Answer:
P_abs = 105120.2 N/m²
Explanation:
We are given;
Specific gravity of oil; ρ_oil = 0.6 g/cm³ = 600 kg/m³
Depth of water; h_w = 21 cm = 0.21 m
Depth of oil; h_o = 35 cm = 0.35 m
From tables specific gravity of water is; ρ_w = 1000 kg/m³
Thus, to get the absolute pressure at the bottom of the container, we will use the formula;
P_abs = (ρ_w × g × h_w) + (ρ_oil × g × h_oil) + P_a
Where P_a is atmospheric pressure with a standard value of 1.01 × 10^(5) N/m²
g is gravitational acceleration = 9.81 m/s²
Thus;
P_abs = (1000 × 9.81 × 0.21) + (600 × 9.81 × 0.35) + (1.01 × 10^(5))
P_abs = 105120.2 N/m²
An object has a mass of 15 kg and is accelerating to the right at 16.3 m/s2. The free-body diagram shows the horizontal forces acting on the object. A free body diagram with 2 forces. The first vector is pointing right, labeled F Subscript a Baseline 250 N. The second vector is shorter pointing left, labeled F Subscript f Baseline. What is the frictional force, Ff, acting on the object?
Answer:
A. -5.5 N
Explanation:
I can confirm that the answer is A.
What is the magnitude and direction of the magnetic force on the bob at the lowest point in its path, if it has a positive 0.250 μC charge and is released from a height of 30.0 cm above its lowest point? The magnetic field strength is 1.50 T.
Answer:
[tex]F=9.09\times 10^{-7}\ N[/tex]
Explanation:
Given that,
Charge, q = 0.250 μC
It is released from a height of 30 cm or 0.03 m
The magnetic field strength is 1.50 T.
First we find the velocity using the conservation of energy as follows :
[tex]mgh=\dfrac{1}{2}mv^2\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 0.3} \\\\v=2.424\ m/s[/tex]
Now, the magnetic force is given by :
[tex]F=qvB\\\\=0.25\times 10^{-6}\times 2.424\times 1.5\\\\=9.09\times 10^{-7}\ N[/tex]
So, the magnetic force is [tex]9.09\times 10^{-7}\ N[/tex]. Since, the bob is at the lowest point, the direction of the magnetic force at the lowest point is upward.
How are the toes of a hen different from that of an eagle?
Answer:
no they different
Explanation:
because hen lives on land and eagle flies in sky it doesnt walk often just it aearch for its prey and it eats there only
Answer:
eheisjsnsndndj
Explanation:
sjdjdj
Help me please!
On the earth, the gravitational field strength is 10 N/kg. On the Moon, the gravitational field strength is 1.6 N/kg.
If an object has a weight of 50 N on earth, what is its weight on the Moon?
A: 1.6 N
B: 5.0 N
C: 8.0 N
D: 80 N
Answer:
This is a way of measuring how much gravity there is. The formula is: weight/mass = gravitational field strength.
Gravitational field strength = Weight/mass unit is N/kg
Weight = mass x gravitational field strength unit is N
On Earth the gravitational field strength is 10 N/kg. Other planets have different gravitational field strengths. The Moon has a gravitational field strength of 1.6 N/kg. You might have seen films of astronauts leaping high on the moon.
Here on Earth, if I jump I am pulled back to ground by gravity. What is my weight? My mass is 80kg and if we multiply by gravitational field strength (10N/kg) - my weight is 800N. Now if I go to the moon, my mass will be the same, 80kg. We multiply that by the moon's gravitational field strength, which is 1.6 N/ kg. That means my weight on the moon is 128N. So I have different weights on the Earth and on the Moon. That's why astronauts can jump high into the air on the moon - they're lighter up there.
Jupiter is a very large planet with strong gravitational field strength of 25 N/ kg. My body is 80kg. If I go to Jupiter my weight is going to be 25 x 80 = 2,000 N. That means I wouldn't be able to get off the ground or stand up straight! I would probably be lying down all the time there. So weight varies depending on which planet you are on. You can find out more yourself by looking up tables of weight on different planets.
Which of the following descriptions best describe a liquid
A
takes the shape and volume of its container
B
matter is made of atoms so tightly packed together that they cannot move around
C
has a definite volume, but takes the shape of its container
Answer:
C
Explanation:
An object that moves in uniform circular motion has a centripetal acceleration of 13 m/s^2 . If the radius of the motion is 0.02m, what is the frequency of motion?
Answer:
f = 3.97 Hz
Explanation:
Given that,
Centripetal acceleration, [tex]a=13\ m/s^2[/tex]
The radius of motion is 0.02 m
The formula for the centripetal acceleration is given by :
[tex]a=\dfrac{v^2}{r}\\\\v=\sqrt{ar} \\\\v=\sqrt{13\times 0.02} \\\\v=0.5\ m/s[/tex]
The speed of an object in a circular path is given by :
[tex]v=\dfrac{2\pi r}{t}[/tex]
t is time period
Also, f=1/t (f is frequency)
[tex]f=\dfrac{v}{2\pi r}\\\\f=\dfrac{0.5}{2\pi \times 0.02}\\\\f=3.97\ Hz[/tex]
Hence, the frequency of motion s 3.97 Hz.
The frequency of the motion is 4.1 Hz.
Linear velocity?The linear velocity of the of the object is calculated as follows;
[tex]a = \frac{v^2}{r} \\\\v^2 = ar\\\\v = \sqrt{ar} \\\\v = \sqrt{13 \times 0.02} \\\\v = 0.51 \ m/s[/tex]
Angular speed of the objectThe angular speed of the object is calculated as follows;
[tex]\omega =\frac{v}{r} \\\\\omega = \frac{0.51}{0.02} \\\\\omega = 25.5 \ rad/s[/tex]
Frequency of motionThe frequency of the motion is calculated as follows;
[tex]\omega = 2\pi f\\\\f = \frac{\omega }{2\pi} \\\\f = \frac{25.5}{2\pi } \\\\f = 4.1 \ Hz[/tex]
Thus, the frequency of the motion is 4.1 Hz.
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An ideal gas occupies 0.4 m3 at an absolute pressure of 500 kPa. What is the absolute pressure if the volume changes to 0.9 m3 and the temperature remains constant?
Answer:
2.22 kPaExplanation:
The new volume can be found by using the formula for Boyle's law which is
[tex]P_1V_1 = P_2V_2[/tex]
Since we are finding the new volume
[tex]V_2 = \frac{P_1V_1}{P_2} \\[/tex]
From the question we have
[tex]V_2 = \frac{0.4 \times 500000}{0.9} = \frac{200000}{0.9} \\ = 222222.2222... \\ = 222222[/tex]
We have the final answer as
2.22 kPaHope this helps you
A 60-W light bulb emits spherical electromagnetic waves uniformly in all directions. If 50% of the power input to such a light bulb is emitted as electromagnetic radiation, what is the radiation intensity at a distance of 2.00 m from the light bulb?
A) 15 W/m2
B) 4.8 W/m2
C) 2.4 W/m2
D) 0.60 W/m2
E) 1.2 W/m2
Answer:
a
Explanation:
A 60-W light bulb emits spherical electromagnetic waves uniformly in all directions. If 50% of the power input to such a light bulb is emitted as electromagnetic radiation, what is the radiation intensity at a distance of 2.00 m from the light bulb
so i did 60(50%)=30÷2=15
The radiation intensity is 15 W / m².
To find the radiation intensity, the values are given as,
Power = 60 W
Distance = 2 m
50% of the power input is emitted as electromagnetic radiation.
What is radiation intensity?The amount of energy emitted per unit solid angle by per unit area of the radiating surface can be said as radiation intensity.
As, when there is a power output, the input power will emit some energy as a kinetic energy and electromagnetic radiation. By the way, the power input emits 50 % as a radiation, so the power input given as,
P = 60 / 2 ( As it was 50 % )
= 30 Watt.
The radiation intensity is,
In = 30 / 2
= 15 W / m².
Thus, the radiation intensity is calculated as, 15 W/ m².
So, Option A is the correct answer.
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a force of 35N is exerted over a cylinder with an area of 5m^2. What pressure,in pascals, will be transmitted in the hydraulic system?
Answer:
The answer is 7 PaExplanation:
The pressure transmitted in the hydraulic system can be found by using the formula
[tex]p = \frac{f}{a} \\ [/tex]
f is the force
a is the area
From the question we have
[tex]p = \frac{35}{5} \\ [/tex]
We have the final answer as
7 PaHope this helps you
Microwave ovens have a plate that rotates at a rate of about 7.0 rev/min. What is this in revolutions per second?
Answer:
The value is [tex]w = 0.1167 \ rev/second[/tex]
Explanation:
From the question we are told that
The rate at which the plate rotates is [tex]w =7.0 \ rev/min[/tex]
Generally the revolution per second is mathematically represented as
[tex]w = \frac{7.0}{60}[/tex]
=> [tex]w = 0.1167 \ rev/second[/tex]
A ball is launched from ground level at 20 m/s at an angle of 40° above the
horizontal. A) How long the ball is in the air? B)What is the maximum
height the ball can reach?
(a) The ball's height y at time t is given by
y = (20 m/s) sin(40º) t - 1/2 g t ²
where g = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve y = 0 for t :
0 = (20 m/s) sin(40º) t - 1/2 g t ²
0 = t ((20 m/s) sin(40º) - 1/2 g t )
t = 0 or (20 m/s) sin(40º) - 1/2 g t = 0
The first time refers to where the ball is initially launched, so we omit that solution.
(20 m/s) sin(40º) = 1/2 g t
t = (40 m/s) sin(40º) / g
t ≈ 2.6 s
(b) At its maximum height, the ball has zero vertical velocity. In the vertical direction, the ball is in free fall and only subject to the downward acceleration g. So
0² - ((20 m/s) sin(40º))² = 2 (-g) y
where y in this equation refers to the maximum height of the ball. Solve for y :
y = ((20 m/s) sin(40º))² / (2g)
y ≈ 8.4 m
A spherical balloon has a radius of 7.15 m and is filled with helium. How large a cargo can it lift, assuming that the skin and structure of the balloon have a mass of 930 kg? Neglect the buoyant force on the cargo volume itself.
Answer:
m = 876.71 kg
Explanation:
This is an exercise of Archimedes' principle, which states that the thrust on a body is equal to the weight of the dislodged liquid
B = ρ g V
therefore the load that the balloon can lift is
B - W_structure - w_load = 0
w_load = B - W_structure
The volume of the balloon is
v = 4/3 π r³
let's substitute
w_carga = rho g 4/3 π r³ - m_structure g
the air density at T = 25ºc is ρ = 1.18 kg / m³
let's calculate
w_load = 1.18 9.8 4/3 π 7.15³ - 930 9.8
w_load = 17705,77 - 9114
w_ load = 8591.77 N
this corresponds to a mass of
w_load = m g
m = w_load / g
m = 8591.77 / 9.8
m = 876.71 kg
7) A moving object is rolling on a surface that is 5 m off the ground. The object is moving at a constant speed of 4 m/s. If the object is 3.2 kg, what is the final energy of the ball after rolling for 10 m, assuming friction is negligible?
156.8 J
131.2 J
182.4 J
25.6 J
8) A spreadsheet application is used to create a computational model of the energy experienced by a pendulum. How do the energy values of the pendulum relate?
The sum of the potential energy and the kinetic energy is always constant.
The sum of the potential energy and the kinetic energy is always 0.
The potential energy is always greater than the kinetic energy.
The kinetic energy is always equal to the potential energy.
131.2 J and The last one on number 8
I gave the same answer and it passed.
7) The final energy of the ball after rolling for 10 m is 182.4 J so, option C is correct.
8) When friction is negligible the total energy is the sum of kinetic and potential energy is constant so, option A is correct.
What is energy?Energy is the ability to perform work in physics. It could exist in several different forms, such as potential, kinetic, thermal, electrical, chemical, radioactive, etc. Additionally, there is heat and work, which is energy being transferred from one body to another.
Given:
A moving object is rolling on a surface that is 5 m off the ground,
The speed of the object, v = 4 m/s,
The mass of the object, m = 3.2 kg,
Calculate the kinetic energy after 10 meters as shown below,
KE = 1/2 × 4² × 3.2
KE = 25.6 J
Calculate the potential energy as shown below,
PE = 3.2 × 9.8 × 5
PE = 156.8 J
Thus, total energy = KE + PE
The total energy = 25.6 + 156.8
The total energy = 182.4 J
8) when there is no resistance. Combined mechanical energy I.e. the total amount of kinetic and potential energy is constant.
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You want to lean your dad's ladder on a smooth wall. If the mass of ladder is 4.42 kg and coefficient
of friction of the floor is 0.53, what is the minimum angle, theta-min at which the ladder does nofip? What
do you think the maximum angle theta-max could be? Sketch and label your free body diagram.
(5 marks)
Answer:
angle minimum θ = 41.3º
Explanation:
For this exercise let's use Newton's second law in the condition of static equilibrium
N - W = 0
N = W
The rotational equilibrium condition, where we place the axis of rotation on the wall
We assume that counterclockwise rotations are positive
fr (l sin θ) - N (l cos θ) + W (l/2 cos θ) = 0
the friction force formula is
fr = μ N
fr = μ W
we substitute
μ m g l sin θ - m g l cos θ + mg l /2 cos θ = 0
μ sin θ - cos θ + ½ cos θ= 0
μ sin θ - ½ cos θ = 0
sin θ / cos θ = 1/2 μ
tan θ = 1/2 μ
θ = tan⁻¹ (1 / 2μ)
θ = tan⁻¹ (1 (2 0.57))
θ = 41.3º
Question 1 of 5
In which way are electromagnetic waves different from mechanical waves?
Answer:
Electromagnetic waves differ from mechanical waves in that they do not require a medium to propagate. This means that electromagnetic waves can travel not only through air and solid materials, but also through the vacuum of space.
Explanation:
Answer:
In which way are electromagnetic waves different from mechanical waves?Electromagnetic waves can travel through empty space
Explanation:
I took A P E X Quiz.
Suppose that an object undergoes simple harmonic motion, and its displacement has an amplitude A = 15.0 cm and a frequency f = 11.0 cycles/s (Hz). What is the maximum speed ( v ) of the object?
A. 165 m/s
B. 1.65 m/s
C. 10.4 m/s
D. 1040 m/s
Answer:
Maximum speed ( v ) = 10.4 m/s (Approx)
Explanation:
Given:
Amplitude A = 15.0 cm = 0.15 m
Frequency f = 11.0 cycles/s (Hz)
Find:
Maximum speed ( v )
Computation:
Angular frequency = 2πf
Angular frequency = 2π(11)
Angular frequency = 69.14
Maximum speed ( v ) = WA
Maximum speed ( v ) = 69.14 x 0.15
Maximum speed ( v ) = 10.371
Maximum speed ( v ) = 10.4 m/s (Approx)
What is the picture called that shows ALL the forces acting on an object?
Answer:
Free Body Diagram
Explanation:
Such picture is called the "free body diagram"
If a person Travels 100 metre due east and then returns to the same place his total displacement is 200. (needed ASAP)
A. True
B. False
Distance is the total path covered by the object
Here, 200 m is the distance covered by the person and NOT the displacement
Displacement of an object is nothing more than the shortest path between the initial and the final point
If the person travelled 100m and came back, his initial and final point will remain the same which means that he will have a displacement of 0 m
For satellite travelling on circular orbit if radius of the orbit increased 4 times then the period of the satellite increased *
2
4
8
none of the above
Answer:
8
Explanation:
Based on the situation above, choose the CORRECT type of error.
The reading from the timer was not accurate because some of
the timer's display was missing and broken."
Answer:
what's your question I can't understand
How many turns are needed in a solenoid of radius 10 cm and length 20 cm for its self-inductance to be 6.0 H?
A) 30
B) 74
C) 500
D) 550
E) 5500
Answer:
B
Explanation:
74 turns are needed in a solenoid of radius 10 cm and length 20 cm for its self-inductance to be 6.0 H
what is solenoid?The solenoid is a type of electromagnet and the main purpose of the solenoid is to generate a controlled magnetic field through a coil wound into a tightly packed helix.
The solenoid is a coil structure of wire, and the plunger is made of soft iron. The magnetic field is formed around it when an electric current passes through it and draws the plunger in.
the solenoid is responsible for converting electrical energy into mechanical work, these were the results of the development of an efficient and greater strength offering magnets which was discovered in the year 1823.
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A 6.0-kilogram cart initially traveling at 4.0 meters per second east accelerates uniformly at 0.50 meter per second squared east for 3.0 seconds. What is the speed of the cart at the end of this 3.0 second interval? A. 1.5 m/s B. 5.5 m/s G. 3.0 m/s D. 7.0 m/s
Answer: 5.5m/s
Explanation:
vf=vi+at
vf= 4.0m/s + (0.50m/s^2)(3.0s)
The speed of the cart at the end of this 3.0 second interval is 5.5 meter per seconds.
Given the following data:
Initial velocity = 4 m/sMass of cart = 6 KgAcceleration = 0.5 [tex]m/s^2[/tex]Time = 3 secondsTo find the speed of the cart at the end of this 3.0 second interval, we would use the first equation of motion;
[tex]V = U + at[/tex]
Where:
U is the initial velocity.V is the final velocity. a is the acceleration. t is the time measured in seconds.Substituting the given parameters into the formula, we have;
[tex]V = 4 + 0.5(3)\\\\V = 4 + 1.5[/tex]
Final velocity, V = 5.5 m/s.
Therefore, the speed of the cart at the end of this 3.0 second interval is 5.5 meter per seconds.
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that delivers oxygen to your body and In the video your blood is compared to a picks up CO2 to be released out when you breath. PLEASE I NEED A ANSWER
A rock rolls down a hill. Which form of energy is this an example of? (2 points)
a
Chemical
b
Electrical
c
Mechanical
d
Thermal
Answer:
c.Mechanical
Explanation:
You are riding a bicycle. If you apply a forward force of 125 N, and you and
the bicycle have a combined mass of 82 kg, what will be the forward
acceleration of the bicycle? (Assume there is no friction.)
I WILL GIVE YOU POINTs
Answer:
1.52g
Explanation:
Given parameters:
Force = 125N
Mass combined = 82kg
Unknown:
Acceleration of the bicycle = ?
Solution:
From Newton second law of motion suggests that:
Force = mass x acceleration
Acceleration = [tex]\frac{force}{mass}[/tex] = [tex]\frac{125}{82}[/tex] = 1.52g
Answer:
The answer that was correct for me was A. 55 N pulling left, and 16 N, 17N p
pulling Right
Explanation:
The components of vector Upper A Overscript right-arrow EndScripts are Ax and Ay (both positive), and the angle that it makes with respect to the positive xaxis is θ. Find the angle θ if the components of the displacement vector Upper A Overscript right-arrow EndScripts are:
(a) Ax = 12 m and Ay = 12 m,
(b) Ax= 19 m and Ay = 12 m, and
(c) Ax = 12 m and Ay = 19 m.
(a) θ = Number____________ Units____
(b) θ = Number____________ Units____
(c) θ = Number ____________Units____
Answer:
(a) θ = 45° = 0.78 rad
(b) θ = 32.27° = 0.56 rad
(c) θ = 57.27° = 1 rad
Explanation:
When a vector is resolved into its rectangular components, the formula for the direction angle of the vector with positive x-axis is given as:
tan θ = Ay/Ax
θ = tan⁻¹(Ay/Ax)
(a)
Ax = 12 m
Ay = 12 m
θ = tan⁻¹(12 m/ 12 m)
θ = tan⁻¹(1)
θ = 45° = 0.78 rad
(b)
Ax = 19 m
Ay = 12 m
θ = tan⁻¹(12 m/19 m)
θ = tan⁻¹(0.6315)
θ = 32.27° = 0.56 rad
(c)
Ax = 12 m
Ay = 19 m
θ = tan⁻¹(19 m/12 m)
θ = tan⁻¹(1.58333)
θ = 57.27° = 1 rad
Which energy transformation occurs after a skydiver reaches terminal velocity? Gravitational potential energy transforms into thermal energy. Gravitational potential energy transforms into kinetic energy. Kinetic energy transforms into thermal energy. Kinetic energy transforms into gravitational potential energy. The answer is A. just took it
Answer:
a
Explanation:
The energy transformation occurs after a skydiver reaches terminal velocity is follows as;
A. Gravitational potential energy transforms into thermal energy.
B. Gravitational potential energy transforms into kinetic energy.
What is the gravitational potential energy?The skydiver, when he is located at a certain height h above the ground, possesses gravitational potential energy, equal to:
U = mgh
where m is the mass of the skydiver, g is the gravitational acceleration and h is the height above the ground.
The skydiver gravitational potential energy decreases as the altitude decreases and his kinetic energy store increases as his speed increases.
When a skydiver jumps out of a plane, the energy transfers take place as;
The skydiver's kinetic energy store increases as their speed increases and the thermal store of the air and the skydiver increases, as there is friction between the skydiver and the air particles.
In the given situation, both options A and B are correct.
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At what speed does a 2000 kg compact car have the same kinetic energy as a 18000 kg truck going 21 km/hr?
Answer:
v = 17.5 m/s = 63 km/h
Explanation:
The general expression for the kinetic energy of one moving object is as follows:[tex]K = \frac{1}{2}*m *v^{2} (1)[/tex]
where m = mass of the object, v= speed of the object.
In order to get the value of the kinetic energy of the truck in Joules, we need to convert km/hr to m/s first, as follows:[tex]21 km/hr * \frac{1 hr}{3600s}*\frac{1000m}{1 km} = 5.83 m/seg (2)[/tex]
Now, replacing (2) and m = 18000 kg in (1), we get:[tex]K = \frac{1}{2}*18000 kg *(5.83m/s)^{2} = 306250 J (3)[/tex]
This value must be the same for the 2000 kg compact car, so we can write:[tex]K = 306250 J = \frac{1}{2}*2000 kg *v^{2} (4)[/tex]
Solving for v, we get:[tex]v = \sqrt{\frac{306250}{1000} (m/s)2} = 17. 5 m/s = 63 km/h (5)[/tex]
A cyclist is riding along at a speed of 20.7 when she decides to apply the brakes which gave a deceleration applied was a rate of -3.4 m/s2 over the span of 7.8 s. What distance does she travel over that period of time.
Answer:
The distance is 58.03 m
Explanation:
Constant Acceleration Motion
It occurs when the velocity of an object changes by an equal amount in every equal period of time.
Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:
[tex]v_f=v_o+at[/tex]
The distance traveled by the object is given by:
[tex]\displaystyle x=v_o.t+\frac{a.t^2}{2}[/tex]
The conditions of the problem state the cyclist has an initial speed of v0=20.7 m/s during t=7.8 seconds and acceleration of -3.4 m/s^2.
The final speed is:
[tex]v_f=20.7+(-3.4)\cdot 7.8[/tex]
[tex]v_f=20.7-26.52[/tex]
[tex]v_f=-5.82\ m/s[/tex]
Note the cyclist has stopped and come back because his speed is negative. Now calculate the distance:
[tex]\displaystyle x=20.7\cdot 7.8+\frac{(-3.4)\cdot 7.8^2}{2}[/tex]
[tex]\displaystyle x=161.46-103.43[/tex]
x=58.03 m
A plane flying at a speed of 59.1 m/s is dropping a package 521 m above the intended target. How long does it take for the package to hit the ground?
A.) 10.3 seconds
B.) 0.1 seconds
C.) 106.3 seconds
D.) 8.8 seconds
Answer:
t = 10.31 seconds which agrees with answer option A)
Explanation:
Notice that the horizontal velocity of the plane imparted to the package, doesn't affect the vertical motion for which the original (vertical velocity) is zero.
Then the equation for the distance travelled by the package is:
d = (1/2) a t^2
in our case, d = 521 m, and a = g (acceleration of gravity 9.8 m/s^2)
then we can solve for t in the equation:
521 = 9.8/2 t^2
t^2 = 521/4.9
t^2 = 106.32 s^2
t = 10.31 seconds