Answers
Answer:
60 deg from Y-axis is 150 deg from X-axis
x = L cos theta = 23,2 cos 150 = -.867 * 23.2 = -20.1
Related Questions
A source of emf is connected by wires to a resistor and electrons flow in the circuit the wire diameter is teh same throughout teh circuit compared to the drift speed of the electrons before entering the source of emf, the drift speed ot eh electrons afte rleaving the source of emf is: ___________
a. faster.
b. slower.
c. the same.
d. either A or B depending on circumstances.
e. any of A, B , or C depending on circumstances.
Answers
I hope that helps
Which quantities below of a solid object on this planet are NOT the same as on Earth?
Choose all
possible answers.
Weight
Mass
Volume
Density
Acceleration when it falls vertically.
Color
Answers
Answer:
Weight, acceleration when it falls vertically, are not same as that of earth.
Explanation:
Weight of the object is given by the product of mass of the object and the acceleration due to gravity of the planet.
So, the weight of object is not same as that on earth.
The mass is defined as the amount of matter contained in the object.
So, the mass of the object is same as that of earth.
The volume of the object is defined as the space occupied by the object.
So, the volume of the object is same as that of earth.
The density is defined as the ratio of mass of the object to its volume.
So, the density of the object is same as that of earth.
The acceleration due to gravity on a planet depends on the mass of planet and radius of planet.
So, the acceleration is not same as that of earth.
The color of the object is its characteristic.
It is same as that of earth.
A charged particle of mass 0.0040 kg is subjected to a magnetic field which acts at a right angle to its motion. If the particle moves in a circle of radius at a speed of what is the magnitude of the charge on the particle
Answers
Complete Question
A charged particle of mass 0.0040 kg is subjected to a 40-T magnetic field which acts at a right angle to its motion. If the particle moves in a circle of radius 0.10 m at a speed of 20m/s, what is the magnitude of the charge on the particle?
Answer:
[tex]q=0.020C[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=0.004kg[/tex]
Magnetic field [tex]B=40-T[/tex]
Radius [tex]r=0.10m[/tex]
Speed [tex]v= 20m/s[/tex]
Generally the equation for Radius of a circular path is mathematically given by
[tex]r=\frac{p}{qB}[/tex]
Where
[tex]p=mv[/tex]
Therefore
[tex]q=\frac{mv}{rB}[/tex]
[tex]q=\frac{0.004kg*20m/s}{0.1*40}[/tex]
[tex]q=0.020C[/tex]
which of the following is correct for solids? a. y=2/3 alpha b.y= 2/3 bita c.y = 3/2 bita d.y=3/2 alpha
Answers
option a ...
please mark hcfhvx
A car starting at rest accelerates at 3m/seconds square How far has the car travelled after 4s?
Answers
Answer:
24 meters
Explanation:
Find the final velocity. 12m/s
d=[final-initial]/2×time
D=(6m/s)×4=24 m/s
the coefficient of linear expansion of aluminum is 24.0 * 10^-6K^-1 and th density of aluminum at ) C is 2.7 *10^3kg/m^3. what is the density of aluminum at 300 C
Answers
Answer:
The density at 300 C is [tex]2641.68 kg/m^3[/tex].
Explanation:
density at 0 degree, d = 2700 kg/m3
coefficient of linear expansion, = 24 x 10^{-6} /K
Let the density at 300 C is d'.
Use the formula of the cubical expansion
[tex]d'= d (1-3\alpha \Delta T)\\\\d'= 2700(1- 3 \times 24\times 10^{-6}\times 300)\\\\d' = 2641.68 kg/m^3[/tex]
Suppose you want to design an air bag system that can protect the driver at a speed 100 km/h (60 mph) if the car hits a brick wall.
Estimate how fast the air bag must inflate to effectively protect the driver. How does the use of a seat belt help the driver?
Answers
When solving question that contains equations and the use mathematical computations, It is always ideal to list the parameters given.
Now, given that:
the speed of the car which is the initial velocity (u) = 100 km/h before it hits the wall.after hitting the wall, the final velocity will be (v) = 0 km/hAssumptions:
Suppose we make an assumption that the distance traveled during the collision of the car with the brick wall (S) = 1 mThat the car's acceleration is also constant.∴
For a motion under constant acceleration, we can apply the kinematic equation:
[tex]\mathsf{v^2 = u^2 + 2as}[/tex]
where;
v = final velocity u = initial velocitya = accelerations = distanceFrom the above equation, making acceleration (a) the subject of the formula:
[tex]\mathsf{v^2 - u^2 =2as }[/tex]
[tex]\mathsf{a = \dfrac{v^2 - u^2 }{2s}}[/tex]
The initial velocity (u) is given in km/h, and we need to convert it to m/s as it has an effect on the unit of the acceleration.
since 1 km/h = 0.2778 m/s
100 km/h = 27.78 m/s
[tex]\mathsf{a = \dfrac{(0)^2 - (27.78)^2 }{2(1)}}[/tex]
[tex]\mathsf{a = \dfrac{- 771.7284 }{2}}[/tex]
a = - 385.86 m/s²
Similarly, from the kinematic equation of motion, the formula showing the relation between time, acceleration, and velocity is;
v = u + at
where;
v = 0
-u = at
[tex]\mathsf{t = \dfrac{-u}{a}}[/tex]
[tex]\mathsf{t = \dfrac{-27.78}{-385.86}}[/tex]
t = 0.07 seconds
An airbag is designed in such a way as to prevent the driver from hitting on the steering wheel or other hard substance that could damage the part of the body. The use of the seat belt is to keep the driver in shape and in a balanced position against the expansion that occurred by the airbag during the collision on the brick wall.
Thus, we can conclude that in order to estimate how fast the airbag must inflate to effectively protect the driver, the airbag must be inflated at 0.07 seconds faster before the collision to effectively protect the driver.
Learn more about the kinematic equation here:
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Wind instruments like trumpets and saxophones work on the same principle as the "tube closed on one end" that we examined in our last experiment. What effect would it have on the pitch of a saxophone if you take it from inside your house (at 76 degrees F) to the outside on a cold day when the outside temperature is 45 degrees F ?
Answers
Answer:
f = v / 4L
the frequency of the instruments is reduced by the decrease in the speed of the wave with the temperature.
Explanation:
In wind instruments the wave speed must meet
v = λ f
λ = v / f
from v is the speed of sound that depends on the temperature
v = v₀ [tex]\sqrt{1+ \frac{T [C]}{273} }[/tex]
where I saw the speed of sound at 0ºC v₀ = 331 m/s the temperature is in degrees centigrade, we can take the degrees Fahrenheit to centigrade with the relation
(F -32) 5/9 = C
76ºF = 24.4ºC
45ºF = 7.2ºC
With this relationship we can see that the speed of sound is significantly reduced when leaving the house to the outside
at T₁ = 24ºC v₁ = 342.9 m / s
at T₂ = 7ºC v₂ = 339.7 m / s
To satisfy this speed the wavelength of the sound must be reduced, so the resonant frequencies change
λ / 4 = L
λ= 4L
v / f = 4L
f = v / 4L
Therefore, the frequency of the instruments is reduced by the decrease in the speed of the wave with the temperature.
During a particular thunderstorm, the electric potential difference between a cloud and the ground is Vcloud - Vground = 4.20 108 V, with the cloud being at the higher potential. What is the change in an electron's electric potential energy when the electron moves from the ground to the cloud?
Answers
Answer:
The electric potential energy is 6.72 x 10^-11 J.
Explanation:
Potential difference, V = 4.2 x 10^8 V
charge of electron, q = - 1.6 x 10^-19 C
Let the potential energy is U.
U = q V
U = 1.6 x 10^-19 x 4.2 x 10^8
U = 6.72 x 10^-11 J
A technician builds an RLC series circuit which includes an AC source that operates at a fixed frequency and voltage. At the operating frequency, the resistance R is equal to the inductive reactance XL. The technician notices that when the plate separation of the parallel-plate capacitor is reduced to one-half its original value, the current in the circuit doubles. Determine the initial capacitive reactance in terms of the resistance R.
Answers
Answer:
Xc = (0.467 - 0.427j)R
Explanation:
Since the resistance in the circuit is R, the reactance of the inductor is XL and the reactance of the capacitor is XC, then the impedance of the circuit is
Z = √[R² + (XL - XC)²]
Since the inductive reactance XL equals the resistance R, we have that
Z = √[R² + (XL - XC)²]
Z = √[R² + (R - XC)²]
Thus, the current in the circuit is thus I = V/Z = V/√[R² + (R - XC)²]
Now, when the plate separation of the parallel plate capacitor is reduced to one-half its original value, the current doubles. Also, when the plate separation is reduced to half, the capacitance doubles since C ∝ 1/d where C is capacitance and d separation between the plates. Since the capacitance doubles, the new reactance XC' is twice the initial reactance XC. So, XC' = 2XC. Thus the new impedance is thus
Z' = √[R² + (R - XC')²]
Z' = √[R² + (R - 2XC)²]
The new current is I' = V/Z' = V/√[R² + (R - 2XC)²]
Since the current doubles, I' = 2I.
V/√[R² + (R - 2XC)²] = 2V/√[R² + (R - XC)²]
1/√[R² + (R - 2XC)²] = 2/√[R² + (R - XC)²]
√[R² + (R - XC)²] = 2√[R² + (R - 2XC)²]
squaring both sides, we have
[R² + (R - XC)²] = 4[R² + (R - 2XC)²]
expanding the brackets, we have
[R² + R² - 2RXC + XC²] = 4[R² + R² - 4RXC + 4XC²]
[2R² - 2RXC + XC²] = 4[2R² - 4RXC + 4XC²]
2R² - 2RXC + XC² = 8R² - 16RXC + 16XC²
collecting like terms, we have
16RXC - 2RXC + XC² - 16XC² = 8R² - 2R²
14RXC - 15XC² = 6R²
15XC² - 14RXC + 6R² = 0
Using the quadratic formula to find XC, we have
[tex]XC = \frac{-(-14R) +/- \sqrt{(-14R)^{2} - 4 X 15 X 6R^{2} } }{2 X 15}\\= \frac{-(-14R) +/- \sqrt{196R^{2} - 360R^{2} } }{30}\\ \\= \frac{14R +/- \sqrt{- 164R^{2} } }{30}\\ \\= \frac{14R +/- 12.81Ri }{30}\\\\= 0.467R +/- 0.427Ri[/tex]
Since it is capacitive, we take the negative part.
So, Xc = (0.467 - 0.427j)R
If the Velocity of the body
is increased to 3v, determine the kinetic energy
Answers
ANSWER; KE=5mv^2 so it is proportional to v^2.
Explanation:So if you triple the velocity you are replacing v with 3v. Then you get (3v)^2=9v^2.
what is the prefix notation of 0.0000738?
Answers
Answer:
7.38 × 10-5
Explanation:
All numbers in scientific notation or standard form are written in the form m × 10n, where m is a number between 1 and 10 ( 1 ≤ |m| < 10 ) and the exponent n is a positive or negative integer.
To convert 0.0000738 into scientific notation, follow these steps:
Move the decimal 5 times to the right in the number so that the resulting number, m = 7.38, is greater than or equal to 1 but less than 10
Since we moved the decimal to the right the exponent n is negative
n = -5
Write in the scientific notation form, m × 10n
= 7.38 × 10-5
Therefore, the decimal number 0.0000738 written in scientific notation is 7.38 × 10-5 and it has 3 significant figures.
Answer = 7.38 × 10-5
A rigid, insulated tank whose volume is 10 L is initially evacuated. A pinhole leak develops and air from the surroundings at 1 bar, 25 C enters the tank until the pressure in the tank becomes 1 bar (assume ideal gas model k=1.4 for the air). Find:
A) final temperature in tank.
B) amount of air that leaks into tank in grams.
C) amount of entropy produced in J/K.
Answers
Answer:
The answer is "[tex]143.74^{\circ} \ C , 8.36\ g, and \ 2.77\ \frac{K}{J}[/tex]"
Explanation:
For point a:
Energy balance equation:
[tex]\frac{dU}{dt}= Q-Wm_ih_i-m_eh_e\\\\[/tex]
[tex]W=0\\\\Q=0\\\\m_e=0[/tex]
From the above equation:
[tex]\frac{dU}{dt}=0-0+m_ih_i-0\\\\\Delta U=\int^{2}_{1}m_ih_idt\\\\[/tex]
because the rate of air entering the tank that is [tex]h_i[/tex] constant.
[tex]\Delta U = h_i \int^{2}_{1} m_i dt \\\\= h_i(m_2 -m_1)\\\\m_2u_2-m_1u_2=h_i(M_2-m_1)\\\\[/tex]
Since the tank was initially empty and the inlet is constant hence, [tex]m_2u-0=h_1(m_2-0)\\\\m_2u_2=h_1m_2\\\\u_2=h_1\\\\[/tex]
Interpolate the enthalpy between [tex]T = 300 \ K \ and\ T=295\ K[/tex]. The surrounding air
temperature:
[tex]T_1= 25^{\circ}\ C\ (298.15 \ K)\\\\\frac{h_{300 \ K}-h_{295\ K}}{300-295}= \frac{h_{300 \ K}-h_{1}}{300-295.15}[/tex]
Substituting the value from ideal gas:
[tex]\frac{300.19-295.17}{300-295}=\frac{300.19-h_{i}}{300-298.15}\\\\h_i= 298.332 \ \frac{kJ}{kg}\\\\Now,\\\\h_i=u_2\\\\u_2=h_i=298.33\ \frac{kJ}{kg}[/tex]
Follow the ideal gas table.
The [tex]u_2= 298.33\ \frac{kJ}{kg}[/tex] and between temperature [tex]T =410 \ K \ and\ T=240\ K.[/tex]
Interpolate
[tex]\frac{420-410}{u_{240\ k} -u_{410\ k}}=\frac{420-T_2}{u_{420 k}-u_2}[/tex]
Substitute values from the table.
[tex]\frac{420-410}{300.69-293.43}=\frac{420-T_2}{{u_{420 k}-u_2}}\\\\T_2=416.74\ K\\\\=143.74^{\circ} \ C\\\\[/tex]
For point b:
Consider the ideal gas equation. therefore, p is pressure, V is the volume, m is mass of gas. [tex]\bar{R} \ is\ \frac{R}{M}[/tex] (M is the molar mass of the gas that is [tex]28.97 \ \frac{kg}{mol}[/tex] and R is gas constant), and T is the temperature.
[tex]n=\frac{pV}{TR}\\\\[/tex]
[tex]=\frac{(1.01 \times 10^5 \ Pa) \times (10\ L) (\frac{10^{-3} \ m^3}{1\ L})}{(416.74 K) (\frac{8.314 \frac{J}{mol.k} }{2897\ \frac{kg}{mol})}}\\\\=8.36\ g\\\\[/tex]
For point c:
Entropy is given by the following formula:
[tex]\Delta S = mC_v \In \frac{T_2}{T_1}\\\\=0.00836 \ kg \times 1.005 \times 10^{3} \In (\frac{416.74\ K}{298.15\ K})\\\\=2.77 \ \frac{J}{K}[/tex]
Which two factors does the power of a machine depend on? А. work and distance B.. force and distance C. work and time D. time and distance?
Answers
[tex]Hello[/tex] [tex]There![/tex]
[tex]AnimeVines[/tex] [tex]is[/tex] [tex]here![/tex]
The answer is...
C. Work and time.
[tex]HopeThisHelps!![/tex]
[tex]AnimeVines[/tex]
Two wires are made of the same material and have the same length but different radii. They are joined end-to- end and a potential difference is maintained across the combination. Of the following quantities that is same for both wires is
A. Potential difference
B. Electric current
C. Current density
D. Electric field
Answers
Answer:
Current
I think The choose (B)
B. Electric current
80 grams of iron at 100°C is dropped into 200 of water at 20°C contained in an iron vessel of mass 50 gram find the resulting temperature.
Answers
Answer:
the resulting temperature is 23.37 ⁰C
Explanation:
Given;
mass of the iron, m₁ = 80 g = 0.08 kg
mass of the water, m₂ = 200 g = 0.2 kg
mass of the iron vessel, m₃ = 50 g = 0.05 kg
initial temperature of the iron, t₁ = 100 ⁰C
initial temperature of the water, t₂ = 20 ⁰C
specific heat capacity of iron, c₁ = 462 J/kg⁰C
specific heat capacity of water, c₂ = 4,200 J/kg⁰C
let the temperature of the resulting mixture = T
Apply the principle of conservation of energy;
heat lost by the hot iron = heat gained by the water
[tex]m_1c_1 \Delta t_1 = m_2c_2\Delta t_2\\\\m_1c_1 (100 - T) = m_2c_2 (T- 20)\\\\0.08 \times 462 (100-T) = 0.2 \times 4,200 (T-20)\\\\36.96 (100-T) = 840 (T-20) \\\\100 - T = 22.72 (T-20)\\\\100-T = 22.72 T - 454.4 \\\\554.4 = 23.72T\\\\T = \frac{554.4}{23.72} \\\\T = 23.37 \ ^0C[/tex]
Therefore, the resulting temperature is 23.37 ⁰C
Can a conductor be given limitless charge
Answers
Answer:
No
Explanation:
You could try to give it enough to fill all valence electrons in all of the atoms in the conductor, but practically this could not be achieved.
1. Consider a 1000 kg car rounding a curve on a flat road of radius 50 m at a speed of
50 km/h (14 m/s).
a. Will the car make the turn if the pavement is dry and the coefficient of static
friction is 0.60?
Answers
Answer:
The car will make the turn perfectly
Explanation:
Given that the centripetal force= mv^2/r
M= mass of the car
v = speed of the car
r= radius
Hence;
F = 1000 × (14)^2/50
F= 3920 N
The frictional force = μmg
μ = coefficient of static friction
m= mass
g = acceleration due to gravity
Frictional force= 0.6 × 1000× 10
Frictional force = 6000 N
The car will not skid off the curve because the frictional force is greater than the centripetal force.
A stone dropped from the top of a 80m high building strikes the ground at 40 m/s after falling for 4 seconds. The stone's potential energy with respect to the ground is equal to its kinetic energy … (use g = 10 m/s 2)
A) at the moment of impact.
B) 2 seconds after the stone is released.
C) after the stone has fallen 40 m.
D) when the stone is moving at 20 m/s.
At the moment of impact both Kinetic Energy and Potential Energy should be 0, right? So it can't be A), right? Or is this wrong? Is it indeed A)? Please show work and explain it well.
Answers
Answer: c. 40m
Explanation:
See picture
A denser object will usually have a ____ index of
refraction.
Answers
Answer:
A denser object will usually have a high index of
refraction.
A denser object will usually have a high index of refraction.
What is index of refraction?The refractive index is the ratio of the speed of light in vacuum and speed of light in any medium.
n = c/v
The density is greater for the denser medium (water, oil, mercury, etc) then the rarer medium (any gas or air).
When a light ray travels through denser medium, its velocity is reduced and the refracted ray bends towards normal.
As, the index of refraction is inversely proportional to the velocity of light in the medium, index of refraction will be high for denser object.
Thus, denser object have high index of refraction.
Learn more about index of refraction.
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which of the following causes the magnetic force between the magnet and the scrap metal?
Answers
Answer:
Alternating current at which when entered into the loop cause it to magnetize
When a ray of light passes through a boundary between two transparent media, it changes direction. What is this phenomenon known as?
Choose the best answer from the choices provided.
a. reflection
b. refraction
c. interference
d. diffraction
e. dispersion
Answers
Answer:
I think Refraction
Explanation:
I think refraction is light which passes through a boundary between 2 transparent media.
A ten loop coil of area 0.23 m2 is in a 0.047 T uniform magnetic field oriented so that the maximum flux goes through the coil. The average emf induced in the coil is
Answers
Answer:
Explanation:
From the question we are told that:
Number of turns [tex]N=10[/tex]
Area [tex]a=0.23m^2[/tex]
Magnetic field [tex]B=0.947T[/tex]
Generally the equation for maximum flux is mathematically given by
[tex]\phi=NBa[/tex]
[tex]\phi=10*0.047*0.23[/tex]
[tex]\phi=0.1081wbi[/tex]
Therefore induced emf
[tex]e= \frac{d\phi}{dt}[/tex]
Since
[tex]t=0[/tex]
Therefore
[tex]e=0[/tex]
A solid conducting sphere of radius ra is placed concentrically inside a conducting spherical shell of inner radius rb1 and outer radius rb2. The inner sphere carries a charge Q while the outer sphere does not carry any net charge. The potential for rb1 < r < rb2 is:________
Answers
Answer:
The right answer is "[tex]\frac{KQ}{r_b_2}[/tex]".
Explanation:
As the outer spherical shell is conducting, so there is no electric field in side from
⇒ [tex]r_b_1 < r < r_b_2[/tex].
So the electric potential at all points inside the conducting shell that from
⇒ [tex]r_b_1<r<r_b_2[/tex]
and will be similar as well as equivalent to the potential on the outer surface of the shell that will be:
⇒ [tex]v=\frac{KQ}{r_b_2}[/tex]
Thus the above is the right solution.
A T-shirt cannon launches a shirt at 5.30 m/s from a platform height of 4.00 m from ground level. How fast (in m/s) will the shirt be traveling if it is caught by someone whose hands are at 5.20 m from ground level (b) 4.00 m from ground level?
Answers
Answer:
(a) the velocity of the shirt is 2.14 m/s
(b) the velocity of the shirt is 5.3 m/s
Explanation:
Given;
initial velocity of the shirt, u = 5.3 m/s
height of the platform above the ground, h = 4.00 m
(a) When the shirt is caught by someone whose hand is 5.20 m from the ground level, the height traveled by the shirt = 5.2 m - 4.0 m = 1.2 m
The velocity at this position is calculated as;
[tex]v^2 = u^2 + 2(-g)h\\\\v^2 = u^2 - 2gh\\\\v^2 = 5.3^2 - (2\times 9.8 \times 1.2)\\\\v^2 = 4.57\\\\v= \sqrt{4.57} \\\\v = 2.14 \ m/s[/tex]
(b) When the shirt is caught by someone whose hand is 4.00 m from the ground level, the height traveled by the shirt = 4.00 m - 4.00 m = 0 m
The velocity at this position is calculated as;
[tex]v^2 = u^2 + 2(-g)h\\\\v^2 = u^2 - 2gh\\\\v^2 = 5.3^2 - (2\times 9.8 \times 0)\\\\v^2 = 28.09\\\\v= \sqrt{28.09} \\\\v = 5.3 \ m/s[/tex]
What is the rate of power flowing through phone if a 2.50V battery produces 1.50A of current?
Answers
Answer:
3.75 watts
Explanation:
power = volts * current
2.5 * 1.5 =
3.75
Automobile A and B are initially 30 m apart travelling in adjacent highway lanes at speeds VA = 14.4 km/hr., VB 23.4 km/hr. at t = 0.0. Knowing that automobile A has a constant acceleration of 0.8 m/s? and automobile B has a constant deceleration of 0.4 m/s2. Automobile A will overtake B after traveling a distance SA: A B. Side view
Answers
Answer:
x = 240 m
Explanation:
This is a kinematics exercise
Let's fix our frame of reference on car A
x = x₀ₐ+ v₀ₐ t + ½ aₐ t²
the initial position of car a is zero
x = 0 + v₀ₐ t + ½ 0.8 t²
for car B
x = x_{ob} + v_{ob} t - ½ a_b t²
car B's starting position is 30 m
x = 30 + v_{ob} t - ½ 0.4 t²
at the point where they meet, the position of the two vehicles is the same
0 + v₀ₐ t + ½ 0.8 t² = 30 + v_{ob} t - ½ 0.4 t²
let's reduce the speeds to the SI system
v₀ₐ = 14.4 km / h (1000 m / 1 km) (1h / 3600s) = 4 m / s
v_{ob} = 23.4 km / h = 6.5 m / s
4 t + 0.4 t² = 30 + 6.5 t - 0.2 t²
0.2 t² - 2.5 t - 30 = 0
t² - 12.5 t - 150 = 0
we solve the quadratic equation
t = [tex]\frac{12.5 \pm \sqrt{12.5^2 + 4 \ 150} }{2}[/tex]
t = [tex]\frac{12.5 \ \pm 27.5}{2}[/tex]
t₁ = 20 s
t₂ = -7.5 s
time must be a positive quantity so the correct result is t = 20 s
let's look for the distance
x = 4 t + ½ 0.8 t²
x = 4 20 + ½ 0.8 20²
x = 240 m
A cylindrical water tank has a height of 20cm and a radius of 14cm. If it is filled to 2/5 of its capacity, calculate.
I. Quantity of water in the tank
II. Quantity of water left to fill the tank to its capacity.
Answers
Answer:
4.926 L Y 7.389 L
Explanation:
first you calculate the tank volume
V = π[tex](14 cm)^{2}[/tex](10 cm = [tex]12315 cm^{3}[/tex]
then you convert to liters
[tex]12315 cm^{3}[/tex] = 12.315 l
then you calculate the liters of water
2/5(12.35 l) = 4.926 l
finally we calculate the amount without water
12.315 l - 4.926 l = 7.389 l
HERE IS MORE INFORMATION ON THE SUBJECT. THEY REMOVED THE
ENGLISH SITE BUT YOU CAN USE TRANSLATOR
LINK: https://gscourses.thinkific.com/courses/fisicai
6. Two astronauts of equal mass step from the top of a rocket on Venus. One slides down a 10- degree ramp while the other slides down a 75-degree ramp. If all friction is ignored, which astronaut reaches the surface of Venus with the lower kinetic energy
Answers
Answer:
Final kinetic energies of both astronauts will be the same.
Explanation:
If we ignore all the friction present between the ramp and the person. Then essentially there is no loss of energy in the system. Hence the initial potential energies of the astronauts must be equal to their final kinetic energies.
Now the potential energy depends upon mass, height and acceleration due to gravity. All these parameters for both the astronauts. Therefore,both astronauts have same initial potential energies.
Similarly, the final kinetic energies of astronauts will also be the same.
The top of a swimming pool is at ground level. If the pool is 2.60 m deep, how far below ground level does the bottom of the pool appear to be located for the following conditions?
a. The pool is completely filled with water.
______m below ground level
b. The pool is filled halfway with water.
______m below ground level
Answers
Answer:
a) [tex]d_g=1.95m[/tex]
b) [tex]d_g'=2.3m[/tex]
Explanation:
From the question we are told that:
Depth [tex]d=2.60[/tex]
a)
Generally the equation for distance to ground is mathematically given by
[tex]\frac{d_g}{d}=\frac{1}{n}[/tex]
Where
[tex]n=\frac{4}{3}[/tex]
Therefore
[tex]d_g=\frac{d}{n}[/tex]
[tex]d_g=\frac{2.6}{4/3}[/tex]
[tex]d_g=1.95m[/tex]
b)
For when The pool is filled halfway with water
[tex]\frac{d_g}{d}=\frac{1}{n}[/tex]
[tex]d_g'=\frac{1.3}{4/3}[/tex]
[tex]d_g'=0.98m[/tex]
Therefore
[tex]d_g'=(1.3+0.98)[/tex]
[tex]d_g'=2.3m[/tex]
