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3 years ago

8

How much force would be required for a ten strand pulley system to lift 1000 newtons?

1 answer:

Pepsi [2]3 years ago

8 0

Answer: 250

Explanation:

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Mount Everest rises to a height of 8,850 m above sea level. At a base camp on the mountain the atmospheric pressure is measured

Mamont248 [21]

Answer:

74.86°C

Explanation:

P₂ = Vapour pressure of water at sea level = 760 mmHg

P₁ = Pressure at base camp = 296 mmHg

T₂ = Temperature of water = 373 K

ΔH°vap for H2O = 40.7 kJ/mol = 40700 J/mol

R = Gas constant = 8.314 J/mol K

From Claussius Clapeyron equation

$ln\frac{P_2}{P_1}=\frac{\Delta H}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\\\Rightarrow ln\frac{760}{296}=\frac{40700}{8.314}\left(\frac{1}{T_1}-\frac{1}{373}\right)\\\Rightarrow ln\frac{760}{296}\times \frac{8.314}{40700}+\frac{1}{373}=\frac{1}{T_1}\\\Rightarrow 0.0028735=\frac{1}{T_1}\\\Rightarrow T_1=347.996\ K$

T₁ = 347.996 K = 74.86°C

∴Water will boil at 74.86°C

4 0

3 years ago

I need help with this problem :(

Reil [10]

Answer:

it tells you that the speed increases until about 20 seconds then keeps a steady pace for 20 seconds then the speed drops and stops at 55 seconds in the process.

7 0

2 years ago

Example 4.6 provides a nice example of the overlap between kinematics and dynamics. It is known that the plane accelerates from

kodGreya [7K]

Answer:

$ax = 2.60m/s^{2}$ , $t = 26.92s$

Explanation:

The acceleration of the plane can be determined by means of the kinematic equation that correspond to a Uniformly Accelerated Rectilinear Motion.

$(vx)f^{2} = (vx)i^{2} + 2ax \Lambda x$ (1)

Where $(vx)f^{2}$ is the final velocity, $(vx)i^{2}$ is the initial velocity, $ax$ is the acceleration and $\Lambda x$ is the distance traveled.

Equation (1) can be rewritten in terms of ax:

$(vx)f^{2} - (vx)i^{2} = 2ax \Lambda x$

$2ax \Lambda x = (vx)f^{2} - (vx)i^{2}$

$ax = \frac{(vx)f^{2} - (vx)i^{2}}{2 \Lambda x}$ (2)

Since the plane starts from rest, its initial velocity will be zero ( $(vx) = 0$ ):

Replacing the values given in equation 2, it is gotten:

$ax = \frac{(70m/s)^{2} - (0m/s)^{2}}{2(940m)}$

$ax = \frac{4900m/s}{2(940m)}$

$ax = \frac{4900m/s}{1880m}$

$ax = 2.60m/s^{2}$

So, The acceleration of the plane is $2.60m/s^{2}$

Now that the acceleration is known, the next equation can be used to find out the time:

$(vx)f = (vx)i + axt$ (3)

Rewritten equation (3) in terms of t:

$t = \frac{(vx)f - (vx)i}{ax}$

$t = \frac{70m/s - 0m/s}{2.60m/s^{2}}$

$t = 26.92s$

<u>Hence, the plane takes 26.92 seconds to reach its take-off speed.</u>

5 0

3 years ago

A family leaves home at 1:00pm and arrives at their vacation spot 200 miles away at 5:00pm on the same day. What is their averag

Shkiper50 [21]

The answer is 40 because you just have to do 200 divided by 5

4 0

3 years ago

If the 20-mm-diameter rod is made of A-36 steel and the stiffness of the spring is k = 55 MN/m , determine the displacement of e

Alinara [238K]

Answer:

σ = 1.09 mm

Explanation:

<u>Step 1:</u> Identify the given parameters

rod diameter = 20 mm

stiffness constant (k) = 55 MN/m = 55X10⁶N/m

applied force (f) = 60 KN = 60 X 10³N

young modulus (E) = 200 Gpa = 200 X 10⁹pa

<u>Step 2:</u> calculate length of the rod, L

$K = \frac{A*E}{L}$

$L = \frac{A*E}{K}$

$A=\frac{\pi d^{2}}{4}$

d = 20-mm = 0.02 m

$A=\frac{\pi (0.02)^{2}}{4}$

A = 0.0003 m²

$L = \frac{A*E}{K}$

$L = \frac{(0.0003142)*(200X10^9)}{55X10^6}$

L = 1.14 m

<u>Step 3:</u> calculate the displacement of the rod, σ

$\sigma = \frac{F*L}{A*E}$

$\sigma = \frac{(60X10^3)*(1.14)}{(0.0003142)*(200X10^9)}$

σ = 0.00109 m

σ = 1.09 mm

Therefore, the displacement at the end of A is 1.09 mm

3 0

3 years ago