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3 years ago

Help pls☹️☹️☹️☹️☹️ anyone

Physics

1 answer:

Alex17521 [72]3 years ago

4 0

Most of the question 1 is chopped off so I'll try my best to answer.

1ai) In a telephone, the sound is converted into electrical signals before being sent through the wire before converted back into sound.

1aii) Sound travels by the collision between particles. In solids, the particles are closely packed (as compared to air molecules) and thus collide with their neighbors quickly after being disturbed, allowing for sound to be passed on at a faster rate.

b) Can't help much on this one.

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Emily uses a rifle to shoot a bullet at a target. The bullet has a mass of 13 grams. The rifle has a mass of 3,500 grams. When s

sp2606 [1]

Force exerted by the bullet = mass * acceleration = 0.013 * 850 = 11.05 Newtons.

the rifle exerts same force in opposite direction so we have

11.05 = 3.5 * a
acceleration = 11.05 / 3.5 = 3.16 m /s^-2

4 0

3 years ago

Read 2 more answers

what does the density of an object tell you about the molecular arrangement of the atoms in an object

Andrews [41]

Answer:

If the density of the object is high its molecular arrangement is compact while if the density is lows its molecular arrangement isnt that compact

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3 years ago

Organic macromolecules called _______ are insoluble in water

NikAS [45]

Answer:

lipids are insoluble in water which is why lipids are often found in biological membranes and other waterproof coverings.

3 0

3 years ago

A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle

tatiyna

Answer:

a) 17.8 m/s

b) 28.3 m

Explanation:

Given:

angle A = 53.0°

sinA = 0.8

cosA = 0.6

width of the river,d = 40.0 m,

the far bank was 15.0 m lower than the top of the ramp h = 15.0 m,

The river itself was 100 m below the ramp H = 100 m,

(a) find speed v

vertical displacement

$-h= vsinA\times t-gt^2/2$

putting values h=15 m, v=0.8

$-15 = 0.8vt - 4.9t^2$ ............. (1)

horizontal displacement d = vcosA×t = 0.6×v ×t

so v×t = d/0.6 = 40/0.6

plug it into (1) and get

$-15 = 0.8\times40/0.6 - 4.9t^2$

solving for t we get

t = 3.734 s

also, v = (40/0.6)/t = 40/(0.6×3.734) = 17.8 m/s

(b) If his speed was only half the value found in (a), where did he land?

v = 17.8/2 = 8.9 m/s

vertical displacement = $-H =v sinA t - gt^2/2$

⇒ $4.9t^2 - 8.9\times0.8t - 100 = 0$

t = 5.30 s

then

d =v×cosA×t = 8.9×0.6×5.30= 28.3 m

3 0

3 years ago

Air enters an adiabatic compressor at 104 kPa and 292 K and exits at a temperature of 565 K. Determine the power (kW) for the co

ladessa [460]

Answer:

$\dot W_{in} = 49.386\,kW$

Explanation:

An adiabatic compressor is modelled as follows by using the First Law of Thermodynamics:

$\dot W_{in} + \dot m \cdot c_{p}\cdot (T_{1}-T_{2}) = 0$

The power consumed by the compressor can be calculated by the following expression:

$\dot W_{in} = \dot m \cdot c_{v}\cdot (T_{2}-T_{1})$

Let consider that air behaves ideally. The density of air at inlet is:

$P\cdot V = n\cdot R_{u}\cdot T$

$P\cdot V = \frac{m}{M}\cdot R_{u}\cdot T$

$\rho = \frac{P\cdot M}{R_{u}\cdot T}$

$\rho = \frac{(104\,kPa)\cdot (28.02\,\frac{kg}{kmol})}{(8.315\,\frac{kPa\cdot m^{3}}{kmol\cdot K} )\cdot (292\,K)}$

$\rho = 1.2\,\frac{kg}{m^{3}}$

The mass flow through compressor is:

$\dot m = \rho \cdot \dot V$

$\dot m = (1.2\,\frac{kg}{m^{3}})\cdot (0.15\,\frac{m^{3}}{s} )$

$\dot m = 0.18\,\frac{kg}{s}$

The work input is:

$\dot W_{in} = (0.18\,\frac{kg}{s} )\cdot (1.005\,\frac{kJ}{kg\cdot K})\cdot (565\,K-292\,K)$

$\dot W_{in} = 49.386\,kW$

5 0

3 years ago