How far moves in 95g ball by applying 200 Newton Force if 14 KJ of energy is transferred

7. A stereo is able to convert 20% of its electrical energy into sound. If 1000 J of sound energy is produced, calculate the ele

Sonja [21]

Answer:

As 20/100 E= 1000

E= 10,000/2

=5000J

3 0

3 years ago

List down the procedures for each swimming stroke 1.Crawl 2.Breaststoke 3.Butterflystroke 4. Backstoke

Elena L [17]

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.

4 0

3 years ago

Light of wavelength 623.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 76.5 cm from the slit.

Dvinal [7]

Answer:

Explanation:

wave length of light λ = 623 x 10⁻⁹ m .

Distance of screen D = 76.5 x 10⁻² m

width of slit = d

Distance on the screen between the second order minimum and the central maximum = 2 λ D / d

1.11 x 10⁻² = (2 x 623 x 10⁻⁹ x 76.5 x 10⁻² )/ d

d = ( 2 x 623 x 10⁻⁹ x 76.5 x 10⁻²) / 1.11 x 10⁻²

= 85872.97 x 10⁻⁹

= 85.87297 x 10⁻⁶

= 85.87 μm

width a of the slit is = 85.87 μm

6 0

3 years ago

the amount of surface area of the block contact with the surface is 2.03*10^-2*m2 what is the average pressure exerted on the su

CaHeK987 [17]

Complete question:

A block of solid lead sits on a flat, level surface. Lead has a density of 1.13 x 104 kg/m3. The mass of the block is 20.0 kg. The amount of surface area of the block in contact with the surface is 2.03*10^-2*m2, What is the average pressure (in Pa) exerted on the surface by the block? Pa

Answer:

The average pressure exerted on the surface by the block is 9655.17 Pa

Explanation:

Given;

density of the lead, ρ = 1.13 x 10⁴ kg/m³

mass of the lead block, m = 20 kg

surface area of the area of the block, A = 2.03 x 10⁻² m²

Determine the force exerted on the surface by the block due to its weight;

F = mg

F = 20 x 9.8

F = 196 N

Determine the pressure exerted on the surface by the block

P = F / A

where;

P is the pressure

P = 196 / (2.03 x 10⁻²)

P = 9655.17 N/m²

P = 9655.17 Pa

Therefore, the average pressure exerted on the surface by the block is 9655.17 Pa

6 0

3 years ago

An object has the acceleration graph shown in (Figure 1). Its velocity at t=0s is vx=2.0m/s. Draw the object's velocity graph fo

timama [110]

Answer:

Explanation:

We may notice that change in velocity can be obtained by calculating areas between acceleration lines and horizontal axis ("Time"). Mathematically, we know that:

$v_{b}-v_{a} = \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

$v_{b} = v_{a}+ \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

Where:

$v_{a}$ , $v_{b}$ - Initial and final velocities, measured in meters per second.

$t_{a}$ , $t_{b}$ - Initial and final times, measured in seconds.

$a(t)$ - Acceleration, measured in meters per square second.

Acceleration is the slope of velocity, as we know that each line is an horizontal one, then, velocity curves are lines with slopes different of zero. There are three region where velocities should be found:

Region I (t = 0 s to t = 4 s)

$v_{4} = 2\,\frac{m}{s} +\int\limits^{4\,s}_{0\,s} {\left(-2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{4} = 2\,\frac{m}{s}+\left(-2\,\frac{m}{s^{2}} \right) \cdot (4\,s-0\,s)$

$v_{4} = -6\,\frac{m}{s}$

Region II (t = 4 s to t = 6 s)

$v_{6} = -6\,\frac{m}{s} +\int\limits^{6\,s}_{4\,s} {\left(1\,\frac{m}{s^{2}} \right)} \, dt$

$v_{6} = -6\,\frac{m}{s}+\left(1\,\frac{m}{s^{2}} \right) \cdot (6\,s-4\,s)$

$v_{6} = -4\,\frac{m}{s}$

Region III (t = 6 s to t = 10 s)

$v_{10} = -4\,\frac{m}{s} +\int\limits^{10\,s}_{6\,s} {\left(2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{10} = -4\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right) \cdot (10\,s-6\,s)$

$v_{10} = 4\,\frac{m}{s}$

Finally, we draw the object's velocity graph as follows. Graphic is attached below.

3 0

3 years ago

How far moves in 95g ball by applying 200 Newton Force if 14 KJ of energy is transferred​

How far moves in 95g ball by applying 200 Newton Force if 14 KJ of energy is transferred