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3 years ago

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The angular quantities: angular displacement, angular velocity, angular acceleration, are defined in a very similar way to the l

inear quantities. The angular displacement represents the angle through which an object has rotated (the difference between two angular positions). The angular velocity equals the angular displacement divided by the elapsed time. The angular acceleration equals the change in the angular velocity divided by the elapsed time. Suppose an object travels through an angular displacement, Δθ, of 50.9 rad over a time of 4.10 s. What is the average angular velocity, ω?

1 answer:

Lena [83]3 years ago

4 0

Answer: [tex]12.415 rad.s^{-1}[/tex]

Explanation: Angular velocity is the rate of change in angular displacement.

We know that:

Angular velocity, $\omega= \frac{\Delta \theta}{t}$ ....................(1)

where:

t= time

$\Delta \theta$ = angular displacement in radians

<u>Given that:</u>

t = 4.10 s

Δθ = 50.9 radian

Putting the respective values in eq. (1)

$\omega = \frac{50.9}{4.10}$

$\omega = 12.415 rad.s^{-1}$

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A 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the cable for the following case

Murljashka [212]

Answer:

The answer is below

Explanation:

A 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the

cable for the following cases:

a. The load moves downward at a constant velocity

b. The load accelerates downward at a rate 0.4 m/s??

C. The load accelerates upward at a rate 0.4 m/s??

Solution:

Acceleration due to gravity (g) = 10 m/s²

a) Given that the mass of the crane (m) is 140 kg. If the load moves downward, the tension (T) is given by:

mg - T = ma

Since the load has a constant velocity, hence acceleration (a) = 0. Therefore:

mg - T = m(0)

mg - T = 0

T = mg

T = 140(10) = 1400 N

T = 1400 N

b) If the load moves downward, the tension (T) is given by:

mg - T = ma

T = mg - ma = m(g - a)

T = 140(10 - 0.4) = 140(9.96) = 134.4

T = 134.4 N

c) If the load moves upward, the tension (T) is given by:

T - mg = ma

T = ma + mg = m(a + g)

T = 140(0.4 + 10) = 140(10.4)

T = 145.6 N

2) To find the distance (s) if the load move from rest (u= 0) and accelerates for 20 seconds (t = 20). We use:

s = ut + (1/2)gt²

s = 0(20) + (1/2)(10)(20)²

s = 2000 m

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3 years ago

in this model, the velocity of the spacecraft at position 2 is A.) equal to B.) greater than C.) less than the velocity of the c

finlep [7]

<h2>1. Right answer: the velocity of the spacecraft at position 2 is <u>greater than</u> the velocity of the craft at position 4. </h2>

This is due the gravity field of the planet (The Earth in this case) is used to accelerate the craft. This is true when in a specific point the direction of the movement of the craft is the same direction of the movement of the planet.

In this case the craft will be “catched” by the Earth’s gravitational field, making the craft to enter a circular orbit.

<h2>2. Right answer: At position 1, the direction of the spacecraft changes because of <u>the gravitational force between Earth and the spacecraft. </u></h2>

As explained in the prior answer, this is the exact and correct point where the trajectory of the spacecraft enters into a circular orbit because of the attraction due gravity of the Earth and therefore changes its direction.

<h2>3. Right answer: Position 3 represents <u>the orbital path or velocity of Earth </u></h2>

Being this the orbital path of the Earth and considering the trajectory of the craft, the condition of accelerating the craft is accomplished.

If the orbital path of the Earth were the opposite from the shown in the figure, the effect on the craft would be braking.

Note all of these is related to the <u>gravitational assistance. </u>

<u>Gravitational assistance</u> is the maneuver in which the energy of the gravitational field of a planet or satellite is used to obtain an acceleration or braking of the probe changing its trajectory.

This maneuver is also called <em>slingshot effect, swing-by</em> or <em>gravity assist</em>. It is a common technique in space for the outer Solar System missions , in order to save costs in the launch rocket and thrusters.

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3 years ago

Read 2 more answers

Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equ

lesantik [10]

Answer:

The magnitude of the force on positive charges will be $\bf{227.06~N}$ and the magnitude of the force on the negative charge is $\bf{302.7~N}$ .

Explanation:

Given:

The value of the charges, $q = 3.5~\mu C$ .

The length of each side of the triangle, $l = 2.9~cm$ .

Consider a equilateral triangle $\bigtriangleup ABC$ , as shown in the figure. Let two point charges of magnitude $q$ are situated at points $A$ and $B$ and another point charge $-q$ is situated at point $C$ .

The value of the force on the charge at point $A$ due to charge at point $C$ is given by

$F_{CA} = \dfrac{kq^{2}}{l^{2}},~~along~CA$

The value of the force on the charge at point $A$ due to charge at point $B$ is given by

$F_{BA} = \dfrac{kq^{2}}{l^{2}},~~along~BA$

The net resultant force on the charge at point $A$ is given by

$~~~~F_{A} = \sqrt{F_{BA}^{2} + F_{CA}^{2} + 2F_{BA}F_{CA}\cos 60^{0}}\\~~~~~= \dfrac{kq^{2}}{l^{2}}\sqrt{2(1 + \cos 60^{0})}\\or, F_{A}= \dfrac{\sqrt{3}kq^{2}}{l^{2}}~~~~~~~~~~~~~~~~~~~~(1)$

The value of the force on the charge at point $B$ due to charge at point $C$ is given by

$F_{CB} = \dfrac{kq^{2}}{l^{2}},~~along~CB$

The value of the force on the charge at point $B$ due to charge at point $A$ is given by

$F_{AB} = \dfrac{kq^{2}}{l^{2}},~~along~AB$

The net resultant force on the charge at point $B$ is given by

$~~~~F_{B} = \sqrt{F_{AB}^{2} + F_{CB}^{2} + 2F_{AB}F_{CB}\cos 60^{0}}\\~~~~~= \dfrac{kq^{2}}{l^{2}}\sqrt{2(1 + \cos 60^{0})}\\or, F_{B}= \dfrac{\sqrt{3}kq^{2}}{l^{2}}~~~~~~~~~~~~~~~~~~~~(2)$

The value of the force on the charge at point $C$ due to charge at point $A$ is given by

$F_{AC} = \dfrac{kq^{2}}{l^{2}},~~along~AC$

The value of the force on the charge at point $C$ due to charge at point $B$ is given by

$F_{BC} = \dfrac{kq^{2}}{l^{2}},~~along~BC$

The net resultant force on the charge at point $C$ is given by

$F_{C} = 2F_{BC} \sin 60^{0}~~along~the~line~perpendicular~to~AB\\~~~~~= \dfrac{2kq^{2}}{l^{2}}\sin 60^{0}~~~~~~~~~~~~~~~~~~~~(3)$

Substitute $3.5~\mu C$ for $q$ , $0.029~m$ for $l$ and $9 \times 10^{9}~Nm^{2}C^{-2}$ for $k$ in equation (1), we have

$F_{A} = \dfrac{\sqrt{3}(9 \times 10^{9}~Nm^{2}C^{-2})(3.5 \times 10^{-6}~C)^{2}}{(0.029~m)^{2}}\\~~~~~= 227.06~N$

Substitute $3.5~\mu C$ for $q$ , $0.029~m$ for $l$ and $9 \times 10^{9}~Nm^{2}C^{-2}$ for $k$ in equation (2), we have

$F_{B} = \dfrac{\sqrt{3}(9 \times 10^{9}~Nm^{2}C^{-2})(3.5 \times 10^{-6}~C)^{2}}{(0.029~m)^{2}}\\~~~~~= 227.06~N$

Substitute $3.5~\mu C$ for $q$ , $0.029~m$ for $l$ and $9 \times 10^{9}~Nm^{2}C^{-2}$ for $k$ in equation (3), we have

$F_{C} = \dfrac{2(9 \times 10^{9}~Nm^{2}C^{-2})(3.5 \times 10^{-6}~C)^{2}}{(0.029~m)^{2}} \sin 60^{0}\\~~~~~= 302.7~N$

8 0

3 years ago

Two skaters collide and grab on to each other on frictionless ice. One of them, of mass 66.0 kg , is moving to the right at 2.00

ivann1987 [24]

Answer:

The velocity of skaters after collision is 0.55 m/s and they are moving to the left

Explanation:

Consider m₁ and m₂ are the masses of two skaters and their initial velocity be v₁ and v₂ respectively.

Consider the velocity along right direction as positive while negative for left direction.

According to the problem,

Mass of first skater, m₁ = 66.0 kg

Mass of second skater, m₂ = 75.0 kg

Velocity of first skater, v₁ = + 2.00 m/s

Velocity of second skater, v₂ = -2.80 m/s

Since, the two skaters grab each other after collision. Hence, they are moving with same final velocity v.

Applying conservation of linear momentum,

Momentum before collision = Momentum after collision

m₁v₁ + m₂v₂ = (m₁+m₂)v

Substitute the suitable values in the above equation.

66 x 2 - 75 x 2.8 = (66 + 75 )v

$v= \frac{-78}{141}$

v = -0.55 m/s

The negative sign denotes that both the skaters are moving in the left direction.

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3 years ago

A person wants to use a lever to lift a dumpster weighing 4200 N.4200 N. The lever is arranged so that it is approximately horiz

natta225 [31]

Answer:

1260 Nm

Explanation:

As one end of the lever makes contact at the point right below the dumpster center of mass and has a distance of 0.3 m to the lever pivot point, plus the fact that the level is also approximately horizontal, we can safely assume that the dumpster gravity is directly perpendicular to the level.

This would make the torque caused by the dumpster gravity on the level is

T = LF = 4200*0.3 = 1260 Nm

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3 years ago