What is the weight of a 700 kg boat in Newton's?

A neighbor's child wants to go to a neighborhood carnival to experience the wild rides. The neighbor is worried about safety as

seraphim [82]

Answer:

No, the ride is not safe

Explanation:

From the diagram attached, it is seen that

$\sum f_{y} = 0$

$\tau cos \theta - W = 0$ ................(1)

$\sum f_{x} = 0$

$-ma + \tau sin \theta = 0$ .......(2)

$a = w^{2} r$

From the diagram, $r = Lsin \theta$

$a = w^{2} L sin \theta$

$w = \frac{2\pi }{T}$

$a = \frac{4\pi ^{2}L sin \theta }{T^{2} }$ ............(3)

Put (3) into (2)

$\tau sin \theta = \frac{4\pi ^{2}m L sin \theta }{T^{2} }$

$\tau = \frac{4\pi ^{2}m L }{T^{2} }$

w = weight of chair + weight of child = 50 + 10 = 60 lb

g = 32 ft/s²

m = w/g = 60/32 = 1.875

L = 30 ft

T = 3 secs

$\tau = \frac{4\pi ^{2}*1.875*30 }{0.3^{2} }$

$\tau = 246.74 lbs$

since 246.74 lbs > 200 lbs, it is not safe because the stationary chair will creak

6 0

3 years ago

Read 2 more answers

(a) What is the cost of heating a hot tub containing 1440 kg of water from 10.0°C to 40.0°C, assuming 75.0% efficiency to take h

BaLLatris [955]

Answer:

a) $E = 6.024\,USD$ , For m kilograms, it is 4184m J., 3600000 joules, b) $i = 88.200\,A$

Explanation:

a) The amount of heat needed to warm water is given by the following expression:

$Q_{needed} = m_{w}\cdot c_{w}\cdot (T_{f}-T_{i})$

Where:

$m_{w}$ - Mass of water, measured in kilograms.

$c_{w}$ - Specific heat of water, measured in $\frac{J}{kg\cdot ^{\circ}C}$ .

$T_{f}$ , $T_{i}$ - Initial and final temperatures, measured in $^{\circ}C$ .

Then,

$Q_{needed} = (1440\,kg)\cdot \left(4184\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (40^{\circ}C - 10^{\circ}C)$

$Q_{needed} = 180748800\,J$

The energy needed in kilowatt-hours is:

$Q_{needed} = 180748800\,J\times \left(\frac{1}{3600000}\,\frac{kWh}{J} \right)$

$Q_{needed} = 50.208\,kWh$

The electric energy required to heat up the water is:

$E = \frac{50.208\,kWh}{0.75}$

$E = 66.944\,kWh$

Lastly, the cost of heating a hot tub is: (USD - US dollars)

$E = (66.944\,kWh)\cdot \left(0.09\,\frac{USD}{kWh} \right)$

$E = 6.024\,USD$

The heat needed to raise the temperature a degree of a kilogram of water is 4184 J. For m kilograms, it is 4184m J. Besides, a kilowatt-hour is equal to 3600000 joules.

b) The current required for the electric heater is:

$i = \frac{Q_{needed}}{\eta \cdot \Delta V \cdot \Delta t}$

$i = \frac{180748800\,J}{0.75\cdot (220\,V)\cdot (3.45\,h)\cdot \left(3600\,\frac{s}{h} \right)}$

$i = 88.200\,A$

7 0

3 years ago

A human services professional has worked with a particular client for a couple of years. The client

zhuklara [117]

Answer:

Respect the client’s decision

Explanation:

just took the test

4 0

3 years ago

How do you work out Potential Difference??? <br> Please can you make it simple :) <br> Thanks.

patriot [66]

Answer:

Potential difference is the work done in moving a positive test charge from infinity to the point in question.

Voltage is an expression of PD. (Joules / Coulomb)

Say that a capacitor has a PD of 5 Volts. The work in moving a positive test charge from the positive plate to the negative plate is -5 Joules/Coulomb or -5 volt. (At the positive plate the positive test charge (1 Coulomb) already has a PD of + 5 Volts.)

7 0

3 years ago

The op amp in this circuit is ideal. R3 has a maximum value of 100 kΩ and σ is restricted to the range of 0.2 ≤ σ ≤ 1.0. a. Calc

Firlakuza [10]

I have attached the circuit image missing in the question.

Answer:

A) The range of vo is; -6.6V≤ vo ≤-1V

B) σ = 0.1861

Explanation:

A) First of all, Let VΔ be the voltage from the potentiometer contact to the ground.

Thus; [(0 - vg)/(2000)] +[(0 - vΔ)/(50,000)] = 0

So, [(- vg)/(2000)] +[(- vΔ)/(50,000)] = 0

Simplifying further; -25 vg - vΔ = 0

From the question, vg = 40mV = 0.04 V

So - 25(0.04) = vΔ

So: vΔ = - 1 V

Now, [vΔ/(σRΔ)] + [(vΔ - 0)/(50,000)] + [(vΔ - vo)/((1 - σ)RΔ))] = 0

So, multiplying each term by RΔ to get; [vΔ/(σ)] + [(vΔ x RΔ)/(50,000)] + [(vΔ - vo)/((1 - σ))] = 0

So RΔ = 100kΩ or 100,000Ω from the question.

So, substituting for RΔ, we get,

[vΔ/(σ)] + [2vΔ] + [(vΔ - vo)/((1 - σ))] = 0

Let's put the value of - 1 for vΔ as gotten before.

So, ( - 1/σ) - 2 + [(-1 - vo)/(1 - σ)] = 0

Now let's make vo the subject of the equation to get;

-1 - vo = (1 - σ)[2 + (1/σ)]

-1 - vo = 2 - 2σ + (1/σ) - 1

-vo = 1 + 2 - 2σ + (1/σ) - 1

-vo = 2 - 2σ + (1/σ)

vo = - 1 (2 - 2σ + (1/σ))

When σ = 0.2; vo = - 1(2 - 0.4 + 5) =

- 1 x 6.6 = - 6.6V

Also when σ = 1;

vo = - 1(2 - 2 + 1) = - 1V

Therefore, the range of vo is;

- 6.6V ≤ vo ≤ - 1V

B) it will saturate at vo = - 7V

So, from;

vo = - 1 (2 - 2σ + (1/σ))

-7 = - 1 (2 - 2σ + (1/σ))

Divide both sides by (-1)

7 = (2 - 2σ + (1/σ))

Now, subtract 2 from both sides to get; 5 = - 2σ + (1/σ)

Multiply each term by α to get;

5σ = - 2σ^(2) + 1

So 2σ^(2) + 5σ - 1 = 0

Solving simultaneously and picking the positive value , we get σ to be approximately 0.1861

8 0

3 years ago