Answer:4
Explanation:
The clips are not affected by magnet strength
3. Your friend says your body is made up of more than 99.9999% empty space. What do you think?
Answer:
I would agree with the statement. it's not just the body, but everything that we see is almost 99.9999% empty space
List down the procedures for each swimming stroke 1.Crawl 2.Breaststoke 3.Butterflystroke 4. Backstoke
Answer: Swimming strokes are techniques that includes arm and leg movements to help push the swimmer against water and propel the swimmer forward.
Explanation:
There are different types of swimming strokes these includes:
--> FRONT CRAWL: This is the fastest of all the techniques. The procedure includes:
• the body is kept flat, facing down and in line with the water surface,
• As the swimmer proceeds with movements, the arms are alternately moved in a PULL (with your palms facing down pull in line with the body) and RECOVERY (with the hand closed to the upper thigh, lift one arm out of the water with a bent elbow) actions.
• As you finish the recovery phase, turn quickly side ways to take in some air.
• With ankles relaxed and flexible, point your toes behind you and kick up-and-down in a continuous motion from your thighs.
BUTTERFLY STROKE: The procedures for this technique includes:
• the body is kept flat, facing down and in line with the water surface.
• the arms are moved in three ways, the catch, pull and recovery movements. The Catch involves the arms being straight, shoulder width apart and palms facing down wards, press down and out against the water with both hands at the same time. The pull involves the hands being pulled towards the body in a semicircular motion. The recovery starts at the end of each pull, the arms are moved out and over the water simultaneously and is thrown forward into the starting position.
• the chin is being raised up at the recovery stage to draw in a breath while looking straight.
• With both legs together and toes pointed, kick downwards at the same time.
• the body is moved in a wave-like manner.
BREASTSTROKE: The procedure for this technique includes;
• the body is kept flat, facing down and in line with the water surface
• the arms are also moved in three ways. In the catch movements, with arms out straight and palms facing downwards, press down and out at the same time. With elevated elbows above the arms, pull hard towards the chest. Then while recovering, to reduce drag when pushing against water, the both palms are joined together Infront of the chest and pushed out until the arms are straight again.
• the head is lifted above water at the end of pulling movement to breath in air.
• bend your knees to bring your heel towards your bottom and make a circular motion outwards with your feet until they return to the starting position.
BACKSTROKE: The procedure for this technique includes
• the body kept flat while backing the water surface. But following the arm movement, it rows from side to side.
• the arms performs alternating and opposite movements. As one arm pulls backwards in the water the other arm recovers above the water.
• taking in air should be alternated with the arm movements.
• the legs are moved up and down in a quick succession to enhance movements.
What is the magnetic field strength at a point 2.5 mm radially from the center of the wire leading to the capacitor
Complete Question
A 8.0 A current is charging a 1.0 -cm-diameter parallel-plate capacitor. What is the magnetic field strength at a point 2.5 mm radially from the center of the wire leading to the capacitor?
Answer:
The magnetic field is [tex]B = 6.4*10^{-4} \ T[/tex]
Explanation:
From the question we are told that
The current is [tex]I = 8.0 \ A[/tex]
The diameter is [tex]d = \ cm = \frac{1}{100} = 0.01 m[/tex]
The position considered is [tex]d = 2.5 \ mm = 0.0025 \ m[/tex]
Generally the magnetic field is mathematically represented as
[tex]B = \frac{\mu_o * I }{ 2 \pi d}[/tex]
Here [tex]\mu_o[/tex] is permeability of free space with value
[tex]\mu_o = 4\pi * 10^{-7} \ N/A^2[/tex]
So
[tex]B = \frac{ 4\pi * 10^{-7} * 8 }{ 2 \pi * 0.0025 }[/tex]
=> [tex]B = \frac{ 2 * 10^{-7} * 8 }{ * 0.0025 }[/tex]
=> [tex]B = 6.4*10^{-4} \ T[/tex]
16. Explain the following observations:
i. Water wets glass surface but does not wet the waxed glass surface.
Answer:
because the cohesive forces within the drops are stronger than the adhesive forces between the drops and the wax. Water wets glass and spreads out on it because the adhesive forces between the liquid and the glass are stronger than the cohesive forces within the water.
Answer:
the adhesive forces between the drops are stronger than cohesive force between the liquid
How is it possible for man made things to move?
Explanation:
Natural things are all the elements present in the world. Whatever is there because of nature, like petrolium and coal which is developed over thousands and millions of year process. There are many more, like even living things do come inside the circle of natural things, because man can not make life we can alter it examples can be found on your dining table, fruits and vegetables. Over the period of time humans have altered the genes of fruits and vegetables to grow the way they want example seedless grapes, seedless banana etc.
QUICK!! What class of leer is shown below?
first class lever
second class lever
third class lever
fourth class lever
Answer:
3rd class lever
HOPE IT HELPS YOU OUT PLEASE MARK IT AS BRAINLIEST AND FOLLOW ME PROMISE YOU TO FOLLOW BACK ON BRAINLY.IN
The class of lever that should be given in the attached image is to be considered the third-class lever.
What is a third-class lever?In a Class Three Lever, the Force should be lies between the Load and the Fulcrum. In the case when the Force is closer to the Load, it should be easier to lift and there is a mechanical advantage.
So based on this, we can say that The class of leer that should be given in the attached image is to be considered as the third-class lever.
Learn more about lever here: https://brainly.com/question/21535944
True or False: The negative oxygen end of one water molecule is attracted
to the positive hydrogen end of another water molecule to form a
hydrogen bond.
Answer: I believe that it is true
Explanation:
I would like to know why this is the correct answer
-Three objects are released to the ground while the mass and net force of each is recorded. Air resistance is assumed to be negligible. Which statement is correct about all three objects?
*The ratio of net force to mass is constant across all objects.
Answer:
The correct answer is the ratio of net force to mass is constant across all objects.
For what angle of incidence at the first mirror will this ray strike the midpoint of the second mirror (which is 28.0 cmcm long) after reflecting from the first mirror
This question is incomplete, the complete question is;
Two plane mirrors intersect at right angles. A laser beam strikes the first of them at a point 11.5 cm from their point of intersection, as shown in the figure.
For what angle of incidence at the first mirror will this ray strike the midpoint of the second mirror (which is 28.0 cmcm long) after reflecting from the first mirror
Answer: angle of incidence is 39.4°
Explanation:
Given that;
two plain mirrors intersect at right angle (90°)
distance d = 11.5 cm
S = 28.0 cm
Now the angle that the reflection ray males with first the mirror equal theta (∅)
so
tan∅ = (S/2) / d
tan∅ = (28/2) / 11.5
tan∅ = 14 / 11.5
tan∅ = 1.2173
∅ = tan⁻¹ (1.2173)
∅ = 50.6°
so angle of incidence = 90° - ∅
= 90° - 50.6°
= 39.4°
Therefore angle of incidence is 39.4°
Ganymede is a moon of Jupiter (mass = 1.90 x 1027 kg). It orbits Jupiter in a circle of radius 1.07 x 109 m. What is Ganymede's period in hours)?
Answer:
T = 171.59 hours
Explanation:
Mass of Ganymede, [tex]M=1.9\times 10^{27}\ kg[/tex]
Radius of the orbit, [tex]R=1.07\times 10^9\ m[/tex]
We need to find Ganymede's period. Let time is T. Using Kepler's third law of motion.
[tex]T^2=\dfrac{4\pi^2}{GM}r^3\\\\T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.9\times 10^{27}}\times (1.07\times 10^9)^3\\\\T=617754.35\ s[/tex]
or
T = 171.59 hours
Hence, Ganymede's period is 171.59 hours.
A frog leaps straight up from the ground and is caught when he reaches his maximum height 2\,\text s2s2, start text, s, end text later by a little girl. We want to find the vertical velocity of the frog at the moment that it left the ground. Which kinematic formula would be most useful to solve for the target unknown
Correct question is;
A frog leaps straight up from the ground and is caught when he reaches his maximum height 2 seconds later by a little girl. We want to find the vertical velocity of the frog at the moment that it left the ground. Which kinematic formula would be most useful to solve for the target unknown? Assume the positive direction is upward and air resistance is negligible.
Answer:
v = u + at
Explanation:
We are given time taken to reach maximum height to be 2 seconds.
We also know that since it's going upwards against gravity, our acceleration due to gravity will be g = -9.8 m/s²
Now, since we have values of g and t and we know that at maximum height vertical component of velocity is zero.
Thus, we can use Newton's first equation of motion to solve for the vertical velocity at which it left the ground.
Thus;
v = u + at
Where v is velocity at maximum height and u is velocity at which it left the ground.
Resistors examples.
Three resistors are connected in series at a common voltage of 300V: R1 = 100ohm, R2 = 200ohm, R3 = 300ohm.
a) calculate the current I in the circuit
b) calculate the ratio of the individual voltages on the resistors, ie: UR1: UR2: UR3
c) verify that: UR1: UR2: UR3 = R1: R2: R3
I will be very happy if someone writes it to me here with an explanation of how they did it.
thank you in advance.
Answer:
Multiply the resistances. Example: Two resistors in parallel, 100 and 200 ohms. 100 x 200 = 20,000
Add the resistances. Example: 100 + 200 = 300
Divide the result in Step 1 by the result in Step 2. This gives you the total resistance. Example: 20,000 / 300 = 66.7 ohms.
Explanation:
Multiply the resistances. Example: Two resistors in parallel, 100 and 200 ohms. 100 x 200 = 20,000
Add the resistances. Example: 100 + 200 = 300
Divide the result in Step 1 by the result in Step 2. This gives you the total resistance. Example: 20,000 / 300 = 66.7 ohms.
Phil is riding a scooter and pushes off the ground with his foot. this causes him to accelerate at 12 m /s. Phil weighs 600 N. how strong was his push off the ground?
Answer:
734.16 kg m/[tex]s^{2}[/tex]
Explanation:
The problem is asking for the Force of pushing off the ground.
The formula of Force is: F = mass x accelerationGiven = Mass: 600 newtons (N)
Acceleration: 12 m/[tex]s^{2}[/tex]
We have to convert the mass into kg first. Remember that 1 kg is equal to 9.80665 newtons.
Let x be the mass in newtons.
Let's convert: [tex]\frac{1 kg}{9.80665 N}[/tex] x [tex]\frac{x}{600 N}[/tex] = [tex]\frac{600}{9.80665}[/tex] = 61.18 kg
Phil's weight is 61.18 kg
Let's go back to finding the force.
F = m x a
F = 61.18 kg x 12 m/[tex]s^{2}[/tex]
F = 734.16 kg m/[tex]s^{2}[/tex]
Potential energy can be converted into kinetic energy, a good example of this is when a pole-vaulter bends the pole during a leap. When the pole is bent the most, does it store elastic or gravitational potential energy? Question 7 options: Only gravitational potential energy because that is more powerful than elastic potential energy. It has neither elastic or gravitational potential energy. The pole stores elastic potential energy when the pole is bent because its shape is change from its natural shape and it will want to go back to its original form, just like a spring or stretched elastic material. The pole stores gravitational potential energy because it is bent from its natural shape when off the ground.
Answer:
The pole stores elastic potential energy when the pole is bent because its shape is change from its natural shape and it will want to go back to its original form, just like a spring or stretched elastic material.
Question:
Why did you lie about being in college?
Answer:
The pole stores gravitational potential energy because it is bent from its natural shape when off the ground.
Explanation: i took the test
Two carts with masses 10.0kg and 2.5 kg move in opposite directions on a frictionless horizontal track with speeds of 6.0m/s and 3.0 m/a respectively. The carts stick together after colliding head on. Find the final speed.
Please help!!
Answer:
4.2
Explanation:
A thin cylindrical shell is released from rest and rolls without slipping down an inclined ramp that makes an angle of 30° with the horizontal. How long does it take it to travel the first 3.1 m?A) 1.4 sB) 1.1 sC) 2.1 sD) 1.6 sE) 1.8 s
Answer:
B ( 1.1 s )
Explanation:
Given that a thin cylindrical shell is released from rest and rolls without slipping down an inclined ramp that makes an angle of 30° with the horizontal.
Let the height = h
Sin 30 = h / 3.1
h = 3.1 Sin 30
The P.E at the top = mgh
P.E = 9.8 × 3.1 Sin 30 × m
P.E = 15.19m
The K.E = P.E
K.E = 1/2mv^2
1/2 × V^2 × m = 15.19m
The mass m will cancel out.
V^2 = 30.38
V = 5.51 m/s
Using third equation of motion to find acceleration
V^2 = U^2 + 2as
Since it starts from rest, initial velocity u = 0
30.38 = 2 × 3.1 × a
30.38 = 6.2a
a = 30.38 / 6.2
a = 4.9 m/s^2
Using equation one of linear motion to calculate time t
V = U + at
5.51 = 4.9t
t = 5.51 / 4.9
t = 1.12 s
Therefore, it will take approximately 1.1 s to travel the first 3.1 m
The correct option is B
What is the velocity of a sound wave that has a frequency of 300 Hz and a wavelength of 10
m?
Answer:
3000 m/s
Explanation:
The velocity of a sound wave is found by the wavelength times the frequency. Therefore, the velocity = wavelength*frequency = (10 m)(300 Hz) = 3000 m/s
I hope this helps! :)
An industrious beaver manages to chew through the base of a wooden tree trunk 18 m tall, causing it to fall over. If the base of the trunk does not slide while it falls over, how fast is the other end of the trunk moving right before it hits the ground
Answer:
Vf = 18.78 m/s
Explanation:
As, the block does not trunk does not slide while falling. hence, we can use third equation of motion to find its velocity before it hits the ground:
2gh = Vf² - Vi²
where,
g = acceleration due to gravity = 9.8 m/s²
h = Height of Trunk = 18 m
Vf = Velocity of trunk just before it hits the ground = ?
Vi = Initial Velocity of the trunk = 0 m/s
Therefore, using these values in the equation, we get:
(2)(9.8 m/s²)(18 m) = Vf² - (0 m/s)²
Vf = √(352.8 m²/s²)
Vf = 18.78 m/s
A student measures the stiffness of a spring by observing how far it stretches with different masses hanging from it. She tries six different masses: 10, 20, 30, 40, 50, 60 grams; and she finds that the spring stretches: 3.1, 6.5, 10.2, 11.1, 16.2, 18.5 centimeters.
a. Based on these observations, what is the best estimate for the spring constant of this spring?
b. With the 60 gram mass attached, the student observes an oscillatory period of 0.886±0.003 seconds. Explain whether this observation is consistent with your answer from part (a).
Answer:
a) k = 3.2 N / m
b) T = 0.27 s
some problem measuring the time of the oscillation.
Explanation:
a) With the student's observations we can make a graph of elongation against applied weight, see attached, with this graph, Hooke's law must comply
F = -k x
where force is the applied weight
F = W = mg
substituting
W = k x
x = 1 / k W (1)
the linear regression of the graph gives
x = 31.2 w + 0.23
the slope of the graph is
m = 31.2 cm/N
if we relate this value with equation 1
1 / k = 131.2
k = 0.032 N / cm
Let's reduce to the SI system
k = 0.032 N / cm (100cm / 1m)
k = 3.2 N / m
This is the best value we can obtain from the spring constant with these experiential data
b) When the student oscillates the system, he has a simple harmonic motion whose angular velocity is
w =[tex]\sqrt\frac{k}{m} {}[/tex]
for the masses of m = 60 gr = 60 10-3 kg = 0.060 kg
w = √ (3.2 / 0.060)
w = 22.82 rad/s
angular velocity is related to frequency and period
w = 2π f = 2π / T
T = 2π / w
T = 2π / 22.82
T = 0.27 s
A great discrepancy is observed between the theoretical value and the experimental value, it is not common erorr so high it can be due to some problem measuring the time of the oscillation.
The two general types of air pollution are _____ and _____.
Answer:
The two general types of air pollution are _____ and _____.
answer: gases and particles
Please Mark me as brainiest and follow
Towards the end of a 400m race, Faisal and Edward are leading and are both running at 6m/s. While Faisal is 72m from the finish line Edward is 100m from the finish line. Realising this and to beat Faisal, Edward decides to accelerate uniformly at 0.2 m/s2 until the end of the race while Faisal keeps on the same constant speed. Does Edward succeed in beating Faisal
Answer:
NO
Explanation:
To determine this, we have to firstly determine the time it took Faisal to complete the race with 72 meters left and then the speed Edward used for the final 100 meters (with both distance occurring at the same point in time).
To determine the time it took Faisal from that onward, we use the formula
Speed = distance/time
Since Faisal maintained his speed at the end of the race, the speed will be 6 m/s while distance at the end to be covered his 72 m. Thus
6 = 72/time
time = 72/6
time = 12 seconds; It will take Faisal 12 seconds to complete the race from that point.
For Edward, he accelerated uniformly (for 0.2 m/s²) after running at a constant speed of 6 m/s. Thus, the time it would have taken him to complete (at a speed of 6 m/s) the final 100 m would be
speed = distance/time
6 = 100/time
time = 100/6
time = 16.67 s.
However, he accelerated uniformly at 0.2m/s², hence to get the speed, the formula; speed = acceleration × time will be used
speed = 0.2 × 16.67
speed = 3.334 m/s
To determine the time it took Edward from 100 meters away,
speed = distance/time
3.334 = 100/time
time = 100/3.334
time = 29.94 seconds
It would have taken Edward 29.94 seconds to complete the race, hence he would not have succeeded in defeating Faisal
A 1,100 kg car comes uniformly to a stop. If the vehicle is accelerating at -1.2 m/s2 , which force is closest to the net force acting on the vehicle?
Answer:
The answer would be 1320 N.
Answer:
the answer would be 1320 , but since it said closet .. its gonne be 1300 N !
Explanation:
when a car is heated , say by taking it to a desert, it expands. Is the increase in length of the car greater than, less than, or equal to its increase in height?
Answer:
ΔL = [tex]\frac{L_{o} }{h_{o} }[/tex] Δh
since the length of the carriages is greater than their height, the length expansion is greater than the height expansion
Explanation:
This is an exercise in thermal expansion, which is described by the expression for one dimension
ΔL = α L₀ ΔT
For the analysis of this exercise we will assume that the sides are independent
Let's look for the condition of the length of the car
ΔL / L₀ = α ΔT
The height delay (h) is
Δh / h₀ = α ΔT
we can match the two expressions
ΔL / L₀ = Δh / h₀
ΔL = [tex]\frac{L_{o} }{h_{o} }[/tex] Δh
since the length of the carriages is greater than their height, the length expansion is greater than the height expansion
Fix this problem please.
Explanation:
Original Kinetic energy = 0.5mv²
= 0.5(5.0kg)(4.0m/s)² = 40J.
New Kinetic energy = 0.5(5.0kg)(10.0m/s)² = 250J.
Hence net work done = 250J - 40J = 210J.
Answer:
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What is the orbital velocity in km/s and period in hours of a ring particle at the outer edge of Saturn's A ring
Answer:
The orbital velocity [tex]v = 16.4 \ km/s[/tex]
The period is [tex]T = 14.8 \ hours[/tex]
Explanation:
Generally centripetal force acting ring particle is equal to the gravitational force between the ring particle and the planet , this is mathematically represented as
[tex]\frac{GM_s * m }{r^2 } = m w^2 r[/tex]
=> [tex]w = \sqrt{ \frac{GM}{r^3} }[/tex]
Here G is the gravitational constant with value [tex]G = 6.67*10^{-11}[/tex]
[tex]M_s[/tex] is the mass of with value [tex]M_s =5.683*10^{26} \ kg[/tex]
r is the is distance from the center of the to the outer edge of the A ring
i.e r = R + D
Here R is the radius of the planet with value [tex]R = 60300 \ km[/tex]
D is the distance from the equator to the outer edge of the A ring with value [tex]D = 80000 \ kg[/tex]
So
[tex]r =80000 + 60300[/tex]
=> [tex]r =140300 \ km = 1.4*10^{8} \ m[/tex]
So
=> [tex]w = \sqrt{ \frac{ 6.67*10^{-11}* 5.683*10^{26}}{[1.4*10^{8}]^3} }[/tex]
=> [tex]w = 1.175*10^{-4} \ rad/s[/tex]
Generally the orbital velocity is mathematically represented as
[tex]v = w * r[/tex]
=> [tex]v = 1.175*10^{-4} * 1.4*10^{8}[/tex]
=> [tex]v = 1.64*10^{4} \ m /s = 16.4 \ km/s[/tex]
Generally the period is mathematically represented as
[tex]T = \frac{2 \pi }{w }[/tex]
=> [tex]T = \frac{2 * 3.142 }{ 1.175 *10^{-4} }[/tex]
=> [tex]T = 53473 \ second = 14.8 \ hours[/tex]
Answer:
The orbital velocity [tex]v = 16.4 \ km/s[/tex]
The period is [tex]T = 14.8 \ hours[/tex]
Explanation:
Generally centripetal force acting ring particle is equal to the gravitational force between the ring particle and the , this is mathematically represented as
[tex]\frac{GM_s * m }{r^2 } = m w^2 r[/tex]
A perfect gas undergoes a constant 100 kPa pressure process with a 20 m3 change in volume. What is the work required to cause this volume change?
Answer:
The work required to cause this volume change is 2 x 10⁶ J
Explanation:
Given;
constant pressure of the gas, P = 100 kPa = 100,000 Pa
change in volume of the gas, ΔV = 20 m³
The work required to cause this volume change is calculated as;
W = PΔV
Substitute the given values and solve for the required work (W).
W = (100,000)(20)
W = 2 x 10⁶ J
Therefore, the work required to cause this volume change is 2 x 10⁶ J
The weight of a person can be represented by a vector that acts...
Which type of change is freezing water an example of?
Answer:
For example, a physical change does not result in the formation of new substances. Physical changes alter only the size, shape, form or matter state of a material. Water boiling, melting ice, tearing paper, freezing water and crushing a can are all examples of physical changes.
Explanation:
A motor exerts a 280 N force and does 1220 J
of work while lifting a box. How far did the
motor lift the box?
We know that:
Work = Force × DistanceSolving for Distance:
Distance = Work / ForceDistance = 1220 J / 280 N Distance = 4.36 mA(n) 131 g ball is dropped from a height
of 61.1 cm above a spring of negligible mass.
The ball compresses the spring to a maximum
displacement of 4.82755 cm.
The acceleration of gravity is 9.8 m/s2.
h
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dellee
Calculate the spring force constant k.
Answer:
26.59 N/m
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 131 g
Extention (e) = 4.82755 cm
Acceleration due to gravity (g) = 9.8 m/s²
Spring constant (K) =?
Next, we shall convert 131 g to Kg. This can be obtained as follow:
1000 g = 1 Kg
Therefore,
131 g = 131 g × 1 Kg / 1000 g
131 g = 0.131 Kg
Thus, 131 g is equivalent to 0.131 Kg.
Next, we shall the force exerted by the ball on the spring. This can be obtained as follow:
Mass (m) = 0.131 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Force (F) =?
F = ma
F = 0.131 × 9.8
F = 1.2838 N
Next, we shall convert 4.82755 cm to metre (m)
This can be obtained as follow:
100 cm = 1 m
Therefore,
4.82755 cm = 4.82755 cm × 1 m / 100 cm
4.82755 cm = 0.0482755 m
Thus, 4.82755 cm is equivalent to 0.0482755 m
Finally, we shall determine the spring constant as follow:
Force (F) = 1.2838 N
Extention (e) = 0.0482755 m
Spring constant (K) =?
F = Ke
1.2838 = K × 0.0482755
Divide both side by 0.0482755
K = 1.2838 / 0.0482755
K = 26.59 N/m
Thus the spring constant is 26.59 N/m
