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3 years ago

What is the primary force that helps suspension bridges use cables to hold their spans up?

Physics

2 answers:

Wittaler [7]3 years ago

7 0

Answer:

a) Tension Force

Explanation:

Since the platform suspended by the strings so here weight of the platform is counter-balanced by the force due to strings.

Here all the molecules of string has tendency to connect with each other so that the internal force between the molecules will help them to bound with each other.

Since we know that this force between the molecules are electromagnetic force due to electron sharing between each other

So this force is known as tension force between the molecules.

So here correct answer force the suspension bridges, cables will exert internal tensile force to counter-balance the weight of the platform.

So correct answer will be

Tension force

Whitepunk [10]3 years ago

3 0

Hello There!

If i remember correctly, it is tension force.

Hope This Helps You!
Good Luck :)

- Hannah ❤

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What do the arrows in this photograph represent?

Anna35 [415]

Answer:

<u>The answer would be D) gravity and air resistance.</u>

4 0

2 years ago

Trying to beat the heat of the last summer, a physics grad student went to the local toy store and purchased a plastic child's s

svp [43]

Answer:

625 piece.

Explanation:

Let n be the required no of piece

200 litre = 200 x 10⁻³ m³ = 200 x 10⁻³ x 10³ kg = 200 Kg

mass of ice piece = n x 30 x /1000

Heat lost by ice pieces = (n x 30) / 1000 x ( 80 + 1 x 16⁰C)

Heat gained by water = 200 x 1 x 9⁰C

Heat lost = Heat gained

(n x 30) /1000 x ( 80 + 16 ) =200 x 9

2.88 n = 1800

n = 625

6 0

4 years ago

Describe and give an example of mutualism.

Slav-nsk [51]

Answer:

Mutualism, commensalism, parasitism, competition, and predation.

Explanation:

mutualism- relationship between two or more organisms where both are benefited. Example-oxpecker with rhino/zebra. They eat bugs off of them which means that they are getting food, while the rhino/zebra are getting cleaned up with pest control.

commensalism- relationship between two organisms where one benefits and the other isn't benefited or harmed. EX- tree frogs use plants as protectioin.he frog is benefited, and the plant is neither harmed nor benefited. Remora fish have a disk on their heads that they use to attach themselves to larger animals for protection. The animals they attach to are neither harmed nor benefited.

parasitism- in a relationship where an organism benefits at the expense of the other. (one is benefited while the other is harmed) ex- fleas and ticks that live on cats and dogs, tape worms that live in people and animals that eat the food which means that the people aren't getting the food or nutrition that they eat. lice, etc

competition- interaction within organisms/species in which both the organisms/species are harmed and is apart of natural selection. Examples may include two males fighting over a mate, animals competing over food, limited habitats that they are fighting over, territory, etc.

predation- the preying of one animal on another. It's where the predator hunts and eats another organism which is its prey. categorized within-(1) carnivory, (2) herbivory, (3) parasitism, and (4) mutualism. Each type of predation can by categorized based on whether or not it results in the death of the prey.ex- owls hunting mice, wolves hunting rabbits, lion hunting gazelle, etc.

7 0

3 years ago

You toss a coin into a wishing well full of i liquid denser than the coin what will happen to the coin

Ksju [112]

If the liquid is denser than the coin, then the coin will eventually
come to rest floating, with part of it above the surface of the liquid.

That's exactly the situation if you drop the coin into mercury.

Density of copper . . . 8.96 gm/cm³
iron . . . 7.87
zinc . . . 7.13
silver . . 10.5
nickel . . .8.91
lead . . . 11.4

Density of mercury . . . 13.53 gm/cm³ !

6 0

4 years ago

A 4.00 kg object is moving at 5.00 m/s NORTH. It strikes a 6.00 kg object that is moving WEST at 2.00 m/s. The objects undergo a

Veronika [31]

We are given that an object is moving north at a speed of 5 m/s, lets call this object 1. We have another object moving west at a speed of 2 m/s, this is object 2. A diagram of the problem is the following:

To determine the loss in kinetic energy we need to determine the difference in kinetic energy before the collision and after the collision:

$\Delta K=K_f-K_0$

The final kinetic energy is:

$K_f=\frac{1}{2}(m_1+m_2)v^2_f$

We use the sum of the masses because the objects are stuck together after the collision. The initial kinetic energy is:

$K_0=\frac{1}{2}m_1v^2_{01}+\frac{1}{2}m_2v^2_{02}$

Substituting we get:

$\Delta K=\frac{1}{2}(m_1+m_2)v^2_f-(\frac{1}{2}m_1v^2_{01}+\frac{1}{2}m_2v^2_{02})$

The only missing variable is the final velocity. To determine the final velocity we will use the conservation of momentum.

We will use the conservation of momentum in the horizontal direction (west) and the conservation of momentum in the vertical direction (north).

In the horizontal direction we have:

$m_1v_{h1}+m_2v_{h2}=(m_1+m_2)v_{hf}$

Since the object 1 has no velocity in the horizontal direction we have that:

$\begin{gathered} m_1(0)+m_2v_{h2}=(m_1+m_2)v_{hf} \\ m_2v_{h2}=(m_1+m_2)v_{hf} \end{gathered}$

Now we solve for the final horizontal velocity:

$\frac{m_2v_{h2}}{\mleft(m_1+m_2\mright)}=v_{hf}$

Now we substitute the values:

$\frac{(6kg)(2\frac{m}{s})}{(4kg+6kg)}=v_{hf}$

Solving the operations we get:

$1.2\frac{m}{s}=v_{hf}$

Now we use the conservation of momentum in the vertical direction, we get:

$m_1v_{v1}+m_2v_{v2}=(m_1+m_2)v_{vf}$

Since the second object has no vertical velocity we get:

$m_1v_{v1}=(m_1+m_2)v_{vf}$

Now w solve for the final vertical velocity, we get:

$\frac{m_1v_{v1}}{\mleft(m_1+m_2\mright)}=v_{vf}$

Now we substitute the values:

$\frac{(4kg)(5\frac{m}{s})}{(4kg+6kg)}=v_{vf}$

Now we solve the operations:

$2\frac{m}{s}=v_{vf}$

Now we determine the magnitude of the final velocity using the following formula:

$v_f=\sqrt[]{v^2_{hf}+v^2_{vf}}$

Substituting the values:

$v_f=\sqrt[]{(1.2\frac{m}{s})^2+(2\frac{m}{s})^2}$

Solving the operations:

$v_f=2.53\frac{m}{s}$

Now we substitute this in the formula for the kinetic energy and we get:

$\Delta K=\frac{1}{2}(4kg+6kg)(2.53\frac{m}{s})^2-(\frac{1}{2}(4kg)(5\frac{m}{s})^2+\frac{1}{2}(6kg)(2\frac{m}{s})^2)$

Solving the operations:

$\Delta K=32J-62J=-30J$

Therefore, there was a loss of 30J of kinetic energy.

6 0

1 year ago