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3 years ago

11

Dawn is trying to find out how much weight she can push across the room. She is really trying to find her

2 answers:

k0ka [10]3 years ago

6 0

♡ <em>Hey sisters its Alejamtz555 here! </em>♡

❆Answer❆

<u><em>◀ D. Muscular Strength ▶</em></u>

<em>Hope this helps! (✿◠‿◠)</em>

mars1129 [50]3 years ago

4 0

Answer:

Muscular strength

Explanation:

She is testing her strength while pushing the weights

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A sensitive gravimeter at a mountain observatory finds that the free-fall acceleration is 0.0055 m/s2 less than that at sea leve

pshichka [43]

Gs*rs^2 = gm*rm^2
<span>rm = rs*√gs/gm </span>
<span>rm = 6370*√9.83/(9.83-0.009) = 6372.92 </span>
<span>mountain observatory is placed at an altitude worth 2920 m asl

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7 0

3 years ago

What is the work behind these answers?

Sati [7]

I think you need to solve them

7 0

3 years ago

A 6,000N is applied to a formula one car that weighs 500kg. What is the car's acceleration?

Vesna [10]

Answer:

<h2>12 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

f is the force

m is the mass

From the question

f = 6000 N

m = 500 kg

We have

$a = \frac{6000}{500} = \frac{60}{5} = 12 \\$

We have the final answer as

<h3>12 m/s²</h3>

Hope this helps you

8 0

3 years ago

A piston-cylinder device contains Helium gas initially at 150 kPa, 20 o C, and 0.5m 3 . The helium is now compressed in a polytr

Molodets [167]

Answer:

Explanation:

Given

$P_1=150 kPa$

$T_1=20^{\circ}C$

$V_1=0.5 m^3$

$T_2=140^{\circ}C$

$P_2=400 kPa$

R for Helium $R=2.076$

$c_v=3.115 kJ/kg-K$

mass of gas $m=\frac{P_1V_1}{RT_1}$

$m=\frac{150\times 0.5}{2.076\times 293}$

$m=0.123 kg$

Similarly $V_2$ can be found

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

$V_2=0.264 m^3$

Work done $W=\int_{V_1}^{V_2}PdV$

$W=\frac{P_2V_2-P_1V_1}{n-1}$

$W=\frac{mR(T_2_T_1)}{n-1}$

Since it is a polytropic Process

therefore $PV^n=c$

$P_1V_1^n=P_2V_2^n$

$(\frac{V_1}{V_2})^n=\frac{P_2}{P_1}$

$(\frac{0.5}{0.264})^n=\frac{400}{150}$

$n=\frac{\ln 2.66}{\ln 1.893}$

$n=1.533$

$W=\frac{0.123\times 2.076(140-20)}{1.533-1}$

$W=57.48 kJ$

From Energy balance

$E_{in}-E_{out}=\Delta E_{system}$

Neglecting kinetic and Potential Energy change

$Q_{in}+W_{in}=change\ in\ Internal\ Energy$

Change in Internal Energy $\Delta U=u_2-u_1$

$\Delta U=mc_v(T_2-T_1)$

$\Delta U=0.123\times 3.115(140-20)$

$\Delta U=45.977 kJ$

$Q_{in}+57.48=45.977$

$Q_{in}=-11.50 kJ$

i.e. Heat is being removed

3 0

3 years ago

g A hydraulic press has a safety feature which consists of a hydraulic cylinder with a piston at one end and a safety valve at t

nlexa [21]

Answer:

58.32 N

Explanation:

Area of a circle = $\pi$ $r^{2}$

where r is the radius of the circle.

The cylinder has a radius of 0.02 m, its area is;

$A_{1}$ = $\pi$ $r^{2}$

= $\frac{22}{7}$ x $(0.02)^{2}$

= $\frac{22}{7}$ x 0.0004

= 1.2571 x $10^{-3}$

Area of the cylinder is 0.0013 $m^{2}$ .

The safety valve has a radius of 0.0075 m, its area is;

$A_{2}$ = $\pi$ $r^{2}$

= $\frac{22}{7}$ x $(0.0075)^{2}$

= $\frac{22}{7}$ x 5.625 x $10^{-5}$

= 1.7679 x $10^{-4}$

Area of the valve is 0.00018 $m^{2}$ .

From Hooke's law, the force on the safety valve can be determined by;

F = ke

$F_{2}$ = 950 x 0.0085

= 8.075 N

Minimum force, $F_{1}$ , required can be determined by;

$\frac{F_{1} }{A_{1} }$ = $\frac{F_{2} }{A_{2} }$

$\frac{F_{1} }{0.0013}$ = $\frac{8.075}{0.00018}$

$F_{1}$ = $\frac{0.0013 *8.075}{0.00018}$

= 58.32

The minimum force that must be exerted on the piston is 58.32 N.

8 0

3 years ago