An alumium rod when measured with a steel scale both being at 25 degree celcius appears to be 1 meter long. If the scale is correct at 0 degree celcius, what is the true length of the rod at 25 degree celcius? What wull be the length of rod at 0 degree celcius?​

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: Ok so We already know that velocity is on the x-axis.

Since acceleration = Force / Mass

Here the Force is downward due to the gravitational pull or we can say it is along y-axis.

Since acceleration is directly proportional to force, so acceleration is also along y-axis. This means that velocity & acceleration are perpendicular to each other.

Example:

Let us assume that an aeroplane is flying parallel to the horizontal plane. The aeroplane will experience the acceleration in several directions. One of them here is the gravitational pull which is perpendicular to the the apparent velocity. So the net velocity & its direction will depend upon the vector sum total of all the forces/acceleration acting on it. Also because of this gravitational pull the aeroplane rotates along with the earth, which is a proof that the force/g experienced by it does not go waste.

Explanation:

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:

C

Explanation:

C is correct

Ok!

151617

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

Answer:

0 m/s

Explanation:

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

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:

[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].

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

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:

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:

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:

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:

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:

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]

(a) The original charge on the 40-pF capacitor is [tex]2 .0 \ \times \ 10^{-8} \ C[/tex].

(b) The charge on each capacitor after the connection is made is [tex]4 .0 \ \times \ 10^{-9} \ C[/tex].

(c)  The potential difference across the plates of each capacitor after the connection is 100 V and 400 V.

The original charge on the 40-pF capacitor is calculated as follows;

[tex]Q = CV\\\\Q = 40 \times 10^{-12} \times 500\\\\Q = 2 .0 \ \times \ 10^{-8} \ C[/tex]

[tex]C = \frac{C_1C_2}{C_1 + C_2} \\\\C = \frac{10 \times 10^{-12} \times 40 \times 10^{-12}}{10\times 10^{-12} \ + \ 40 \times 10^{-12}} \\\\C = 8 \times 10^{-12} \ F[/tex]

[tex]Q = Q_1 = Q_2\\\\Q = 8 \times 10^{-12} \ \times \ 500\\\\Q = 4 \times 10^{-9} \ C[/tex]

The potential difference across the plates of each capacitor after the connection is calculated as follows;

[tex]V = \frac{Q}{C} \\\\V_1 = \frac{Q}{C_1} \\\\V_1 = \frac{4 \times 10^{-9}}{40 \times 10^{-12}} \\\\V_1 = 100 \ V\\\\V_2 = \frac{Q}{C_2} \\\\V_2 = \frac{4 \times 10^{-9}}{10 \times 10^{-12} } \\\\V_2 = 400 \ V[/tex]

Learn more about charge of a capacitor here: https://brainly.com/question/13578522

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.

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

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:

Sunlight is the greatest source of UV radiation. Man-made ultraviolet sources include several types of UV lamps, arc welding, and mercury vapour lamps.

Explanation:

Answer:

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

Explanation:

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

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:

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:

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