 J.

The work done by the normal force exerted by the table is 0.

The work done by the force of gravity is 0.

The work done by the net force on the block is 26.65 J.

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A. The acceleration of the ball while it is in flight?

magnitude is 0 m/s² (magnitude is zero)

B. The velocity of the ball when it reaches its maximum height is 0 m/s (magnitude is zero)

C. The initial velocity of the ball 8.036 m/s upward

D. The maximum height reached by the ball is 3.29 m

Considering that the ball came to rest after 1.64s, it means the entire acceleration of the flight is zero as the ball was not moving in any form again.

At maximum height, the velocity of the ball is zero as it no longer has magnitude to keep going upwards. Hence the ball begins to ball down.

v = u - gt (since the ball is going against gravity)

0 = u - (9.8 × 0.82)

0 = u - 8.036

Collect like terms

u = 0 + 8.036

u = 8.036 m/s upward

h = ½gt²

h = ½ × 9.8 × 0.82²

h = 3.29 m

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Rank from largest to smallest, To rank items as equivalent, overlap them is Pd>Pa>Pb>Pc.

What is quantum particles ?

A quantum particle has a complex wave function that is square-integrable and has the form (x1,... xN), where ||2 is the probability density of finding N particles at positions x1,x2,... xN. These places are taken to be inside a square box of size d, with side length L, periodic boundary conditions, and position V.

What is potential energy ?

Potential energy is a form of energy that can be stored but is affected by the way a system's component elements are assembled.  A steel ball is more energetic when it is raised above the ground than when it is lowered.

Therefore, rank from largest to smallest, To rank items as equivalent, overlap them is Pd>Pa>Pb>Pc.

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The speed of the boy and his friend at the bottom of the slope is 16.52 m/s.

Apply the principle of conservation of energy,

E(up) - E(friction) = E(bottom)

mg sin(15) + ¹/₂(M + m)u² - μ(M + m)cos 15 = ¹/₂(M + m)v²

[tex]v = \sqrt{2[\frac{mgd \ sin15 \ + \frac{1}{2}(M + m)u^2 \ -\mu (M + m)g cos\ 15 }{M + m}] }[/tex]

where;

u = √2gh

u = √(2gL sin15)

u = √(2 x 9.8 x 28 x sin 15)

u = 11.92 m/s

[tex]v = \sqrt{2[\frac{(30)(9.8)(70) \ sin15 \ + \frac{1}{2}(30 + 50)(11.92)^2 \ - 0.12 (30 + 50)9.8 cos\ 15 }{30 + 50}] }\\\\v = 16.52 \ m/s[/tex]

Thus, the speed of the boy and his friend at the bottom of the slope is 16.52 m/s.

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The bigger object will move some distance because the initial and final momentum of the colliding particles are not zero.

The term momentum has to do with the product of mass and velocity. We know that during a collision, momentum is conserved. This implies that the momentum before collision is equal to the momentum after collision. Thus, the total momentum of the system is constant.

Given the fact that the initial and final momentum of the colliding particles are not zero, the  the big object is not going to stop immediately but must move some distance towards the smaller object.

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According to mathematics, the planet's angle is stated as

dY=704 degrees.

We may demonstrate this by using Kepler's third law, which asserts that a planet's orbit squared is a function of cubed radius.

The equation for the period is often expressed numerically as

[tex](periodX / periodY)^2 = (radius X / radius Y)^3[/tex]

Therefore

(pX / pY)^2 = 4^3

(pX / pY)^2 = 64

[tex]\sqrt{(pX / pY )^2}= \sqrt{64}[/tex]

pX / pY=8

In conclusion, planet Y travels 8 times further than planet X does in the same amount of time since one orbit on planet X takes 8 times longer to complete.

planet Y travels ;

dY=8 * 88.0

dY= 704 degrees

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Answer:

Yes.

Explanation:

When you move your hand from the cold water to the “warmer” (room temp) water, one hand feels warm.

As you move your hand from the warm water to the “colder” (room temp) water, that hand feels colder.

Although both hands experience the last bowl of water at the same temperature, your brain senses two

separate sensations. So the water feels “warm” or “cold” relative to the water your hand was in previously.

The greater the difference in temperature, the easier it is to sense a difference.

The kind of equation that can be used to differentiate the kind of separatrix that shows change on motion is

H = 2g/l.

A simple pendulum can be defined as the equipment that displays an oscillatory motion when a mass is tied on a rope and is suspended from it.

The various movements that occur using a simple pendulum is translational ( side to side) or continuous circle (oscillatory motion).

The equation that show that a change from one type of motion to another is H = 2g/l.

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Answer:

Wc = 7.84    weight of crown

Ww = 7.84 - 6.86 = .98       weight of water displaced

Density = 7.84 / .98 = 8     crown is 8 X that of water

Since gold has a density of 19.3 that of water the crown is certainly not 100 percent (if any) gold  

The power of the motor in watts is 4000 W.

Power (P) = Work (W) / Time (t)

P = 4000 J / 1 min

P = 4000 J/s

1 J/s = 1 W

P = 4000 W

What is power?

Power is a measure of energy transfer over time. In physics, it is the rate of energy transfer, usually measured in Watts (W) or Joules per second (J/s). Power is the rate at which work is done or energy is converted from one form to another. It can be calculated by multiplying the force applied to an object by the speed at which it moves. Power is related to work, as work is equal to the amount of energy transferred, divided by time. Power is an important concept in physics as it is used to calculate the energy required to perform a task.

Therefore, The power of the motor in watts is 4000 W.

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Answer: A treasure chest full of silver and gold coins is being lifted from a pirate ship to the shore using two ropes as shown in the figure. The mass of the treasure chest is 75.6 kg. The tension in Rope A is 7.42x10^2 N, and the tension Rope B carries is 7.52x10^2 N. Then, the tension in rope C is 376N

Explanation: To find the correct answer, we have to know more about the Basic forces that acts upon a body.

                 [tex]T_A=7.42*10^2N\\T_B=7.54*10^2N[/tex]

                 [tex]T_C=T_B sin30\\thus,\\T_C=7.52*10^2*0.5=376N[/tex].

Thus, we can conclude that the tension in the rope c will be 376N.

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376N is the tension in rope C.

In order to determine the right response, we must have a better understanding of the fundamental forces that affect a body.

                              [tex]T_c=T_Bsin30\\T_c=7.52*10*2*0.5=376N[/tex]

Thus, we can infer that the rope's c tension will be 376N.

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Answer: A train traveling initially at 16 m/s is under constant acceleration of 2 m/2. At a distance of 720m it will travel in 20 s, and the final velocity will be 56m/s.

Explanation: To find the answer, we need to know about uniformly accelerated motion.

                     [tex]u=16m/s\\a=2 m/s^2\\t=20s[/tex]

                      [tex]S=ut+\frac{1}{2}at^2[/tex]

                      [tex]S=(16*20)+\frac{2*20^2}{2} =720 m.[/tex]

                    [tex]v^2=u^2+2aS\\thus,\\v=\sqrt{u^2+2aS} =\sqrt{16^2+(2*2*720)} =56 m/s.[/tex]

Thus, we can conclude that, a train traveling initially at 16 m/s is under constant acceleration of 2 m/2. At a distance of 720m it will travel in 20 s, and the final velocity will be 56m/s.

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A train experiencing constant acceleration of 2 m/s^2 is moving at an initial speed of 16 m/s. It will go 720 meters in 20 seconds, with a final velocity of 56 m/s.

Understanding uniformly accelerated motion is necessary in order to determine the solution.

                         [tex]S=ut+\frac{1}{2}at^2 \\S=720m\\where,\\u=16m/s, a=2m/s^2,t=20s[/tex]

                          [tex]v^2-u^2=2aS\\v=\sqrt{u^2+2aS} \\v=56m/s[/tex]

Thus, we may say that a train moving at 16 m/s initially experiences constant acceleration of 2 m/2. It will go 720 meters in 20 seconds, with a final velocity of 56 meters per second.

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The magnitude of the acceleration of block m2 is 0.5 m/s² and the tension in the cord if the surface is frictionless is 0.02 N.

The acceleration of bock m2 is calculated from Net force exerted by the pulley.

F - T = m2a

F - m1a = m2a

F = m2a + m1a

F = a(m2 + m1)

a = F/(m1 + m2)

a = (0.03) / (0.04 + 0.02)

a = 0.5 m/s²

T = m1a

T = 0.04 x 0.5

T = 0.02 N

Thus, the magnitude of the acceleration of block m2 is 0.5 m/s² and the tension in the cord if the surface is frictionless is 0.02 N.

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Answer:

See below

Explanation:

To change C to F°

  F = 9/5 C   + 32

     = 9/5 ( 45) + 32 = 113° F

To change F to C°   <= you could use the same equation...or re-earrange to:

C = 5/9 (F-32)

   = 5/9 ( 98.6 - 32) = 37° C

Similarly

100 C = 212 ° F

 to change C °  to  K    add 273.15   =  373.15 K    

A wave in which the particles of the medium move in the same direction as the wave is a longitudinal wave.

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The speed of the satellite in a circular orbit around the Earth is 1.32 x 10⁵ m/s.

v = √(GM/r)

where;

v = √(6.67 x 10⁻¹¹ x 5.98 x 10²⁴/3.57 x 6.37x 10³)

v = 1.32 x 10⁵ m/s

Thus, the speed of the satellite in a circular orbit around the Earth is 1.32 x 10⁵ m/s.

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(a) The  acceleration of the ball while it is in flight has a magnitude of 9.81 m/s2 in downward direction.

(b) The velocity of the ball when it reaches its maximum height is zero.

(c) The initial velocity of the ball is 17.36 m/s.

(d) The maximum height it reaches is 15.36 m.

The acceleration of the ball while it is in flight has a magnitude of 9.81 m/s2 in downward direction.

The velocity of the ball decreases as the ball moves upwards and eventually becomes zero at maximum height.

v = u - gt

at maximum height, final velocity, v = 0

0 = u - gt

u = gt

u = 9.81 x 1.77

u = 17.36 m/s

h = ut - ¹/₂gt

h = 17.36(1.77) - ¹/₂(9.81)(1.77²)

h = 15.36 m

Thus, the  acceleration of the ball while it is in flight has a magnitude of 9.81 m/s2 in downward direction.

The velocity of the ball when it reaches its maximum height is zero.

The initial velocity of the ball is 17.36 m/s.

The maximum height it reaches is 15.36 m.

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The final temperature of the system will be equal to the initial temperature, and which is 373K. The work done by the system is 409.8R Joules.

To find the answer, we need to know about the thermodynamic processes.

                                [tex]dU=dQ-dW[/tex]

                                  [tex]dU=0[/tex],

                             [tex]dU=C_pdT=0\\\\thus, dT=0\\\\or , T=constant\\\\i.e, T_1=T_2[/tex]

                          [tex]T_1=T_2=100^0C=373K[/tex]

                               [tex]W=RT*ln(\frac{V_2}{V_1} )=373R*ln(\frac{4}{\frac{4}{3} })\\ \\W=409.8R J[/tex]

Where, R is the universal gas constant.

Thus, we can conclude that, the final temperature of the system will be equal to the initial temperature, and which is 373K. The work done by the system is 409.8R Joules.

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The system's final temperature will be 373K, which is the same as its starting temperature. The system exerts 409.8R Joules of work.

We need to understand the thermodynamic processes in order to locate the solution.

                    [tex]dU=dQ-dW[/tex]     , It is 0 here.

                  [tex]dU=C_pdT=0\\dT=0\\T=constant\\T_1=T_2=373K[/tex]

As a result, the system's final temperature will be equal to its starting temperature.

                  [tex]W=RT ln(\frac{V_2}{V_1} )=373R*ln(3 )\\W=409.8R Joules[/tex]

R is the gaseous universal constant.

Thus, we can draw the conclusion that the system's end temperature will be 373K, the same as its starting temperature. The system exerts 409.8R Joules of work.

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The work done during expansion is 2.73×10³J and the heat transferred to the surrounding is -2.73×10³ J

The thermodynamic process which takes place in constant temperature of the system is called isothermal process.

n= number of moles

R= universal gas constant

V2= volume of the system after expansion

V1= volume of the system before expansion

= 2.73×10³ J.

Thus, we can conclude that a) the amount of work done is 2.73×10³J and b) the amount of heat transferred to the surrounding is -(2.73×10³J).

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Answer:

pictures please

Explanation:

I need a picture so I can tell you