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Likurg_2 [28]
3 years ago
7

6) If a mass of an object is decreased to half and acting force is reduced by quarter the acceleration of its motion

Physics
1 answer:
Zepler [3.9K]3 years ago
7 0

Answer:

Decreases to half.

Explanation:

From the question given above, the following data were obtained:

Initial mass (m₁) = m

Initial force (F₁) = F

Initial acceleration (a₁) =?

Final mass (m₂) = ½m

Final force (F₂) = ¼F

Final acceleration (a₂) =?

Next, we shall determine a₁. This can be obtained as follow:

F₁ = m₁a₁

F = ma₁

Divide both side by m

a₁ = F / m

Next, we shall determine a₂.

F₂ = m₂a₂

¼F = ½ma₂

2F = 4ma₂

Divide both side by 4m

a₂ = 2F / 4m

a₂ = F / 2m

Finally, we shall determine the ratio of a₂ to a₁. This can be obtained as follow:

a₁ = F / m

a₂ = F / 2m

a₂ : a₁ = a₂ / a₁

a₂ / a₁ = F/2m ÷ F/m

a₂ / a₁ = F/2m × m/F

a₂ / a₁ = ½

Cross multiply

a₂ = ½a₁

From the illustrations made above, the acceleration of the car will decrease to half the original acceleration

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artcher [175]
<h2>Answer:</h2>

0.126m

<h2>Explanation:</h2>

According to Hooke's law, the force (F) acting on a spring to cause an extension or compression (e) is given by;

F = k x e            -------------------(i)

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k = the spring's constant.

From the question, the force acting on the spring is the weight(W) of the mass. i.e

F = W               -----------------------(ii)

<em>But;</em>

W = m x g;

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m = mass of the object

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<em>From equation (ii), it implies that;</em>

F = W = m x g

<em>Now substitute F = m x g into equation(i) as follows;</em>

F = k x e

m x g = k x e      ------------------(iii)

<em>From the question;</em>

m = m1 = 3.5kg

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3.5 x 10 = 278 x e

35 = 278e

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e = 35/278

e = 0.126m

Therefore, the distance the spring is stretched from its unstretched length (which is the same as the extension of the spring) is 0.126m

3 0
3 years ago
In one scene in the movie The Godfather II, a solid gold phone is passed around a large table for everyone to see. Suppose the v
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3 years ago
I need the solution to this
posledela

Answer:

He could jump 2.6 meters high.

Explanation:

Jumping a height of 1.3m requires a certain initial velocity v_0. It turns out that this scenario can be turned into an equivalent: if a person is dropped from a height of 1.3m in free fall, his velocity right before landing on the ground will be v_0. To answer this equivalent question, we use the kinematic equation:

v_0 = \sqrt{2gh}=\sqrt{2\cdot 9.8\frac{m}{s^2}\cdot 1.3m}=5.0\frac{m}{s}

With this result, we turn back to the original question on Earth: the person needs an initial velocity of 5 m/s to jump 1.3m high, on the Earth.

Now let's go to the other planet. It's smaller, half the radius, and its meadows are distinctly greener. Since its density is the same as one of the Earth, only its radius is half, we can argue that the gravitational acceleration g will be <em>half</em> of that of the Earth (you can verify this is true by writing down the Newton's formula for gravity, use volume of the sphere times density instead of the mass of the Earth, then see what happens to g when halving the radius). So, the question now becomes: from which height should the person be dropped in free fall so that his landing speed is 5 m/s ? Again, the kinematic equation comes in handy:

v_0^2 = 2g_{1/2}h\implies \\h = \frac{v_0^2}{2g_{1/2}}=\frac{25\frac{m^2}{s^2}}{2\cdot 4.9\frac{m}{s^2}}=2.6m

This results tells you, that on the planet X, which just half the radius of the Earth, a person will jump up to the height of 2.6 meters with same effort as on the Earth. This is exactly twice the height he jumps on Earth. It now all makes sense.

6 0
3 years ago
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yes  ( true)

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t = 22.22m/s × s2/1.15m
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6 0
3 years ago
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