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Arisa [49]
3 years ago
13

9. Two moving objects will change their motion if which of the following is true? a.There is no friction on the objects. b. The

objects collide with each other. c.One object is heavier than the other. d.The objects are moving in space.​
Physics
1 answer:
adoni [48]3 years ago
8 0

Answer:

lol

Explanation:

lolllll i need points sorry

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You guys bad news... i failed nnn :c
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Read 2 more answers
A 910 kg car is approaching a loop-the-loop. The loop has a diameter of 50 m. Determine the minimum speed the car must have at t
scoundrel [369]

Answer:

The minimum speed the car must have at the top of the loop to not fall = 35 m/s

Explanation:

Anywhere else on the loop, the speed needed to keep the car in the loop is obtained from the force that keeps the body in circular motion around the loop which has to just match the force of gravity on the car. (Given that frictional force = 0)

mv²/r = mg

v² = gr = 9.8 × 25 = 245

v = 15.65 m/s

But at the top, the change in kinetic energy of the car must match the potential energy at the very top of the loop-the-loop

Change in kinetic energy = potential energy at the top

Change in kinetic energy = (mv₂² - mv₁²)/2

v₁ = velocity required to stay in the loop anywhere else = 15.65 m/s

v₂ = minimum velocity the car must have at the top of the loop to not fall

And potential energy at the top of the loop = mgh (where h = the diameter of the loop)

(mv₂² - mv₁²)/2 = mgh

(v₂² - v₁²) = 2gh

(v₂² - (15.65)²) = 2×9.8×50

v₂² - 245 = 980

v₂² = 1225

v₂ = 35 m/s

Hence, the minimum speed the car must have at the top of the loop to not fall = 35 m/s

4 0
3 years ago
During combustion reactions, explain why the energy of the reactants must exceed the total energy of the products​
maxonik [38]

Answer:

In these reactions the products are higher in energy than the reactants. ... This barrier is due to the fact that to make CO2 and H2O we have to break 4 carbon-hydrogen bonds and some ...

Explanation:

8 0
3 years ago
If the mass of the ladder is 12.0 kgkg, the mass of the painter is 55.0 kgkg, and the ladder begins to slip at its base when her
Marysya12 [62]

Answer:

 μ = 0.336

Explanation:

We will work on this exercise with the expressions of transactional and rotational equilibrium.

Let's start with rotational balance, for this we set a reference system at the top of the ladder, where it touches the wall and we will assign as positive the anti-clockwise direction of rotation

          fr L sin θ - W L / 2 cos θ - W_painter 0.3 L cos θ  = 0

          fr sin θ  - cos θ  (W / 2 + 0,3 W_painter) = 0

          fr = cotan θ  (W / 2 + 0,3 W_painter)

Now let's write the equilibrium translation equation

     

X axis

        F1 - fr = 0

        F1 = fr

the friction force has the expression

       fr = μ N

Y Axis

       N - W - W_painter = 0

       N = W + W_painter

       

we substitute

      fr = μ (W + W_painter)

we substitute in the endowment equilibrium equation

     μ (W + W_painter) = cotan θ  (W / 2 + 0,3 W_painter)

      μ = cotan θ (W / 2 + 0,3 W_painter) / (W + W_painter)

we substitute the values ​​they give

      μ = cotan θ  (12/2 + 0.3 55) / (12 + 55)

      μ = cotan θ  (22.5 / 67)

      μ = cotan tea (0.336)

To finish the problem, we must indicate the angle of the staircase or catcher data to find the angle, if we assume that the angle is tea = 45

       cotan 45 = 1 / tan 45 = 1

the result is

    μ = 0.336

5 0
4 years ago
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