Answer:
The pressure of water progressively increases as the depth of the water increases. The pressure increases as the depth of a point in a liquid increases. The walls of the vessel in which liquids are held are likewise subjected to pressure. The sideways pressure exerted by liquids increases as the liquid depth increases.
A boat is able to move at 7.6 m/s in still water. If the boat is placed on the south shore of a river (water current of 3.4 m/s [SE]), and the captain wants to head straight across to the north shore:
a) In what direction should the captain point the boat?
b) Calculate the time it will take to cross (the river is 212.0 m from the south to the north shore).
Answer:
I don't get it rewrite please
What is a downside of nonprofit fitness centers?
They offer fewer luxury services than for-profit health centersThey offer fewer luxury services than for-profit health centers , ,
Very few people live near a nonprofit fitness center.Very few people live near a nonprofit fitness center. , ,
Members tend to be wealthy and lack diversity. Members tend to be wealthy and lack diversity. , ,
Membership costs more than it does in for-profit centers.Membership costs more than it does in for-profit centers. , ,
they offer fewer luxury services because they lack the money to provide them.
The image shows a facility that converts the energy of moving water into
electrical energy. What is one advantage of using this technology in place of a
coal-burning power plant?
A. It is fueled by a nonrenewable resource.
B. It causes no harm to ecosystems.
оооо
C. It produces more water pollution.
D. It does not emit greenhouse gases.
Answer:
D. It does not emit greenhouse gases.
Explanation:
D. It does not emit greenhouse gases.
calculate the opportunity acost of the production of good x at each point .what law dose the ternd in those value exhibit ?
Answer:
a
Explanation:
5. An electron is moving north in a region where
the magnetic field is south. The magnetic force
exerted on the electron is:
A.Zero
B. up
C. down
D. east
E. west
Answer:
eastlove india
What is not one of the main uses of springs?
A. Car suspension
B. Bike suspension
C. The seasons
D. Clock making
A man on top of a tall building 60m high projects a body vertically upwards with an initial velocity of 20m/s
(1) find the maximum height above the ground reached by the body.
(2) the velocity with which the body hits the ground.
Answer:
1) 65 m
2) 40 m/s downward
Explanation:
Using for both questions the kinematic equation
v² = u² + 2as
and ignoring air resistance.
1) h = 60 + √(20²/(2(9.8))) = 64.51753...
2) v = √(20² + 2(9.8)(60)) = 39.69886...
which sources of associated with the line emission spectrum?
1) electric fire
2) neon street sign
3) red traffic light
4) sun
Answer:
Answer if this question neon street sign
Explanation:
may this answer is helpful for you
when 999mm is added to 100m ______ is the result
Answer:
what, 100.999m
Explanation:
convert 999 mm into meters, which is 0.999m and add that to a 100 m and that will make the total 100.999 m
The result of the addition of the two values is equal to 100.999 meters.
Given the following data:
Value 1 = 999 millimetersValue 2 = 100 metersTo determine the result of the addition of the two values:
First of all, we would convert the value in millimeter (mm) to meter (m) as follows:
Conversion:
1 millimeter = 0.001 meter
999 millimeter = X meter
Cross-multiplying, we have:
[tex]X = 0.001 \times 999[/tex]
X = 0.999 meter.
For the result:
[tex]Result = 0.999 +100[/tex]
Result = 100.999 meters.
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a model car moves round a circular path of radius 0.3m at 2 revolutions per secs what is its angular speed, the period of the car and the speed of the car
Answer:
a) T = 0.5 s
b) v = 1.2π m/s ≈ 3.77 m/s
Explanation:
It makes two revolutions in one second so makes one revolution in ½ second
circumference of the circle is
C = 2πr = 0.6π m
which it traverses in one time period
0.6π m / 0.5 s = 1.2π m/s
To solve this, we must be knowing each and every concept related to speed and its calculations. Therefore, the angular speed of a model car moves round a circular path of radius 0.3m at 2 revolutions per secs is 3.77 m/.
What is speed?Speed may be defined as the distance traveled by an item in the amount of time it requires to travel that distance. In other words, it measures how rapidly an item travels but does not provide direction.
Speed may be calculated in Science. The speed equation is a scientific formula that is used to calculate various types of speed.
Mathematically, the formula for speed can be given as
speed= distance/time
Values that are given
Time period= 0.5 s
Circumference = 2πr = 0.6π m
substituting all the given values in the above equation, we get
speed =0.6π m / 0.5 s
On calculations, we get
= 1.2π m/s
=3.77 m/s
Therefore, the angular speed of a model car moves round a circular path of radius 0.3m at 2 revolutions per secs is 3.77 m/.
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A rubber ball and a steel ball are dropped from the same height onto a concrete floor. They have the same mass, and lets assume that they both rebound to the same height after hitting the ground. How would the average forces experienced by the balls compare. Group of answer choices The steel ball would experience the greater average force The average forces would be the same The rubber ball would experience the greater average force
Answer:
The average forces would be the same
Explanation:
Both have the same velocity on impact as they fell from the same height.
Both have the same velocity after the bounce because they reach the same height.
Both have the same mass
Both will thus experience the same impulse because both have the same change in momentum.
Therefore both experience the same average force.
If the horizontal range and the max height of a body projected at an angle titre to the horizontal is K and Q respectively. Show that the muzzle velocity Vo is given by: Vo=√[2gQ + K²/8Q]
Explanation:
Recall that
[tex]K = \dfrac{v_0^2\sin2\theta}{g}\:\:\:\:\:\:\:\:\:(1)[/tex]
and
[tex]Q = \dfrac{v_0^2\sin^2\theta}{2g}\:\:\:\:\:\:\:\:\:(2)[/tex]
From Eqn(2), we can write
[tex]\sin\theta = \sqrt{\dfrac{2gQ}{v_0^2}}\:\:\:\:\:\:\:\:\:(3)[/tex]
Using the identity [tex]\sin\theta = 2\sin\theta \cos\theta[/tex], we can rewrite Eqn(1) as
[tex]\dfrac{gK}{2v_0^2} = \sin\theta \cos\theta[/tex]
Squaring the above equation, we get
[tex]\dfrac{g^2K^2}{4v_0^4} = \sin^2\theta \cos^2\theta[/tex]
[tex]\:\:\:\:\:\:\:\:\:=\sin^2\theta(1 - \sin^2\theta)\:\:\:\:\:\:\:(4)[/tex]
Use Eqn(3) on Eqn(4) and we will get the following:
[tex]\dfrac{g^2K^2}{4v_0^4} = \dfrac{2gQ}{v_0^2}(1 - \dfrac{2gQ}{v_0^2})[/tex]
This simplifies to
[tex]\dfrac{gK^2}{8v_0^2Q} = 1 - \dfrac{2gQ}{v_0^2}[/tex]
Rearranging this further, we get
[tex]1 = \dfrac{2gQ}{v_0^2} + \dfrac{gK^2}{8v_0^2Q}[/tex]
Putting [tex]v_0^2[/tex] to the left side, we get
[tex]v_0^2 = 2qQ + \dfrac{gK^2}{8Q}[/tex]
Finally, taking the square root of the equation above, we get the expression for the muzzle velocity [tex]v_0[/tex] as
[tex]v_0 = \sqrt{2gQ + \dfrac{gK^2}{8Q}}[/tex]
Help again! Thank you all xoxo
Answer:
Amount of charge
Explanation:
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Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 55.3 kg, down a theta= 79.6º slope at constant acceleration a=-4.3 m/s2, as shown in Figure (here we assume the positive direction is going down the slope. So the given acceleration is a negative value, it means its direction is going up the slope, slowing down as it moving downward). So, the coefficient of friction between the sled and the snow is 0.100. How many Joules of work is done by the tension in the rope as the sled moves 2.1 m along the hill? Use g= 10 m/s2.
The tension in the rope is doing a work of 1662.544 joules as the sled moves 2.1 meters along the hill.
In this case, we need to construct the Free Body Diagram of the sled-victim System in order to determine what Forces are doing Work. Then, we construct the respective Energy equation by Newton's Laws of Motion, Work-Energy Theorem and definition of Work.
Given that system experiments an uniform Acceleration, we must solve the resulting model for the work done by the Tension in the rope.
From the Free Body Diagram (see image attached), we see that both Weight of the sled and Friction between sled and snow are doing work in favor of gravity, whereas Tension forces is against gravity. Normal force is not doing work as its direction is perpendicular to the direction of motion. The energy equation of this system is:
[tex]-W_{T} + \mu\cdot m\cdot g \cdot s\cdot \cos \theta + m\cdot g\cdot s\cdot \sin \theta = m\cdot a\cdot s[/tex] (1)
Where:
[tex]W_{T}[/tex] - Work done by tension, in joules.
[tex]m[/tex] - Mass of the sled-victim system, in kilograms.
[tex]\mu[/tex] - Coefficient of kinetic friction, no unit.
[tex]g[/tex] - Gravitational acceleration, in meters per square second.
[tex]s[/tex] - Travelled distance, in meters.
[tex]\theta[/tex] - Slope angle, in sexagesimal degrees.
[tex]a[/tex] - Net acceleration of the sled-victim system, in meters per square second.
If we know that [tex]\mu = 0.100[/tex], [tex]m = 55.3\,kg[/tex], [tex]g = 10\,\frac{m}{s^{2}}[/tex], [tex]s = 2.1\,m[/tex], [tex]\theta = 79.6^{\circ}[/tex] and [tex]a = -4.3\,\frac{m}{s^{2}}[/tex], then the work done by the tension in the rope is:
[tex]-W_{T} + \mu\cdot m\cdot g \cdot s\cdot \cos \theta + m\cdot g\cdot s\cdot \sin \theta = m\cdot a\cdot s[/tex]
[tex]W_{T} = \mu\cdot m\cdot g \cdot s\cdot \cos \theta + m\cdot g\cdot s\cdot \sin \theta -m\cdot a\cdot s[/tex]
[tex]W_{T} = (0.100)\cdot \left(55.3\,kg\right)\cdot \left(10\,\frac{m}{s^{2}} \right)\cdot (2.1\,m)\cdot \cos 79.6^{\circ} + \left(55.3\,kg\right)\cdot \left(10\,\frac{m}{s^{2}} \right)\cdot (2.1\,m)\cdot \sin 79.6^{\circ} - (55.3\,kg)\cdot \left(-4.3\,\frac{m}{s^{2}} \right) \cdot (2.1\,m)[/tex]
[tex]W_{T} = 1662.544\,J[/tex]
The tension in the rope is doing a work of 1662.544 joules as the sled moves 2.1 meters along the hill.
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Someone help me with these questions please!
Answer:
a 25 and b 25
2. 26
60n
An electron is moving at speed of 6.3 x 10^4 m/s in a circular path of radius of 1.7 cm inside a solenoid the magnetic field of the solenoid is perpendicular to the plane of the electron's path. Find its relevatn motion.
Answer:
Here, m=9×10
−31
kg,
q=1.6×10
−19
C,v=3×10
7
ms
−1
,
b=6×10
−4
T
r=
qB
mv
=
(1.6×10
−19
)(6×10
−4
)
(9×10
−31
)×(3×10
7
)
=0.28m
v=
2πr
v
=
2πm
Bq
=
2×(22/7)×9×10
−31
(6×10
−4
)×(1.6×10
−19
)
=1.7×10
7
Hz
Ek=
2
1
mv
2
=
2
1
×(9×10
−31
)×(3×10
7
)
2
J
=40.5×10
−17
J=
1.6×10
−16
40.5×10
−17
keV
=2.53keV
Calculate the magnetic field and its direction at point P, which is 2.0 cm away from the top wire and 4.0 cm from the bottom wire. Assume both wires are infinitely long and each carries a current of 1.5 A.
Answer:
Explanation : B = μ₀i / 2πr
B1 = (2×10^-7) × 1.5/0.02
= 15 μT
B2 = (2×10^-7 ) × 1.5/0.04
= 7.5 μT
total field = 15 μT + 7.5 μT
= 22.5 μT
find the velocity of the object for all relevent times find the position of the object for all relevent times a softball is popped up vertically velocity of 32 m/s
Answer:
whats the formula
Explanation:
a bullet is dropped from the same height when another bullet is fired horizontally. they will hit the ground
Answer:
it will drop simultaneously
Two long straight wires are suspended vertically. The wires are connected in series, and a current from a battery is maintained in them. What happens to the wires? What happens if the battery is replaced by an a-c source?
Answer:
(i) When a battery is connected inseries to two long parallel wires, the currents in the two wires will be in opposite directions. Due to which a force of repulsion will be acting between them and they are moving further apart.
(ii) When a battery is connected in parallel to two long parallel wires, the currents in the two wires will be in same direction. Due to it, a force of attraction will be acting between them and they are coming closer to each other.
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Explanation:
Hope it will helps you lot!
A magnifying glass produces a maximum angular magnification of 5.4 when used by a young person with a near point of 20 cm. What is the maximum angular magnification obtained by an older person with a near point of 65 cm
Explain why liquid particles at a high pressure would need more
energy to change to a gas than liquid particles at a low pressure.
Answer:
Liquids evaporate faster as they heat up and more particles have enough energy to break away. The particles need energy to overcome the attractions between them. ... At this point the liquid is boiling and turning to gas. The particles in the gas are the same as they were in the liquid they just have more energy.
As a skydiver accelerates downward, what force increases? A. Gravity B. Thrust C. Air resistance D. Centripetal
Answer:
(A) Gravity is you're answer.
Explanation:
When an object or human is falling at an increased rate, The force of gravity is taking place.
Select the correct answer. Which of Newton's laws explains why your hands get red when you press them hard against a wall? A. Newton's law of gravity B. Newton's first law of motion C. Newton's second law of motion D. Newton's third law of motion
Answer:
D newton third law
Explanation:
good luck
If there is no slipping, a frictional force must exist between the wheels and the ground. In what direction does the frictional force from the ground on the wheels act
Answer:
tire advances to the right, the friction force must be directed to the left, that is, opposing the possible movement of the tire.
Explanation:
For the movement of the wheel to be composed of a rotating part and a translational part, it is necessary that there be a static friction force between the floor and the tire.
As the tire advances to the right, the friction force must be directed to the left, that is, opposing the possible movement of the tire.
A very long straight wire carries a 12 A current eastward and a second very long straight wire carries a 14 A current westward. The wires are parallel to each other and are 42 cm apart. Calculate the force on a 6.4 m length of one of the wires.
Answer: 5.12x10∧-4N
Explanation:
Force = I B L
L = 6.4m
Let Current (I) I₁ = I₂= 14A
Distance of the wire = 42cm = 0.42m
BUT
B = μ₀I / 2πr
=(2X10∧-7 X 12) / 0.42
B =5.714×10∧-6T
Force = I B L
Force = 14x [5.714×10-6]×6.4
Force = 5.12x10∧-4N
What is the final velocity if you have an initial velocity of 5 m/s with an acceleration of 3 m/s2 over a 4 second interval?
Answer: 17 m/s
Explanation:
Using the formula Vf=Vi+at
Vf=5m/s+3m/s^2(4s)
Vf= 17m/s
Which of the following exhibit the Tyndall Effect?
A flashlight beam through air in a room.
A laser beam through salt water.
Car headlights on a foggy night.
A laser beam through a soda drink.
Answer:
Probably all but (a)
The Tyndall Effect is caused by dispersion of the incident light by the individual molecules in the liquid.
Salt water and a foggy night will cause dispersion of the incident light.
A soda drink may also cause this dispersion, but and not sure.
A train moving west with an initial velocity of 20 m/s accelerates at 4 m/s2 for 10 seconds. During this tim
moves a distance of
meters.
Answer:
400m
Explanation:
[tex]x = v_{0}t + \frac{at^{2} }{2} \\x = 20*10+\frac{4*10x^{2} }{2} = 400m[/tex]The car has a mass of 0·50 kg. The boy
now increases the speed of the car to 6·0
ms-1 . The total radial friction between
the car and the track has a maximum
value of 7.0 N. Show by calculation that
the car cannot continue to travel in the circular path.
Answer:
A solenoid is a type of electromagnet, the purpose of which is to generate a controlled magnetic field through a coil wound into a tightly packed helix. The coil can be arranged to produce a uniform magnetic field in a volume of space when an electric current is passed through it.
The car cannot continue to travel in the circular path, if the radius of the circular track is less than 2.57 m.
What is meant by centripetal force ?Centripetal force is described as the force applied to a body that is travelling in a circular motion and is pointed in the direction towards the center of the circular path.
Here,
Mass of the car, m = 0.5 kg
Velocity of the car, v = 6 m/s
Radial friction between the car and the track, f = 7 N
The necessary centripetal force for the car to execute the circular motion is provided by the maximum radial frictional force between the car and the track.
So, the condition that the car cannot continue to travel in the circular path is that the centripetal force required is greater than the maximum radial friction.
So,
mv²/r > f
0.5 x 6²/r > 7
Therefore, the radius of the circular track,
r < 18/7
r < 2.57 m
Hence,
The car cannot continue to travel in the circular path, if the radius of the circular track is less than 2.57 m.
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