I need help!!!!!!!!!!!pleaseeeeeee

I Need Help!!!!!!!!!!!pleaseeeeeee

Answers

Answer 1
A claim is the second one
And the second one is the the 1 and 2

Related Questions

I'm asking for a quick favor. I'm trying to understand an equation that has to do with Projectile motion, Bernoulli's principle, and Magnus Effect. Basically focused on understanding air resistance on a projectile. I would like to discuss this privately rather than have it on this public forum. I'll give you 100 of my points if you help.

Answers

Explanation:

Projectile motion, Bernoulli's principle, and Magnus Effect.

Sure I would be happy to discuss projectile motion!

I'll do it if you mark brainliest :) I need the points thanks

Why might video games increase creativity while the use of cell phones, the internet, or computers do not?

Answers

Explanation:

Video games are developed around a structure that is unique from your usual media, where there is a sense purpose in mind when playing a game, as decided by the player, and it allows them to explore creative options in order to solve scenarios, depending on the genre. In the use of phones, internet, or computers, this structure is more rare and diverse from video games, which does not lean more toward a creative purpose to build from. That idea gives people the inspiration and discover skills they never knew they even had.


Name the four layers of the atmosphere (in order starting at the bottom

Answers

Answer:

Troposphere, stratosphere, mesosphere and thermosphere. The next region is the exosphere, but that region is 500+ km from the Earth's surface.

[tex]\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}[/tex]

Actually Welcome to the Concept of the Atmosphere.

The four layers of Atmosphere starting from bottom are as follows:

1.) Troposphere - The troposphere is the lowest layer of our atmosphere. Starting at ground level, it extends upward to about 10 km (6.2 miles or about 33,000 feet) above sea level.

2.) Stratosphere - The next layer up is called the stratosphere. The stratosphere extends from the top of the troposphere to about 50 km (31 miles) above the ground.

3.) Mesosphere - Above the stratosphere is the mesosphere. It extends upward to a height of about 85 km (53 miles) above our planet. Most meteors burn up in the mesosphere.

4.) Thermosphere - The layer of very rare air above the mesosphere is called the thermosphere. High-energy X-rays and UV radiation from the Sun are absorbed in the thermosphere, raising its temperature to hundreds or at times thousands of degrees.

PLEASE CLICK ON THIS IMAGE I NEED HELP

Answers

Answer:

Second option

Explanation:

"Uniform" pretty much means the same thing happens.

an egg is Free Falling Down from a nest in a tree (neglect air resistance)​

Answers

We are to calculate the acceleration.

Answer:

-9.8 m/s²

Explanation:

Since the egg is in a free fall, it means that the force due to gravity will be equal to the normal force.

Now,

Force due to gravity: F_g = mg

Normal force; F_n = ma

Thus;

mg = ma

m will cancel out to get

g = a

Since it is a free fall motion, then gravity is negative;

-g = a

g has a constant value of 9.8 m/s². Thus;

a = -9.8 m/s²

You pull with a force of 295 N on a rope that is attached to a block of mass 22 kg, and the block slides across the floor at a constant speed of 1.6 m/s. The rope makes an angle of 35 degrees with the horizontal. What is the net force on the block

Answers

Answer:

Fnet = 0

Explanation:

Since the block slides across the floor at constant speed, this means that it's not accelerated.According Newton's 2nd Law, if the acceleration is zero, the net force on the sliding mass must be zero.This means that there must be a friction force opposing to the horizontal component of the applied force, equal in magnitude to it:

       [tex]F_{appx} = F_{app} * cos \theta = 295 N * cos 35 = 242 N (1)[/tex]

In the vertical direction, the block is not accelerated either, so the sum of the normal force and the vertical component of the applied force, must be equal in magnitude to the force of gravity on the block:

      [tex]F_{appy} = F_{app} * cos \theta = 295 N * sin 35 = 169 N (2)[/tex]

⇒    169 N + Fn = Fg = 216 N  (3)

This means that there must be a normal force equal to the difference between Fappy and Fg, as follows:Fn = 216 N - 169 N = 47  N (4)

At an amusement park there are 200-kg bumper cars A, B, and C that have riders with masses of 55 kg, 90 kg, and 42.5 kg respectively. Car A is moving to the right with a velocity vA = 2 m/s and car C has a velocity vC = 1.5 m/s to the left, but car B is initially at rest. The coefficient of restitution between each car is 0.8. Determine the final velocity of each car, after all impacts, assuming car A hits car B before car C does. Assume positive sign denoting forward motion and negative sign denoting backward motion.

Answers

Answer:

Vb = 0.334 m/s

Va = -1.265 m/s

Vc = 1.424 m/s

Explanation:

Favorite Answer

Initial momentum = 255(2) – 242.5(1.5) = 146.25

Final momentum = 255Va + 290Vb + 242.5 Vc = 146.25

Vb - Va = 0.8(2) = 1.6

Vc - Vb = 0.8(1.5) = 1.2

Va = Vb -1.6

Vc = Vb + 1.2

255(Vb -1.6) + 290Vb + 242.5(Vb + 1.2) = 146.25

255 Vb – 408 + 290 Vb + 242.5 Vb + 291 = 146.25

787.5 Vb = 263.25

Vb = 0.334 m/s

Va = Vb -1.6 = 0.334 – 1.6 = -1.265 m/s

Vc = Vb + 1.2 = 0.224 + 1.2 = 1.424 m/s

A wave with a frequency of 5Hz travels a distance of 40mm in 2 seconds.What is the speed of the wave​

Answers

Answer:

20mm per second

Explanation:

Who watching all star draft? Luka better get picked first ong

Answers

Na dame Dolla fo surrrreeeeee

why type of volcano is built almost entirely from ejected lava fragments

Answers

Answer:

Shield volcanoes

Explanation:

a car moved 120km to the north. what is its displacement?

Answers

120Km. They are the same

A 7.80 g bullet has a speed of 620 m/s when it hits a target, causing the target to move 6.30 cm in the direction of the bullet's velocity before stopping. (a) Use work and energy considerations to find the average force (in N) that stops the bullet. (Enter the magnitude.) N (b) Assuming the force is constant, determine how much time elapses (in s) between the moment the bullet strikes the target and the moment it stops moving. s

Answers

Answer:

[tex]23796.19\ \text{N}[/tex]

[tex]0.0002032\ \text{s}[/tex]

Explanation:

F = Force

s = Displacement = 6.3 cm

m = Mass of bullet = 7.8 g

v = Velocity of bullet = 620 m/s

t = Time taken

Work done is given by

[tex]W=Fs[/tex]

Kinetic energy is given by

[tex]K=\dfrac{1}{2}mv^2[/tex]

Using work energy considerations we get

[tex]Fs=\dfrac{1}{2}mv^2\\\Rightarrow F=\dfrac{1}{2s}mv^2\\\Rightarrow F=\dfrac{1}{2\times 0.063}\times 7.8\times 10^{-3}\times 620^2\\\Rightarrow F=23796.19\ \text{N}[/tex]

The average force that stops the bullet is [tex]23796.19\ \text{N}[/tex].

Force is given by

[tex]F=m\dfrac{v-u}{t}\\\Rightarrow t=m\dfrac{v-u}{F}\\\Rightarrow t=7.8\times 10^{-3}\times \dfrac{620}{23796.19}\\\Rightarrow t=0.0002032\ \text{s}[/tex]

The time taken to stop the bullet is [tex]0.0002032\ \text{s}[/tex].

A physics student sits in a chair. The chair pushes up on the student's body. Identify the other force of the interaction force pair.

Answers

Answer:

The other force is the weight of the student.

Explanation:

With respect to Newton's third law of motion, for the student to sit and balance on the chair, there must be two equal and opposite forces involved. The student applies his/ her weight on the chair which acts downwards, while the chair applies an equal but opposite force to the weight of the student.

The force applied by the chair on the student's body is counter balanced by the student's weight. Note that, if the weight of the student is greater than the opposing force from the chair, the chair would collapse.

The earth rotates through one complete revolution every 24 hours. Since the axis of rotation is perpendicular to the equator, you can think of a person standing on the equator as standing on the edge of a disc that is rotating through one complete revolution every 24 hours. Find the angular and linear velocity of a person standing on the equator. The radius of earth is approximately 4000 miles.

Answers

Answer:

ω = 7.27 x 10⁻⁵ rad/s

v = 467.99 m/s

Explanation:

First, we will find the angular velocity of the person:

[tex]Angular\ Velocity = \omega = \frac{Angular\ Distance}{Time}[/tex]

Angular distance covered = 1 rotation = 2π radians

Time = (24 h)(3600 s/ 1 h) = 86400 s

Therefore,

[tex]\omega = \frac{2\pi\ rad}{86400\ s}[/tex]

ω = 7.27 x 10⁻⁵ rad/s

Now, for the linear velocity:

[tex]v = r\omega[/tex]

where,

v = linear velocity = ?

r = radius of earth = (4000 miles)(1609.34 m/1 mile) = 6437360 m

Therefore,

[tex]v = (6437360\ m)(7.27\ x\ 10^{-5}\ rad/s)[/tex]

v = 467.99 m/s

The chemical equation for the decomposition of potassium chlorate into potassium chloride and oxygen gas is
KCIO: |_ KCI + ___ 02
Which coefficients correctly balances the equation?
A) 4.4,3
B 3,3,2
C) 2,2,3
D
2, 1,3
E The equation is already balanced.

Answers

Answer:

Option C. 2, 2, 3

Explanation:

__KClO₃ —> __ KCl + __O₂

The above equation can be balance as illustrated below:

KClO₃ —> KCl + O₂

There are 2 atoms of O on the right side and 3 atoms on the left side. It can be balance by writing 2 before KClO₃ and 3 before O₂ as shown below:

2KClO₃ —> KCl + 3O₂

There are 2 atoms of K on the left side and 1 atom on the right side. It can be balance by writing 2 before KCl as shown below:

2KClO₃ —> 2KCl + 3O₂

Thus, the equation is balanced. The coefficients are: 2, 2, 3

A vertical piston cylinder assembly contains 10.0kg of a saturated liquid-vapor water mixture with initial quality of 0.85 The water receives energy by heat transfer until the temperature reaches 320*C. The piston has a mass of 204kg and area of 0.005m2. Atmospheric pressure of 100kPa acts on the top side of the piston. Local gravitational acceleration is 9.81m/s2 Calculate the amount of heat transfer between the water and the surroundings in kJ. Enter a numeric value only. 6735.66

Answers

Answer:

Explanation:

From the given information:

At state 1:

Initial Quality [tex]= x_1 = 0.85[/tex]

mass = 10.0 kg

At state 2:

Temperature [tex]T_2 = 320^0[/tex]

mass of the piston [tex]m_p = 204 \ kg[/tex]

area of the piston [tex]A_p = 0.00 5 \ m^2[/tex]

Atmospheric pressure [tex]P_{atm}= 100 \ kPa = 100 \times 10^3 \ Pa[/tex]

Gravitational acceleration = 9.81 m/s²

[tex]\mathbf{P= P_1=P_2}[/tex], This is because there exists no restriction to the movement of the piston and provided the process is frictionless. So, the process 1-2 is regarded as constant.

To calculate the applying force balance over the piston by using force balance in the vertical direction:

[tex]\mathbf{P_{AP} = P_{atmA_p} + m_pg}[/tex]

(100 × 10³)×0.005 + 204 × 9.31 = P × 0.05

P = 500248 Pa

P = 500.25 kPa

At state 1:

[tex]\mathbf{P_1 = P = 500.25 \ kPa}[/tex]

[tex]x_1 = 0.85[/tex]

Hence, this is a saturated mixture of liquid and vapor

Using the steam tables at 500.25 kPa

[tex]V_f = 1.093 \times 10^{-3} \ m^3/kg \\ \\ V_g = 0.375 \ m^3/kg \\ \\ U_f = 639.72 \ kJ/kg \\ \\ U_g = 2560.72 \ kJ/kg[/tex]

Specific volume at state 1 is given as:

[tex]V_1 = [ V_f +x_1(v_g -v_f) ] \ at \ 500.25 \ kPa \\ \\ V_1 = 0.319 \ m^3/kg[/tex]

volume at state 1 is given by:

[tex]V_1 = mV_1 = 10 \times 0.319 \\ \\ V_1 = 3.19 \ m^3[/tex]

Similarly, the specific internal energy is:

[tex]U_1 = [U_f +x_1 (U_o-Uf)] \ at \ 500.25 \ kPa[/tex]

[tex]U_1 = 639.72 +0.82 (2560.72 -639.72)[/tex]

[tex]U_1 = 2272.57 \ kJ/kg[/tex]

At state 2:

[tex]P = P_1 = P_2 = 500.25 \ kPa \\ \\ T_2 = 320^0 \ C[/tex]

Using steam tables at P = 500.25 kPa and T = 320° C

[tex]V_2 = 0.541 \ m^3/kg \\ \\ U_2 = 2835.08 \ kJ/kg[/tex]

[tex]V_2 = mV-2 = 10 \times V_2 = 5.41 \ m^3[/tex]

[tex]\text{Now; Applying the 1st law of thermodynamics to the system}[/tex]

[tex]_1Q_2 -_1W_2 = \Delta V =m(u_2-u_1) \\ \\ where;\ _1W_2 = P(V_2-V_1) \\ \\ _1Q_2 -P(V_2-V_1) = m(u_2-u_1) \\ \\ _1Q_2 - 500.25(5.91 -3.19) = 10( 2835.08 -2272.57) \\ \\ \mathbf{ _1Q_2 = 6735.66 \ kJ}[/tex]

4. When you are holding a book, energy is stored between the book and the Earth.
This type of energy is called
potential energy.
A. Elastic potential energy
B. Chemical potential energy
C. Gravitational potential energy
D. Kinetic energy

Answers

Answer:

gravitational potential energy

Pleaseee I need help!!

Answers

Answer:

R1=3.333 Ohms

R2=10 Ohms

R3=16.666 Ohms

Explanation:

30 total

30=R1+R2R3

30=1(x):3(x):5(x)

x=3.3333

R1=3.333 Ohms

R2=10 Ohms

R3=16.666 Ohms

Neuroticism, psychoticism, and extraversion are the three dominant personality traits according to
A. Paul Costa
B. Hans Eysenck
C. Gordon Allport
D. Robert McCrae

Answers

The answer is b I just took the test

Neuroticism, psychoticism, and extraversion are the three dominant personality traits according to Hans Eysenck.

What is Hans Eysenck's theory?

Eysenck's theory of personality is based on three logical attributes namely:  

a) Introversion vs. extroversion Extroversion leads to sociable life , while introversion causes need of being alone and limited interactions

b) Neuroticism vs. stability – Neuroticism leads to anxiousness and an overactive sympathetic nervous system while stability leads to emotional stability.

c) Nsychoticism vs. socialization. - psychoticism  leads to independent thinking, and hostility. While socialization leads to co-operative and conventional behaviour.  

Therefore Neuroticism, psychoticism, and extraversion are the three dominant personality traits according to Hans Eysenck.

To know more about Hans Eysenck's theory follow

https://brainly.com/question/11912668

#SPJ2

A 50 kg mass is sitting on a frictionless surface. An unknown constant force called force A pushes the mass for 2 seconds until the mass reaches a velocity of 3 m/s. If the 50 kg mass is now pushed by an unknown force B and reaches the velocity of 3 m/s in 4 seconds, compare the impulse delivered to the mass when acted upon by force A with the impulse delivered to the mass when acted on by force B? *

A) The impulse delivered to the mass when acted upon by force A is greater
B) The impulse delivered to the mass when acted upon by force B is greater
C) The impulse is the same in each case
D) We need to know the value of force A and force B in order to determine this

Answers

Answer:

aaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

A 20g bullet moving at 200m/s hits a bag of sand and comes to rest in 0.011s, calculate the momentum of the bullet just before hitting the bag​

Answers

Answer:

momentum = mass * velocity ... kg•m/s

acceleration = v / t = 200 m/s / 11 ms ... m/s^2

force = mass * acceleration = .02 kg * (200 m/s / 11 ms) N

Explanation:

A tank is is half full of oil that has a density of 900 kg/m3. Find the work W required to pump the oil out of the spout. (Use 9.8 m/s2 for g. Assume r = 15 m and h = 5 m.) W = 1.59 J

Answers

Answer:

3.9 × 10^7 J

Explanation:

Given that a tank is is half full of oil that has a density of 900 kg/m3. Find the work W required to pump the oil out of the spout. (Use 9.8 m/s2 for g. Assume r = 15 m and h = 5 m.) W = 1.59 J

Solution

Since the tank is half full, the height = 2.5m

Pressure = density × gravity × height

Pressure = 900 × 9.8 × 2.5

Pressure = 22050 Pascal

The cross sectional area of the pump will be area of a circle.

A = πr^2

A = π × 15^2

A = 706.858 m^2

Using the formula

Density = mass/volume

Mass = density × volume

Mass = 900 × 706.86 × 2.5

Mass = 1590.435

Energy = mgh

Energy = 1590.435 × 9.8 × 2.5

Energy = 38965657.8 J

Since the work done = energy

Therefore, the work done = 3.9 × 10^7 J

****PLEASE HELP**** WILL MARK BRAINLIEST
Assuming that voltage remains constant, what happens to the current in a
wire if the length of the wire increases?
O A. The current decreases.
OB. The current alternates between high and low values.
O C. The current increases.
O D. The current is not affected by a change in wire length.

Answers

Answer:

The Current decreases

Explanation:

HOPE THIS HELPS!

3. If a spring extends by 3 cm when a 4 N weight is suspended from it, find the extension
when the weight is changed to
(a) 8 N
(b) 10 N
(c) 14 N

Answers

8N - 6cm
10N - 7.5cm
14N - 10.5cm

Which word best completes the sentence?

Select the word from the drop-down menu

He is quite
Choose...
despite never having left his smalL TOWEN

Answers

Answer:

it’s cosmopolitan

Explanation:

k12

We intend to measure the open-loop gain (LaTeX: A_{open}A o p e n ) of an actual operational amplifier. The magnitude of LaTeX: A_{open}A o p e n is in the range of 106 V/V. However, the signal generator in measurement setup can supply minimal voltage of 1 mV, and the oscilloscope used at amplifier output can measure maximal voltage level of 10 V. Can you design a simple measurement setup using this signal generator and oscilloscope, and accurately measure the LaTeX: A_{open}A o p e n

Answers

Answer:

voltage divider,  R₂ = 1000 R₁

measuring the output in the resistance R₁

Explanation:

Let's analyze the situation, in an op amp in open gain loop, the gain is maximum G = 10⁶ V / V

in this case the signal generator gives a minimum wave of 1 10⁻³ V, after passing through the amplified it becomes 10³ V which saturates the oscilloscope.

To solve this problem we must use a simple voltage divider, for this we use the fact that in a series circuit the voltage is the sum of the voltages of each element.

If we use two resistors whose relationship is

            R₂ / R₁ = 10³

            R₂ = 1000 R₁

When measuring the output in the resistance R₁ we have the desired divider, with a tolerance range, for the minimum output of the generator (1 10⁻³V) we have a reading of V = 1 V in the oscilloscope, for which we can use voltage up to 10V on the generator

An object which is dropped from a certain height has zero (0) initial velocity.

Answers

Answer:

0 m/s

Explanation:

if an object is dropped we know the initial velocity is zero when in free fall

When individuals improve their aerobic endurance, which body systems are affected?

Answers

When individuals improve their aerobic endurance, which body systems are affected?

Anwser: your heart and lungs

A cyclist exerts a 15.0 N force while riding 251 m in 30.0 s. What power does the cyclist develop?

Answers

Answer:

P=126W

Explanation:

Sorry if im wrong!

Answer:

125.5 watts

Explanation:

P=work/time

work=F*d

P=(F*d)/t

P=(15*251)/30

P=125.5 watts

A 125 kg mail bag hangs by a vertical rope 3.3 m long. A postal worker then displaces the bag to a position 2.2 m sideways from its original position, always keeping the rope taut.
1) What horizontal force is necessary to hold the bag in the new position?
2) As the bag is moved to this position, how much work is done by the rope?
3) As the bag is moved to this position, how much work is done by the worker?

Answers

Answer:

1) the required horizontal force F is 1095.6 N

2) W = 0 J { work done by rope will be 0 since tension perpendicular }

3) work is done by the worker is 1029.4 J

Explanation:

Given that;

mass of bag m = 125 kg

length of rope [tex]l[/tex] = 3.3 m

displacement of bag d = 2.2 m

1) What horizontal force is necessary to hold the bag in the new position?

from the figure below; ( triangle )

SOH CAH TOA

sin = opp / hyp

sin[tex]\theta[/tex] = d / [tex]l[/tex]

sin[tex]\theta[/tex] = 2.2/ 3.3

sin[tex]\theta[/tex] = 0.6666

[tex]\theta[/tex] = sin⁻¹ ( 0.6666 )

[tex]\theta[/tex]  = 41.81°

Now, tension in the string is resolved into components as illustrated in the image below;

Tsin[tex]\theta[/tex] = F  

Tcos[tex]\theta[/tex] = mg

so

Tsin[tex]\theta[/tex] / Tcos[tex]\theta[/tex] = F / mg

sin[tex]\theta[/tex] / cos[tex]\theta[/tex] = F / mg

we know that; tangent = sine/cosine

so

tan[tex]\theta[/tex] = F / mg

F = mg tan[tex]\theta[/tex]

we substitute

Horizontal force F = (125kg)( 9.8 m/s²) tan( 41.81° )

F = 1225 × 0.8944

F = 1095.6 N

Therefore, the required horizontal force F is 1095.6 N

2)  As the bag is moved to this position, how much work is done by the rope?

Tension in the rope and displacement of mass are perpendicular,

so, work done will be;

W = Tdcos90°

W = Td × 0

W = 0 J { work done by rope will be 0 since tension perpendicular }

3) As the bag is moved to this position, how much work is done by the worker

from the diagram in the image below;

SOH CAH TOA

cos = adj / hyp

cos[tex]\theta[/tex]  = ([tex]l[/tex] - h) / [tex]l[/tex]

we substitute

cos[tex]\theta[/tex]  = ([tex]l[/tex] - h) / [tex]l[/tex]  = 1 - h/[tex]l[/tex]

cos[tex]\theta[/tex] = 1 - h/[tex]l[/tex]

h/[tex]l[/tex] = 1 - cos[tex]\theta[/tex]

h = [tex]l[/tex]( 1 - cos[tex]\theta[/tex] )

now, work done by the worker against gravity will be;

W = mgh = mf[tex]l[/tex]( 1 - cos[tex]\theta[/tex] )

W = mf[tex]l[/tex]( 1 - cos[tex]\theta[/tex] )

we substitute

W = (125 kg)((9.8 m/s²)(3.3 m)( 1 - cos41.81° )

W = 4042.5 × ( 1 - 0.745359 )

W = 4042.5 × 0.254641

W = 1029.4 J

Therefore,  work is done by the worker is 1029.4 J

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