Answer:
Podemos decir que la presión que se aplica sobre un liquido encerrado en un tanque es de 5000 Pa.
Explanation:
Answer:
Podemos decir que la presión que se aplica sobre un liquido encerrado en un tanque es de 5000 Pa.
Explanation:
Please help I need this done within 30 mins
It may be thinner and more dense? I’m not too experienced in the study of Earth’s crust. However, I know enough to remember that the earths crust is thin.
How far did you travel in 10 hours if you drove at a constant speed of 5km/hr? *
Answer:
you drove 50km
Explanation:
10×5 hope this helps
Answer:
50 Km
Explanation:
This is how far you have got on your journey if traveling like this.
Please Mark as Brainliest
Hope this Helps
plz answer fast the question
Answer:
Angle of incidence = 20°
Angle of reflection = 20°
Explanation:
Applying,
The first Law of Refraction: The incident ray, the reflected ray and the normal at the point of incidence all lies in the plane.
From the diagram,
Angle of incidence = 90-70
Angle of incidence = 20°
From the law of reflection,
Angle of incidence = Angle of reflection
Therefore,
Angle of reflection = 20°
True or False. A person who is nearsighted cannot see objects that are close to them clearly.
Answer:
false
Explanation:
hope it works
Hai điện tích điểm Q1 = 8 C, Q2 = –6
C đặt tại hai điểm A, B cách nhau 0,1
m trong không khí. Tính cường độ điện
trường do hai điện tích này gây ra tại
điểm M, biết MA = 0,2 m
Answer:
English please
Explanation:
I don't understand the question
Write down the chemical formula and the ratio in which the four elements combine with each other in the compound Sodium- bicarbonate
Answer: The chemical formula is [tex]NaHCO_{3}[/tex] and the ratio in which the four elements combine with each other in the compound Sodium- bicarbonate is 1 : 1 : 1 : 3.
Explanation:
The chemical formula of sodium bicarbonate is [tex]NaHCO_{3}[/tex].
In this compound, there are 1 sodium atom, 1 hydrogen atom, 1 carbon atom and three oxygen atoms present.
Therefore, the ratio of atoms is 1 : 1 : 1 : 3
Thus, we can conclude that the chemical formula is [tex]NaHCO_{3}[/tex] and the ratio in which the four elements combine with each other in the compound Sodium- bicarbonate is 1 : 1 : 1 : 3.
A girl and her bicycle have a total mass of 40.0 kg. At the top of the hill her speed is 5.0 m/s, and her speed doubles as she rides down the hill. The hill is 10.0 m high and 100 m long. How much kinetic energy and potential energy is lost to friction
Answer:
The kinetic energy and potential energy lost to friction is 2,420 J.
Explanation:
Given;
total mass, m = 40 kg
initial velocity of the girl, Vi = 5 m/s
hight of the hill, h = 10 m
length of the hill, L = 100 m
initial kinetic energy of the girl at the top hill:
[tex]K.E_{i} = \frac{1}{2} mv_i^2 = \frac{1}{2} \times 40 \times (5)^2\\\\K.E_{i} = 500 \ J[/tex]
initial potential energy of the girl at the top hill:
[tex]P.E_{i} = mgh_i = 40 \times 9.8 \times 10\\\\P.E_{i}= 3920 \ J[/tex]
Total energy at the top of the hill:
E = 500 J + 3920 J
E = 4,420 J
At the bottom of the hill:
final velocity = double of the initial velocity = 2 x 5 m/s = 10 m/s
hight of the hill = 0
final kinetic energy of the girl at the bottom of the hill:
[tex]K.E_{f} = \frac{1}{2} mv_f^2 \\\\K.E_f = \frac{1}{2} \times 40 \times (10)^2 = 200 0 \ J[/tex]
final potential energy of the girl at the bottom of the hill:
[tex]P.E_f = mgh_f = 40 \times 9.8 \times 0 = 0[/tex]
Based on the principle of conservation of energy;
the sum of the energy at the top hill = sum of the energy at the bottom hill
The energy at the bottom hill is less due to energy lost to friction.
[tex]E_{friction} \ + E_{bottom}= E_{top}\\\\E_{friction} = E_{top} - E_{bottom}\\\\E_{friction} = 4,420 \ J - 2,000 \ J\\\\E_{friction} = 2,420 \ J[/tex]
Therefore, the kinetic energy and potential energy lost to friction is 2,420 J.
1 point
Q.29. A stone has a weight of 5.7 N.
The gravitational field strength g is 10
N/kg.What is the mass of the stone?
O A 0.57 kg
Answer:
weight/mass = gravitational field strength
Given :
Weight of stone = 5.7 N
Gravitational field strength (g) = 10 N/kg
Taking Mass of stone x
=> 5.7/x = 10
x = 10 * 5.7
x = 57 kg
Therefore mass of stone is 57 kg
The speed of a 2.0-kg object changes from 30 m/s to 40 m/s during a 5.0-second time interval.
During this same time interval, the velocity of the object changes its direction by 90°. What is the
magnitude of the average total force acting on the object during this time interval?
a. 30 N
b. 20 N
c. 15 N
d. 40 N
e. 10 N
Which is the correct answer?
Answer:
F = 2 * 30 / 5 = 12 N to stop forward motion
F = 2 * 40 / 5 = 16 N to accelerate to 90 degrees
(12^2 + 16^2)^1.2 = 20 N average force applied
The magnitude of the average total force acting on the object during this time interval is 20 N.
The given parameters:
Mass of the object, m = 2.0 kgInitial velocity, u = 30 m/sFinal velocity, v = 40 m/sTime of motion, t = 5.0 sThe magnitude of the average total force acting on the object during this time interval is calculated as follows;
[tex]F = \frac{mv }{t} \\\\F_1 = \frac{2(40)}{5} \\\\F_1 = 16\ N\\\\F_2= \frac{2(30)}{5} \\\\F_2 = 12 \ N\\\\F = \sqrt{F_1^2 + F_2^2} \\\\F = \sqrt{16^2 + 12^2} \\\\F = 20 \ N[/tex]
Learn more about resultant force here: https://brainly.com/question/25239010
Define Potential Energy
Begin by defining potential energy in your own words within one concise eight word sentence
Answer:
potential energy is a type of energy an object has because of it's position
7. An electric train moving at 20km/hrs
. Accelerates to a speed of 30km/hrs. in
20 sec, find the distance travelled in meters during the period of
acceleration
Answer
NB:
- speed, U is measure in m/s
- acceleration, a is measured in m/s²
-time t in seconds , s
Therefore conversation must be made
Speed U = 20km/hrs
=20km÷1hr
But 20km= 20×1000=20000m
1hr= 1×60min×60sec=3600s
U=20000÷3600=5.56m/s
a=30km/hrs
=30km÷1hr
But 30km=30×1000=30000
1hr=3600s
a=30000÷3600=8.33m/s²
From the equation of motion
S=Ut + ½ at².
Where s= distance
S = 5.56m/s × 20s + ½(8.33m/s²)(20s)²
S = 1777.3m
A cement block accidentally falls from rest from the ledge of a 81.5-m-high building. When the block is 14.0 m above the ground, a man, 2.00 m tall, looks up and notices that the block is directly above him. How much time, at most, does the man have to get out of the way
Answer:
0.41s
Explanation:
solve for t and y
,........
A car starts from rest. If its acceleration is 1.5 m/s2 in 1.5 seconds, then calculate the
distance travelled by it.
Answer:
1.6875 m
Explanation:
Here, we are given that,
Initial velocity (u) = 0 (as it starts from rest)Acceleration (a) = 1.5 m/s²Time taken (t) = 1.5 sThen,
Distance (s) = ?By using the second equation of motion,
★ s = ut + ½at²
s is distanceu is initial velocityt is timea is acceleration⇒ s = 0(1.5) + ½ × 1.5 × (1.5)²
⇒ s = ½ × 1.5 × 2.25
⇒ s = ½ × 1.5 × 2.25
⇒ s = ½ × 3.375
⇒ s = 1.6875 m
∴ Distance travelled by it is 1.6875 m.
Calculate the rms speed of helium atoms near the surface of the Sun at a temperature of about 5100 K. Express your answer to two significant figures and include the appropriate units.
Answer:
[tex]V_{rms}=5.6*10^3m/s[/tex]
Explanation:
From the question we are told that:
Temperature [tex]T=5100K[/tex]
Generally the equation for RMS Speed is mathematically given by
[tex]V_{rms}=\sqrt{\frac{3kT}{m}}[/tex]
Where
[tex]K=Boltzman's constant[/tex]
[tex]K=1.38*10^{-23}[/tex]
And
[tex]M=molecular mass[/tex]
[tex]M=4*1.67*10^{-27}[/tex]
[tex]V_{rms}=\sqrt{\frac{3(1.38*10^{-23})5100}{4*1.67*10^{-27}}}[/tex]
[tex]V_{rms}=5.6*10^3m/s[/tex]
7. A car is travelling along a road at 30 ms when a pedestrian steps into the road 55 m ahead. The
driver of the car applies the brakes after a reaction time of 0.5 s and the car slows down at a rate of
10 ms. What happens?
Answer: Car collide with man
Explanation:
Given
Speed of car is [tex]u=30\ m/s[/tex]
Distance of the man from the car is [tex]s=55\ m[/tex]
Reaction time [tex]t_r=0.5\ s[/tex]
Rate of deceleration [tex]a_d=-10\ m/s^2[/tex]
Distance traveled in the reaction time [tex]d_o=30\times 0.5=15\ m[/tex]
Net effective distance to cover [tex]d=55-15=40\ m[/tex]
Distance required to stop the car
[tex]\Rightarrow v^2-30^2=2(-10)(s)\\\Rightarrow 0-900=-20s\\\Rightarrow s=45\ m[/tex]
Require distance is more than that of net effective distance. Hence, car collides with the man.
E=kq/r^2 chứng minh điện thế V=kq/r từ mối liên hệ giữa điện trường E và điện thế V
Answer:
hindi ko maintindihan teh
I need help with physics question.
(D)
Explanation:
Assuming that the charge q is moving perpendicular to the magnetic field B, the magnitude of the force experienced by the charge is
F = qvB = (2.9×10^-17 C)(4.0×10^5 m/s)(1.7T)
= 2.0×10^-11 N
Two identical loudspeakers 2.0 m apart are emitting sound waves into a room where the speed of sound is 340 m/sec. John is standing 5.0m in front of one of the speakers, perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound. What is the lowest possible frequency of sound for which this is possible?
Answer: The lowest possible frequency of sound for which this is possible is 212.5 Hz.
Explanation:
It is known that formula for path difference is as follows.
[tex]\Delta L = (n + \frac{1}{2}) \times \frac{\lambda}{2}[/tex] ... (1)
where, n = 0, 1, 2, and so on
As John is standing perpendicular to the line joining the speakers. So, the value of [tex]L_{1}[/tex] is calculated as follows.
[tex]L_{1} = \sqrt{(2)^{2} + (5)^{2}}\\= 5.4 m[/tex]
Hence, path difference is as follows.
[tex]\Delta L = (5.4 - 5) m = 0.4 m[/tex]
For lowest frequency, the value of n = 0.
[tex]\Delta L = (0 + \frac{1}{2}) \times \frac{\lambda}{2} = \frac{\lambda}{4}[/tex]
[tex]\lambda = 4 \Delta L[/tex]
where,
[tex]\lambda[/tex] = wavelength
The relation between wavelength, speed and frequency is as follows.
[tex]\lambda = \frac{\nu}{f}\\4 \Delta L = \frac{\nu}{f}\\[/tex]
where,
[tex]\nu[/tex] = speed
f = frequency
Substitute the values into above formula as follows.
[tex]f = \frac{\nu}{4 \Delta L}\\f = \frac{340}{4 \times 0.4 m}\\= 212.5 Hz[/tex]
Thus, we can conclude that the lowest possible frequency of sound for which this is possible is 212.5 Hz.
Four identical masses m are evenly spaced on a frictionless 1D track. The first mass is sent at speed v toward the other three. As the masses collide they stick together. Mass 1 sticks to 2, then 1 2 sticks to 3, then 1 2 3 sticks to 4. When the combined 1 2 3 mass strikes mass 4, by what percentage does the speed decrease in %
Answer:
The speed decreases 75%.
Explanation:
Since no friction present, assuming no external forces acting during the three collisions, total momentum must be conserved.For the first collission, only mass 1 is moving before it, so we can write the following equation:[tex]p_{i} = p_{f} = m*v_{o} (1)[/tex]
Since both masses are identical, and they stick together after the collision, we can express the final momentum as follows:[tex]p_{f1} = 2*m*v_{1} (2)[/tex]
From (1) and (2) we get:v₁ = v₀/2 (3)Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1 and 2 combined together before colliding with mass 3 is just v₁, so the initial momentum prior the second collision (p₁) can be expressed as follows:[tex]p_{1} = 2*m*v_{1} = 2*m*\frac{v_{o} }{2} = m*v (4)[/tex]
Since after the collision the three masses stick together, we can express this final momentum (p₂) as follows:[tex]p_{2} = 3*m*v_{2} (5)[/tex]
From (4) and (5) we get:v₂ = v₀/3 (6)Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1, 2 and 3 combined together before colliding with mass 4 is just v₂, so the initial momentum prior the third collision (p₂) can be expressed as follows:[tex]p_{2} = 3*m*v_{2} = 3*m*\frac{v_{o} }{3} = m*v (7)[/tex]
Since after the collision the four masses stick together, we can express this final momentum (p₃) as follows:[tex]p_{3} = 4*m*v_{3} (8)[/tex]
From (7) and (8) we get:v₃ = v₀/4This means that after the last collision, the speed will have been reduced to a 25% of the initial value, so it will have been reduced in a 75% regarding the initial value of v₀.16. The sum of kinetic energies in an object.
17. The essential device in power plants that convert mechanical
energy to electricity.
18. The device that converts electricity back to mechanical energy
19. The only EM wave that is seen by naked eye.
20. A device that converts light to electricity.
A lens with a focal length of 15 cm is placed 45 cm in front of a lens with a focal length of 5.0 cm .
Required:
How far from the second lens is the final image of an object infinitely far from the first lens?
Answer:
the required distance is 6 cm
Explanation:
Given the data in the question;
f₁ = 15 cm
f₂ = 5.0 cm
d = 45 cm
Now, for first lens object distance s = ∝
1/f = 1/s + 1/s' ⇒ 1/5 = 1/∝ + 1/s'
Now, image distance of first lens s' = 15cm
object distance of second lens s₂ will be;
s₂ = 45 - 15 = 30 cm
so
1/f₂ = 1/s₂ + 1/s'₂
1/5 = 1/30 + 1/s'₂
1/s'₂ = 1/5 - 1/30
1/s'₂ = 1 / 6
s'₂ = 6 cm
Hence, the required distance is 6 cm
The distance of the final image from the first lens will be is 6 cm.
What is mirror equation?The mirror equation expresses the quantitative connection between object distance (do), image distance (di), and focal length (fl).
The given data in the problem is;
f₁ is the focal length of lens 1= 15 cm
f₂ s the focal length of lens 2= 5.0 cm
d is the distance between the lenses = 45 cm
From the mirror equation;
[tex]\frac{1}{f} = \frac{1}{s} +\frac{1}{s'} \\\\ \frac{1}{5} = \frac{1}{\alpha} +\frac{1}{s'} \\\\[/tex]
If f₁ is the focal length of lens 1 is 15 cm then;
[tex]s'=15 cm[/tex]
f₂ s the focal length of lens 2= 5.0 cm
s₂ = 45 - 15 = 30 cm
From the mirror equation;
[tex]\frac{1}{f_2} = \frac{1}{s_1} +\frac{1}{s_2'} \\\\ \frac{1}{5} = \frac{1}{30} +\frac{1}{s_2'} \\\\ \frac{1}{s_2'}= \frac{1}{5} -\frac{1}{30} \\\\ \frac{1}{s_2'}= \frac{1}{6} \\\\ \rm s_2'= 6 cm[/tex]
Hence the distance of the final image from the first lens will be is 6 cm.
To learn more about the mirror equation refer to the link;
https://brainly.com/question/3229491
Celestial Events, such as rise, set or transit times are represented by the intersection of various diagonal lines (and loops) with the horizontal and vertical lines, this will allow us to determine what about the Celestial Event?
a) Distance
b) Latitude
c) Time and Date
d) Gamma Rays
Answer:
C) time and date
Explanation:
Celestial event is an astronomical phenomenon. This involves the conjunction of one or more celestial objects such as lunar and solar eclipse or meteor shower. The intersecting horizontal and vertical lines allow the astrologists to determine the time and date of the celestial event.
Which conclusion can be made based on the information in the table?
Wave speed and wavelengths can vary inversely to produce the same frequency.
O Frequency and wave speed can vary directly to produce the same wavelength.
O Wavelengths and frequency can vary inversely to produce the same wave speed.
O Frequency and wavelengths can vary directly to produce the same wave speed.
Mark this and return
Save and Exit
Next
Sul
Previous Activity
Answer:
The correct option is (b).
Explanation:
The relation between the wavelength and frequency is given by :
[tex]\lambda=\dfrac{v}{f}[/tex]
Where
v is the wave speed
f is the frequency of a wave
It is clear from the above equation that the wavelengths and frequency can vary inversely to produce the same wave speed.
A rocket at fired straight up from rest with a net upward acceleration of 20 m/s2 starting from the ground. After 4.0 s, the thrusters fail and the rocket continues to coast upward with insignificant air resistance. (a) What is the maximum height reached by the rocket
Answer:
The maximum height reached by the rocket is 486.53 m
Explanation:
Given;
initial velocity of the rocket, u = 0
acceleration of the rocket, a= 20 m/s²
duration of the rocket first motion, t = 4 s
The distance traveled by the rocket before its thrust failed
h₁ = ut + ¹/₂at²
h₁ = 0 + ¹/₂ x 20 x 4²
h₁ = 160 m
The second distance moved by the rocket is calculated as follows;
The velocity of the rocket before its thrust failed;
v = u + at
v = 0 + 20 x 4
v = 80 m/s
This becomes the initial velocity for the second stage
At maximum height, the final velocity = 0
[tex]v_f^0 = v_i^2 - 2gh_2\\\\0 = (80)^2 - (2 \times 9.8)h_2\\\\0 = 6400 - 19.6h_2\\\\19.6h_2 = 6400\\\\h_2 = \frac{6400}{19.6} \\\\h_2 = 326.53 \ m[/tex]
The maximum height reached by the rocket = h₁ + h₂
= 160 + 326.53
= 486.53 m
A 2000 kg car experiences a braking force of 10000N and skids to a 14 m stop. What was the speed of the car just before the brakes.
Answer:
V = 11.83 m/s
Explanation:
Given the following data;
Mass = 2000 kg
Force = 10000N
Distance = 14 m
To find the final velocity of the car;
First of all, we would determine the acceleration of the car;
Acceleration = force/mass
Acceleration = 10000/2000
Acceleration = 5 m/s²
Next, we would use the third equation of motion to find the final velocity;
[tex] V^{2} = U^{2} + 2aS [/tex]
Where;
V represents the final velocity measured in meter per seconds.
U represents the initial velocity measured in meter per seconds.
a represents acceleration measured in meters per seconds square.
S represents the displacement measured in meters.
Substituting into the equation, we have;
V² = 0² + 2*5*14
V² = 0 + 140
V = √140
V = 11.83 m/s
Kaseem is taking his bicycle for a ride. His bicycle is a system, and its main purpose is to provide transportation. What is the main input into this system? What is the desired output of this system?
which team won the champions league in 2020 2021
Answer:
Chelsea F.C
Explanation:
Chelsea F.C
Soccer
Are you aware of human rights violation happening in the community and explain
Answer:
Individuals who commit serious violations of international human rights or humanitarian law, including crimes against humanity and war crimes, may be prosecuted by their own country or by other countries exercising what is known as “universal jurisdiction.”
1 airplane
travel due north at 300 km while another airplane travels Due South and 300 km are there speed the same or their velocities the same
Answer:
Explanation:
Speed is scalar and velocity is vector. Vector values imply direction as well as magnitude. Therefore, speed and velocity are not the same. The speeds of these 2 planes are the same at 300km/hr, but the velocity of the plane traveling north is +300km/hr while the velocity of the plane traveling south is -300km/hr if we define north as positive and south as negative.
Please help I need this done
