Ask question

Ask question

All categories

3 years ago

15

What stellar formaoasxcscjc

1 answer:

ExtremeBDS [4]3 years ago

8 0

I need to answer this with 20 characters so hey wasssup

You might be interested in

A manager at a local bank analyzed the relationship between monthly salary and three independent variables: length of service (m

Readme [11.4K]

Answer:

a

Explanation:

8 0

3 years ago

A constant force of 8N acting on an object displaces it through a distance of 3.0 m in the direction of force. Calculate work-do

sweet-ann [11.9K]

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{Work\:done=24\:J}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline{\bold{Given :}}} \\ \tt: \implies Constant \: force(F) = 8 \: N \\ \\ \tt: \implies Displacement(s) = 3 \: m \\ \\ \red{\underline{\bold{To \: Find : }}} \\ \tt: \implies Work \: Done(W.D) = ?$

• <u>According to given question</u> :

$\green{ \star} \tt \: \theta \: = 0 \degree \: \: \: \: (Angle \: between \: force \: and \: displacement) \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies Work \: Done = FS \: cos \: \theta \\ \\ \tt: \implies Work \: Done = 8 \times 3 \times cos \:0 \degree \\ \\ \green{ \circ} \tt \: cos \: 0 \degree = 1 \\ \\ \tt: \implies Work \: Done =24 \times 1 \\ \\ \green{\tt: \implies Work \: Done =24 \: J}$

5 0

3 years ago

A silver wire 2.6 mm in diameter transfers a charge of 420 C in 80 min. Silver contains 5.8 x 10- free electrons per cubic meter

kifflom [539]

Answer:

a). 87.5 mA or $87.5 x10^{-3}$ A

b). 1.78 $\frac{m}{s}$

Explanation:

$d=2.6 mm \\Q=420C\\t=80min\\n=5.8x10^{28} \\q=1.6x10^{-19}$

n the number of free electrons is 28 in text reference and if they don't give q is take as the charge of electron.

a).

$I=\frac{Q}{t}\\ I= \frac{420 C}{80 min}*\frac{1min}{60 s} =\frac{420 C}{4800s}\\ I=87.5 x10^{-3}$ A

b).

$I=n*abs (q)*V_{d}*A$

$A= \pi * (\frac{d}{2})^{2} \\A=\pi (*\frac{2.6x10^{-3} m}{2})^{2} \\A=5.309x10^{-6}$

$V_{d} =\frac{I}{n*abs(q)*A} \\V_{d}=\frac{87.5 x10^{-2} }{5.8x^10{28} *1.6x^{-19} *5.3x^{6} }\\V_{d}=1.78 \frac{m}{s}$

8 0

3 years ago

A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oF. The air undergoes a process to a state where the

vladimir2022 [97]

Answer:

The work and heat transfer for this process is = 270.588 kJ

Explanation:

Take properties of air from an ideal gas table. R = 0.287 kJ/kg-k

The Pressure-Volume relation is <em>PV</em> = <em>C</em>

<em>T = C </em> for isothermal process

Calculating for the work done in isothermal process

<em>W</em> = <em>P</em>₁<em>V</em>₁ $ln[\frac{P_{1} }{P_{2} }]$

= <em>mRT</em>₁ $ln[\frac{P_{1} }{P_{2} }]$ [∵<em>pV</em> = <em>mRT</em>]

= (5) (0.287) (272.039) $ln[\frac{2.0}{1.0}]$

= 270.588 kJ

Since the process is isothermal, Internal energy change is zero

Δ<em>U</em> = $mc_{v}(T_{2} - T_{1} ) = 0$

From 1st law of thermodynamics

Q = Δ<em>U </em>+ <em>W</em>

= 0 + 270.588

= 270.588 kJ

4 0

3 years ago

2 ways weak nuclear forces and gravity are alike <br><br> PLEASE HELP ):

shtirl [24]

They are attractive
They don’t depend on charge

7 0

3 years ago