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2 years ago

A 900 kg steel beam is supported by the two ropes shown in (Figure 1) . Calculate the tension in the rope.

1 answer:

8 0

Let T1 and T2 be tension in ropes1 and 2 respectively.
<span>since system is stationary (equilibrium), considering both ropes + beam as a system </span>

<span>for horizontal equilibrium (no movement in that direction, so resultant force must be zero horizontally) </span>
<span>T1sin(20) = T2sin(30) </span>
<span>=> T1 = T2sin(30) / sin(20) </span>

<span>for vertical equilibrium, (no movement in this direction, so resultant force must be zero vertically) </span>
<span>T1cos(20) + T2cos(30) = mg </span>

<span>m = 900kg, substituting for T1 </span>
<span>T2sin(30)*cos(20)/sin(20) + T2cos(30) = 900g </span>
<span>2.328*T2 = 900*9.8 </span>
<span>T2 = 3788.65N </span>
<span>so T1 from (1) </span>
<span>T1 = 5535.21N</span>

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A metal sphere with radius R1 has a charge Q1. Take the electric potential to be zero at an infinite distance from the sphere.

Airida [17]

Answer:

Part A : E = $\frac{1}{4\pi}$ ε₀ Q₁/R₁² Volt/meter

Part B : V = $\frac{1}{4\pi}$ ε₀ Q₁/R₁ Volt

Explanation:

Given that,

Charge distributed on the sphere is Q₁

The radius of sphere is R

₁

The electric potential at infinity is 0

The space around a charge in which its influence is felt is known in the electric field. The strength at any point inside the electric field is defined by the force experienced by a unit positive charge placed at that point.

If a unit positive charge is placed at the surface it experiences a force according to the Coulomb law is given by

F = $\frac{1}{4\pi}$ ε₀ Q₁/R₁²

Then the electric field at that point is

E = F/1

E = $\frac{1}{4\pi}$ ε₀ Q₁/R₁² Volt/meter

Part B

The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against electric forces.

Thus, the electric potential at the surface of the sphere of radius R₁ and charge distribution Q₁ is given by the relation

V = $\frac{1}{4\pi}$ ε₀ Q₁/R₁ Volt

4 0

1 year ago

The velocity of a car increases from 2.0 m/s to 16.0 m/s in a time period of 3.5 s. What was the average acceleration?

dangina [55]

Answer:

the acceleration is a=3ms^-2

Explanation:

8 0

1 year ago

Read 2 more answers

Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Initially ball A is mov

frez [133]

Answer:

Speed of ball A after collision is 3.7 m/s

Speed of ball B after collision is 2 m/s

Direction of ball A after collision is towards positive x axis

Total momentum after collision is m×4·21 kgm/s

Total kinetic energy after collision is m×8·85 J

Explanation:

<h3>If we consider two balls as a system as there is no external force initial momentum of the system must be equal to the final momentum of the system</h3>

Let the mass of each ball be m kg

v $_{1}$ be the velocity of ball A along positive x axis

v $_{2}$ be the velocity of ball A along positive y axis

u be the velocity of ball B along positive y axis

Conservation of momentum along x axis

m×3·7 = m× v $_{1}$

∴ v $_{1}$ = 3.7 m/s along positive x axis

Conservation of momentum along y axis

m×2 = m×u + m× v $_{2}$

2 = u + v $_{2}$ → equation 1

<h3>Assuming that there is no permanent deformation between the balls we can say that it is an elastic collision</h3><h3>And for an elastic collision, coefficient of restitution = 1</h3>

∴ relative velocity of approach = relative velocity of separation

-2 = v $_{2}$ - u → equation 2

By adding both equations 1 and 2 we get

v $_{2}$ = 0

∴ u = 2 m/s along positive y axis

Kinetic energy before collision and after collision remains constant because it is an elastic collision

Kinetic energy = (m×2² + m×3·7²)÷2

= 8·85×m J

Total momentum = m×√(2² + 3·7²)

= m× 4·21 kgm/s

3 0

2 years ago

lidiya [134]

Answer:

ma= ma

m⋅a = m⋅a

And equivalently:

am=ma

a⋅m = m⋅a

Explanation:

Question

Assuming this question "Similar to what you see in your textbook, you can generally omit the multiplication symbol as you answer questions online, except when the symbol is needed to make your meaning clear. For example, 1*10^5 is not the same as 110^5 . When you need to be explicit, type * (Shift + 8) to insert the multiplication operator. You will see a multiplication dot (⋅) appear in the answer box. Do not use the symbol x. For example, for the expression ma,

typing m⋅a would be correct, but mxa would be incorrect".

Solution to the problem

For this case we want to write a expression for ma, and based on the previous info we can write:

ma= ma

m⋅a = m⋅a

And equivalently:

am=ma

a⋅m = m⋅a

But is not correct do this:

mxa=mxa

axm = mxa

8 0

2 years ago

Read 2 more answers

A hot air balloon of total mass M (including passengers and luggage) is moving with a downward acceleration of magnitude a. As i

LUCKY_DIMON [66]

Answer:

The fraction of mass that was thrown out is calculated by the following Formula:

M - m = (3a/2)/(g²- (a²/2) - (ag/2))

Explanation:

We know that Force on a moving object is equal to the product of its mass and acceleration given as:

F = ma

And there is gravitational force always acting on an object in the downward direction which is equal to g = 9.8 ms⁻²

Here as a convention we will use positive sign with acceleration to represent downward acceleration and negative sign with acceleration represent upward acceleration.

Case 1:

Hot balloon of mass = M

acceleration = a

Upward force due to hot air = F = constant

Gravitational force downwards = Mg

Net force on balloon is given as:

Ma = Gravitational force - Upward Force

Ma = Mg - F (balloon is moving downwards so Mg > F)

F = Mg - Ma

F = M (g-a)

M = F/(g-a)

Case 2:

After the ballast has thrown out,the new mass is m. The new acceleration is -a/2 in the upward direction:

Net Force is given as:

-m(a/2) = mg - F (Balloon is moving upwards so F > mg)

F = mg + m(a/2)

F = m(g + (a/2))

m = F/(g + (a/2))

Calculating the fraction of the initial mass dropped:

$M-m = \frac{F}{g-a} - \frac{F}{g+\frac{a}{2} }\\M-m = F*[\frac{1}{g-a} - \frac{1}{g+\frac{a}{2} }]\\M-m = F*[\frac{(g+(a/2)) - (g-a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{g+(a/2) - g + a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{(3a/2)}{g^{2}-\frac{a^{2}}{2}-\frac{ag}{2}} ]$

5 0

2 years ago