An object is moving with constant non-zero velocity. Which of thw following statements about it must be true

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

The answer is b I just took the test

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

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

Answer:

Second option

Explanation:

"Uniform" pretty much means the same thing happens.

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

Answer:

1237 J

Explanation:

The work done by the hose on the balloon is the work done by a spring which is

W = 1/2k(x₀² - x₁²) where k = spring constant = 112 N/m, x₀ = 4.70 m and x₁ = 0 m.

Substituting the values of the variables into the equation, we have

W = 1/2k(x₀² - x₁²)

W = 1/2 × 112 N/m((4.70 m)² - (0 m)²)

W = 56 N/m(22.09 m² - 0 m²)

W = 56 N/m(22.09 m²)

W = 1237.04 J

W ≅ 1237 J

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.

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]

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

Answer:

0 m/s

Explanation:

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

Answer:

"Cannot be determined," if that's an answer choice. It depends on the velocities of both objects, since momentum=mass*velocity.

Explanation:

Answer:

The Current decreases

Explanation:

HOPE THIS HELPS!

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.

Answer:

(a) T = 18309.63 N = 18.31 KN

(b) T = 16314.03 N = 16.314 KN

Explanation:

(a)

The tension in an elevator while moving upward with some acceleration is given by the following formula:

[tex]T = m(g+a)\\[/tex]

where,

T = Tension = ?

m = mass = 1663 kg

g = acceleration due to gravity = 9.81 m/s²

a = acceleration of elevator = 1.2 m/s²

Therefore,

[tex]T = (1663\ kg)(9.81\ m/s^2 + 1.2\ m/s^2)\\[/tex]

T = 18309.63 N = 18.31 KN

(b)

Constant velocity means no acceleration. So, in that case, the tension will be equal to the weight of the elevator:

[tex]T = mg\\T = (1663\ kg)(9.81\ m/s^2)\\[/tex]

T = 16314.03 N = 16.314 KN

Answer:

gravitational potential energy

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

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

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.

Answer:

There isnt enough in your question to answer the question bro, like we need a picture or something bro.

Explanation:

You don't have a image attached

Answer:

2km

Explanation:

Given data

We are told that the direction traveled are

North>>>East>>>South

Hence the displacement is defined as the distance away from the initial position is

Initial position =18km

FInal position = 16km

The displacement = 18-16= 2km

Hence the displacement is 2km

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:

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

Anwser: your heart and lungs

Answer:

Fnet = 0

Explanation:

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

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

⇒    169 N + Fn = Fg = 216 N  (3)

Answer:

a).[tex]$3.99 \times 10^5 \ v/m$[/tex]

b). 2.9925 kV

Explanation:

Given :

For mass spectrometer

The magnetic field = B

B = 0.105 T

a).  Given speed, v = [tex]$3.8 \times 10^6 \ m/s$[/tex]

We known

[tex]$\frac{E}{B}=v$[/tex]

∴ [tex]$E= 3.8 \times 10^6 \times 0.105$[/tex]

      [tex]$=3.99 \times 10^5 \ v/m$[/tex]

b). Now spectrometer, d = 0.75 cm

                                      [tex]$d=0.75 \times 10^{-2} \ m$[/tex]

We known

[tex]$E=\frac{V}{d}$[/tex]

[tex]$V = E\times d$[/tex]

[tex]$V = 3.99 \times 10^5 \times 0.75 \times 10^{-2}$[/tex]

[tex]$V = 2.9925 \times 10^3 \ V$[/tex]

   = 2.9925 kV

Answer:

C

Explanation:

C is correct

Ok!

151617

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

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