Ask question

Ask question

All categories

4 years ago

6

Secondclass lever short note

2 answers:

Artist 52 [7]4 years ago

7 0

Answer:

second

Explanation:

seraphim [82]4 years ago

6 0

Answer:

wow

Explanation:

You might be interested in

100 POINTS HELP!!!!!!!!!!!!!

ivanzaharov [21]

Answer:

it is whatever the temperature is at 5(I cant seem to see it clearly)

Explanation:

4 0

3 years ago

Read 2 more answers

A boy is pulling a box on the ground by a string to the right. What four forces he using?

kirill115 [55]

The strong force in his body

3 0

3 years ago

One kg of air contained in a piston-cylinder assembly undergoes a process from an initial state whereT1=300K,v1=0.8m3/kg, to a f

lutik1710 [3]

Answer:

1. Yes, it can occur adiabatically.

2. The work required is: 86.4kJ

Explanation:

1. The internal energy of a gas is just function of its temperature, and the temperature changes between the states, so, the internal energy must change, but how could it be possible without heat transfer? This process may occur adiabatically due to the energy balance:

$U_{2}-U_{1}=W$

This balance tell us that the internal energy changes may occur due to work that, in this case, si done over the system.

2. An internal energy change of a gas may be calculated as:

$du=C_{v}dT$

Assuming $C_{v}$ constant,

$U_{2}-U_{1}=W=m*C_{v}(T_{2}-T_{1})$

$W=0.72*1*(420-300)=86.4kJ$

8 0

4 years ago

A standard basketball (mass = 624 grams; 24.3 cm in diameter) is held fully under water. a. Calculate the buoyant force and weig

galina1969 [7]

Answer:

Explanation:

Mass = 624 gm = .624 kg

weight = .624 x 9.8

= 6.11 N

Radius of ball = 12.15 x 10⁻² cm

volume of ball

= 4/3 x 3.14 x ( 12.15 x 10⁻²)³

= 7509.26 x 10⁻⁶ m³

Buoyant force = weight of displaced water

= 7509.26 x 10⁻⁶ x 10³ x 9.8

= 73.59 N

b ) Since buoyant force exceeds the weight of the ball , it will float .

c )

Let volume v sticks out while floating .

Volume under water

= 7509.26 x 10⁻⁶ - v

its weight

= (7509.26 x 10⁻⁶ - v ) x 10³ x 9.8

For floating

(7509.26 x 10⁻⁶ - v ) x 10³ x 9.8 = .624 x 9.8 ( weight of ball )

(7509.26 x 10⁻⁶ - v ) x 10³ = .624

7.509 - v x 10³ = .624

v x 10³ = 7.509 - .624

v x 10³ = 6.885

v = 6.885 x 10⁻³ m³

fraction

= v / total volume

= 6.885 x 10⁻³ / 7.51 x 10⁻³

91.67 %

8 0

3 years ago

The intensity of the sound from a certain source is measured at two points along a line from the source. The points are separate

Artemon [7]

Answer:

The source is at a distance of 4.56 m from the first point.

Solution:

As per the question:

Separation distance between the points, d = 11.0 m

Sound level at the first point, L = 66.40 dB

Sound level at the second point, L'= 55.74 dB

Now,

$L = 10log_{10}\frac{I}{I_{o}}$

$I = I_{o}10^{\frac{L}{10}} = I_{o}10^{0.1L} = 10^{- 12}\times 10^{0.1\times 66.40} = 10^{- 5.36}$

$L' = 10log_{10}\frac{I'}{I_{o}}$

$I' = I_{o}10^{\frac{L'}{10}} = 10^{- 12}\times 10^{0.1\times 55.74} = 10^{- 6.426}$

where

$I_{o} = 10^{- 12} W/m^{2}$

I = Intensity of sound

Now,

$I = \frac{P}{4\pi R^{2}}$

Similarly,

$I' = \frac{P}{4\pi (R + 11.0)^{2}}$

Now,

$\frac{I}{I'} = \frac{(R + 11.0)^{2}}{R^{2}}$

$\frac{10^{- 5.36}}{10^{- 6.426}} = \frac{(R + 11.0)^{2}}{R^{2}}$

$R ^{2} + 22R + 121 = 11.64R^{2}}$

$10.64R ^{2} - 22R - 121 = 0$

Solving the above quadratic eqn, we get:

R = 4.56 m

8 0

3 years ago