Answer:
DEVICE E
Explanation:
QUICK!! What class of leer is shown below?
first class lever
second class lever
third class lever
fourth class lever
Answer:
3rd class lever
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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
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
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 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
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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
The two general types of air pollution are _____ and _____.
Answer:
The two general types of air pollution are _____ and _____.
answer: gases and particles
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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
A(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.
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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
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.
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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