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sergij07 [2.7K]
3 years ago
9

Why is the mechanical advantage of using a single pulley always 1? Assume there’s no friction. A. The input force is in a differ

ent direction than the output force. B. The input force is less than the output force. C. The input force is greater than the output force. D. The input force is equal to the output force.
Physics
2 answers:
RoseWind [281]3 years ago
8 0

Answer:

D) The input force is equal to the output force.

Explanation:

I got this question on a test and this was the right answer

~Please mark as the brainliest :)

STALIN [3.7K]3 years ago
3 0

The answer  is D. The input force is equal to the output force.

I just did test and got it right

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Select the phrases that correctly describe density. Density is a chemical property. Density relates a mass to its volume. Densit
sp2606 [1]

Answer:

Density relates a mass to its volume.

Density varies with temperature

Density determines if a substance floats or sinks.

Density may have units of grams per milliliter (g/mL)

Explanation:

Density D is a characteristic property of a substance or material and is defined as the relationship between the mass m of a body or substance and the volume V it occupies:

D=\frac{m}{V}

This means the density is inversely proportional to the volume.

On the other hand, density is a scalar quantity and according to the International System of Units its unit is D=\frac{kg}{m^{3}} , although it can be also expressed in \frac{g}{ml}.

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5 0
3 years ago
Suppose that an object is moving along a vertical line. Its vertical position is given by the equation L(t) = 2t3 + t2-5t + 1, w
Tatiana [17]

Answer:

The average velocity is

266\frac{m}{s},274\frac{m}{s} and 117\frac{m}{s} respectively.

Explanation:

Let's start writing the vertical position equation :

L(t)=2t^{3}+t^{2}-5t+1

Where distance is measured in meters and time in seconds.

The average velocity is equal to the position variation divided by the time variation.

V_{avg}=\frac{Displacement}{Time} = Δx / Δt = \frac{x2-x1}{t2-t1}

For the first time interval :

t1 = 5 s → t2 = 8 s

The time variation is :

t2-t1=8s-5s=3s

For the position variation we use the vertical position equation :

x2=L(8s)=2.(8)^{3}+8^{2}-5.8+1=1049m

x1=L(5s)=2.(5)^{3}+5^{2}-5.5+1=251m

Δx = x2 - x1 = 1049 m - 251 m = 798 m

The average velocity for this interval is

\frac{798m}{3s}=266\frac{m}{s}

For the second time interval :

t1 = 4 s → t2 = 9 s

x2=L(9s)=2.(9)^{3}+9^{2}-5.9+1=1495m

x1=L(4s)=2.(4)^{3}+4^{2}-5.4+1=125m

Δx = x2 - x1 = 1495 m - 125 m = 1370 m

And the time variation is t2 - t1 = 9 s - 4 s = 5 s

The average velocity for this interval is :

\frac{1370m}{5s}=274\frac{m}{s}

Finally for the third time interval :

t1 = 1 s → t2 = 7 s

The time variation is t2 - t1 = 7 s - 1 s = 6 s

Then

x2=L(7s)=2.(7)^{3}+7^{2}-5.7+1=701m

x1=L(1s)=2.(1)^{3}+1^{2}-5.1+1=-1m

The position variation is x2 - x1 = 701 m - (-1 m) = 702 m

The average velocity is

\frac{702m}{6s}=117\frac{m}{s}

5 0
3 years ago
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