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3 years ago

8. Fig. 4.1 shows a heavy ball B of weight W suspended from a fixed beam by two ropes P and Q.

Physics

1 answer:

mart [117]3 years ago

3 0

Answer:

The resultant tension of the two ropes is approximately 42.4 N

The length of the line representing the resultant tension is approximately 8.48 cm

Please find included with the answer the scale drawing created with Microsoft Word

Explanation:

The given parameters are;

The tension in rope P, $T_P$ = 30 N

The tension in rope Q, $T_Q$ = 30 N

The angle the rope, 'P', makes with the horizontal = 45°

The angle the rope, 'Q', makes with the horizontal = 45°

The scale factor of the scale diagram, S.F. = 5.0 N/cm

By the resolution of forces at equilibrium, we have;

The sum of the vertical forces, $\Sigma F_y$ = $T_P_y$ + $T_Q_y$ + W = 0

∴ W = -( $T_P_y$ + $T_Q_y$ )

W = -(30 × sin(45°) + 30 × sin(45°)) = -42.4264068712

The weight of the heavy ball, W ≈ 42.4 N acting downwards

The sum of the horizontal forces, $\Sigma F_x$ = $T_P_x$ + $T_Q_x$ = 0

The length of the resultant force, W = W/(S.F.) ≈ 42.4 N/(5.0 N/cm) = 8.48 cm

The drawing of the vectors using the scale factor of 5.0 N/cm is created using Microsoft Word is included

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Answer:

p=m/V

Explanation:

M= Mass

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3 years ago

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A gazelle is running at a constant speed of 19.3 m/s toward a motionless hidden cheetah. At the instant the gazelle passes the c

mafiozo [28]

Answer:

the animals are 26.2 meters apart.

Explanation:

Let's define t = 0s as the moment when the cheetah starts accelerating.

The gazelle moves with constant velocity, thus, it is not accelerating, then the acceleration of the gazelle is:

a₁(t) = 0m/s^2

where I will use the subscript "1" to refer to the gazelle and "2" to refer to the cheetah.

for the velocity of the gazelle we just integrate over time to get:

v₁(t) = V0

where V0 is the initial speed of the gazelle, which we know is 19.3 m/s

v₁(t) = 19.3 m/s

To get the position of the gazelle we integrate again:

p₁(t) = ( 19.3 m/s)*t + P0

where P0 is the position of the gazelle at t = 0s, let's define P0 = 0m

p₁(t) = ( 19.3 m/s)*t

The equations that describe the motion of the gazelle are:

a₁(t) = 0m/s^2

v₁(t) = 19.3 m/s

p₁(t) = ( 19.3 m/s)*t

Now let's do the same for the cheetah.

We know that its acceleration is 7.1 m/s^2

then:

a₂(t) = 7.1 m/s^2

for the velocity of the cheetah we integrate:

v₂(t) = (7.1 m/s^2)*t + V0

where v0 is the initial velocity of the cheetah, which we know its zero.

v₂(t) = (7.1 m/s^2)*t

Finally, for the position equation we integrate again, and remember that we have defined the initial position for the gazelle as zero, then the same happens for the cheetah.

p₂(t) = (1/2)*(7.1 m/s^2)*t^2

The equations for the cheetah are:

a₂(t) = 7.1 m/s^2

v₂(t) = (7.1 m/s^2)*t

p₂(t) = (1/2)*(7.1 m/s^2)*t^2

Now, we want to find the distance between both animals when the speed of the cheetah is 19.3 m/s, then first we need to solve:

v₂(t) = (7.1 m/s^2)*t = 19.3 m/s

t = (19.3 m/s)/(7.1 m/s^2) = 2.72s

Now, to find the distance between the two animals, we just compute the difference between the position equations for t = 2.72s

Distance = p₁(2.72s) - p₂(2.72s)

= ( 19.3 m/s)*2.72s - (1/2)*(7.1 m/s^2)*(2.72s)^2

= 26.2 m

So the animals are 26.2 meters apart.

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3 years ago

6 A simple ideal Rankine cycle with water as the working fluid operates between the pressure limits of 4 MPa in the boiler and 2

makvit [3.9K]

To solve this problem we will start by differentiating the values in each of the states of matter. Subsequently through the thermodynamic tables we will look for the values related to the entropy, enthalpy and respective specific volumes. Through the relationship of Power defined as the product between mass and enthalpy and mass, specific volume and pressure, we will find the energetic values in the two states investigated. We will start defining the states

State 1

$T_1 = 700\°C$

$P_1 = 4 Mpa$

From steam table

$h_1 =3906.41 KJ/Kg$

$s_1 = 7.62 KJ/Kg.K$

Now

$s_1 = s_2 = 7.62 KJ/Kg.K$ <em>As 1-2 is isentropic</em>

State 2

$P_2 = 20 Kpa$

$s_2 = 7.62 KJ/Kg \cdot K$

From steam table

$h_2 = 2513.33 KJ/Kg$

PART A) The power produced by turbine is the product between the mass and the enthalpy difference, then

$Power = m \times (h_1-h_2)$

$P = (50)(3906.41 - 2513.33)$

$P = 69654kW$

b) Pump Work

State 3

$P_3 = 20 Kpa$

$\upsilon= 0.001 m^3/kg$

The Work done by the pump is

$W= m\upsilon \Delta P$

$W = (50)(0.001)(4000-20)$

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3 years ago

In 2017, the company SpaceX became the first private company to send supplies to the International Space Station with a reusable

pav-90 [236]

Answer:

Approximately $3.98\; \rm m \cdot s^{-2}$ .

Assumption: air resistance on the rocket is negligible. Take $g = \rm 9.81\; m \cdot s^{-2}$ .

Explanation:

By Newton's Second Law of Motion, the acceleration of the rocket is proportional to the net force on it.

$\displaystyle \text{Acceleration} = \frac{\text{Net Force}}{\text{Mass}}$ .

Note that in this case, the uppercase letter $\rm M$ in the units stands for "mega-", which is the same as $10^6$ times the unit that follows. For example, $\rm 1\; Mg = 10^6\; g$ , while $\rm 1\; MN = 10^6\; N$ .

Convert the mass of the rocket and the thrust of its engines to SI standard units:

The standard unit for mass is kilograms: $\displaystyle m = \rm 552\; Mg = 552 \times 10^6\; g \times \frac{1\; \rm kg}{10^3\; g} = 552 \times 10^3 \; kg$ .
The standard for forces (including thrust) is Newtons: $\text{Thrust} = \rm 7.61 \; MN = 7.61 \times 10^6\; N$ .

At launch, the velocity of the rocket would be pretty low. Hence, compared to thrust and weight, the air resistance on the rocket would be pretty negligible. The two main forces that contribute to the net force of the rocket would be:

Thrust (which is supposed to go upwards), and
Weight (downwards due to gravity.)

The thrust on the rocket is already known to be $\rm 7.61 \times 10^6\; N$ . Since the rocket is quite close to the ground, the gravitational acceleration on it should be approximately $9.81\; \rm m \cdot s^{-2} = 9.81 \; N \cdot kg^{-1}$ . Hence, the weight on the rocket would be approximately $9.81\; \rm N \cdot kg^{-1} \times 552 \times 10^3\; kg = 5.41412\times 10^6\; N$ .

The magnitude of the net force on the rocket would be

$\begin{aligned}&\text{Thrust} - \text{Weight} \\ &= 7.61 \times 10^6\; \rm N - 5.41412\times 10^6\; N \\ &\approx 2.19 \times 10^6\; \rm N\end{aligned}$ .

Apply the formula $\displaystyle \text{Acceleration} = \frac{\text{Net Force}}{\text{Mass}}$ to find the net force on the rocket. To make sure that the output (acceleration) is in SI units (meters-per-second,) make sure that the inputs (net force and mass) are also in SI units (Newtons for net force and kilograms for mass.)

$\begin{aligned}\displaystyle &\text{Acceleration} \\ &= \frac{\text{Net Force}}{\text{Mass}} \\ &= \frac{2.19 \times 10^6\; \rm N}{552 \times 10^3\; \rm kg} \\ &\approx \rm 3.98\; \rm m \cdot s^{-2}\end{aligned}$ .