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3 years ago

The amount of gravity between 1kg of lead and Earth is ___ the amount of gravity between 1kg of marshmallows and Earth.

1 answer:

3 0

The amount of gravity between 1 kg of lead and Earth is the same as the amount of gravity between 1 kg of marshmallows and Earth.

Answer: Option C

<u>Explanation:</u>

According to universal law of gravity, the gravitational force is directly proportionate to the product of masses of the interacting objects and inversely proportionate to its distance squared between them.

Since the mass of Earth is very high compared to the mass of the other objects, be it lead or marshmallows, and the distance will be mostly same. Thus, there will be no change in the amount of gravity acting between them. Thus, the amount of gravity between 1 kg of lead and Earth is the same as the amount of gravity between 1 kg of marshmallows and Earth.

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3 years ago

25. IF THE ROADWAY IS WET AND THE CAR STARTS TO SKID, YOU SHOULD: A. Slow down by pumping the brakes quickly and firmly B. Slow

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If the roadway is wet and the cars starts to skid, one should slow down by shifting to lower gear. The correct option is B.

<h3>What are gears?</h3>

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For a car to avoid skidding on the wet roads, one must drive the car slowly. This can be done by shifting the gear to the lower one. This will help car to keep balance when brakes are applied on wet roads.

Thus, the correct option is B.

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2 years ago

I.Name two commonly used thermometric liquids.

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3 years ago

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Nataly_w [17]

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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

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3 years ago