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Mamont248 [21]
3 years ago
15

The first game in volleyball is played until a team reaches 25 points.

Physics
2 answers:
Zinaida [17]3 years ago
7 1
I don't understand this question
andreyandreev [35.5K]3 years ago
7 0

Answer: The answer to your question is true. Please mark me as brainliest!

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Andre, whose mass is 77.1 kg, sits on a diving board above a dunk tank. If he is sitting 0.90 m above the water, what is his tot
Ganezh [65]

Answer:

680 J

Explanation:

Mechanical energy = potential energy + kinetic energy

ME = PE + KE

ME = mgh + ½ mv²

ME = (77.1 kg) (9.8 m/s²) (0.90 m) + ½ (77.1 kg) (0 m/s)²

ME = 680 J

8 0
3 years ago
Read 2 more answers
If two items are equal in size, the denser one will
Talja [164]
Be heavier

density=mass÷volume

if two items have the same size they have the same volume so the heavier one will be the denser one
6 0
3 years ago
A plate drops onto a smooth floor and shatters into three pieces of equal mass.Two of the pieces go off with equal speeds v at r
Firlakuza [10]

Answer:

Speed of the this part is given as

v_3 = \sqrt2 v

Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate

Explanation:

As we know by the momentum conservation of the system

we will have

P_1 + P_2 + P_3 = P_i

here we know that

P_1 = P_2

the momentum of two parts are equal in magnitude but perpendicular to each other

so we will have

P_1 + P_2 = \sqrt{P^2 + P^2}

P_1 + P_2 = \sqrt2 mv

now from above equation we have

P_3 = -(P_1 + P_2)

mv_3 = -(\sqrt 2 mv)

v_3 = \sqrt2 v

Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate

6 0
3 years ago
A concert loudspeaker suspended high off the ground emits 34 W of sound power. A small microphone with a 1.0 cm2 area is 44 m fr
rjkz [21]

Answer:

<u>Part A</u>

I = 1.4 mW/m²  

<u>Part B</u>

β = 91.46 dB

Explanation:

<u>Part A</u>

Sound intensity is the power per unit area of sound waves in a direction perpendicular to that area. Sound intensity is also called acoustic intensity.

For a spherical sound wave, the sound intensity is given by;

                                            I = \frac{P}{A}

                                            I = \frac{P}{4\pi r^{2}}

Where;

P is the source of power in watts (W)

I is the intensity of the sound in watt per square meter (W/m2)

r is the distance r away

Given:

P = 34 W,

A = 1.0 cm²

r = 44 m

The sound intensity at the position of the microphone is calculated to be;

                                     I = \frac{34}{4\pi (44)^{2}}

                                     I = \frac{34}{4\pi (44)^{2}}

                                     I = 0.0013975 W/m²

                                 ≈  I = 0.0014 W/m² = 1.4 × 10⁻³ W/m²

                                     I = 1.4 mW/m²

The sound intensity at the position of the microphone is 1.4 mW/m².

<u>Part B</u>

Sound intensity level or acoustic intensity level is the level of the intensity of a sound relative to a reference value.  It is a a logarithmic quantity. It is denoted by β and expressed in nepers, bels, or decibels.

Sound intensity level is calculated as;  

                                    β = 10log_{10}\frac{I}{I_{0}}  dB

Where,

β is the Sound intensity level in decibels (dB)

I is the sound intensity;

I₀ is the reference sound intensity;

By pluging-in, I₀ is 1.0 × 10⁻¹² W/m²

           ∴        β = 10log_{10}\frac{1.4 * 10^{-3} W/m^{2}}{1.0 * 10^{-12} W/m^{2}}

                      β = 10log_{10} (1.4 * 10^{9})

                      β = 91.46 dB

The sound intensity level at the position of the microphone is 91.46 dB.                

4 0
3 years ago
A concave spherical mirror has a radius of curvature of magnitude 20.0 cm. (a) Find the location of the image for object distanc
Tems11 [23]

Answer:

Answered

Explanation:

The radius of curvature of the mirror R = 20 cm

then the focal length f = R/2 = 10 cm

(a) From mirror formula

 1/f = 1/di + /1do

then the image distance

    di = fd_o / d_o - f

  = (10)(40) / 40-10

 = 30.76 cm

since the image distance is positive so the image is real

ii) when the object distance d_0=20 cm

 di = 10×20/ 20-10

= 20

Hence, the image must be  real

iii)when the object distance d_0 = 10

di = 10×10 / 10-10 =  ∞ (infinite)

the image will be formed at ∞

here also image will be real but diminished.

7 0
3 years ago
Read 2 more answers
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