A group of a few thousand stars that are loosely held together by gravity is called a

What do you need to calculate the mechanical advantage of a block and tackle

Ira Lisetskai [31]

In a block and tackle, some friction in the pulleys will reduce the mechanical advantage of the machine. To include friction in a calculation of the mechanical advantage of a block and tackle, divide the weight of the object being lifted by the weight necessary to lift it.

Hope this helps

7 0

2 years ago

Read 2 more answers

Two air track carts move along an air track towards each other. Cart A has a mass of 450 g and moves toward the right with a spe

Blizzard [7]

Answer:

The right answer is D) the total momentum of the system is 0.047 kg · m/s toward the right.

Explanation:

Hi there!

The total momentum of the system is given by the sum of the momentum vectors of each cart. The momentum is calculated as follows:

p = m · v

Where:

p = momentum.

m = mass.

v = velocity.

Then, the momentum of the system will be the momentum of cart A plus the momentum of cart B (let´s consider the right as the positive direction):

mA · vA + mB · Vb

0.450 kg · 0.850 m/s + 0.300 kg · (- 1.12 m/s) = 0.047 kg · m/s

The right answer is D) the total momentum of the system is 0.047 kg · m/s toward the right.

6 0

2 years ago

You attach a meter stick to an oak tree, such that the top of the meter stick is 2.27 meters above the ground. later, an acorn f

Alexandra [31]

The acorn was at a height of 4.15 m from the ground before it drops.

The acorn takes a time t to fall through a distance h₁, which is the length of the scale. When the acorn reaches the top of the scale, its velocity is u.

Calculate the speed of the acorn at the top of the scale, using the equation of motion,

$s=ut+ \frac{1}{2} at^2$

Since the acorn falls freely under gravity, its acceleration is equal to the acceleration due to gravity g.

Substitute 2.27 m for s (=h₁), 0.301 s for t and 9.8 m/s² for a (=g).

$s=ut+ \frac{1}{2} at^2\\ (2.27 m)=u(0.301s)+\frac{1}{2}(9.8m/s^2)(0.301s)^2\\ u=\frac{1.8261m}{0.301s} =6.067m/s$

If the acorn starts from rest and reaches a speed of 6.067 m/s at the top of the scale, it would have fallen a distance h₂ to achieve this speed.

Use the equation of motion,

$v^2=u^2+2as$

Substitute 6.067 m/s for v, 0 m/s for u, 9.8 m/s² for a (=g) and h₂ for s.

$v^2=u^2+2as\\ (6.067m/s)^2=(0m/s)^2+2(9.8m/s^2)h_2\\ h_2=\frac{(6.067m/s)^2}{2(9.8m/s^2)} =1.878 m$

The height h above the ground at which the acorn was is given by,

$h=h_1+h_2=(2.27 m)+(1.878 m)=4.148 m$

The acorn was at a height 4.15m from the ground before dropping down.

3 0

2 years ago

A particle vibrates in Simple Harmonic Motion with amplitude. What will be its displacement in one time-period? A 2A 4A 0

yan [13]

'Displacement' is the distance and direction between the starting point and
ending point, regardless of the path followed to get there.

A particle that's executing simple harmonic motion is always in the same place
where it was one time period ago, and where it will be later after another time
period has passed.

So its displacement during exactly one time period is exactly zero.

3 0

3 years ago

A 1.0-kg block and a 2.0-kg block are pressed together on a horizontal frictionless surface with a compressed very light spring

egoroff_w [7]

Answer:

4. both blocks will both have the same amount of kinetic energy.

Explanation:

When the blocks are released free from the compression force, the spring exerts equal and opposite force on each block but the block with heavier (double) mass will attain slower ( half ) speed as compared to the lighter block according to the law of inertia. This works in synchronization to energy conservation.

Spring force is given as:

$F=k.\Delta x$

where: $\Delta x=$ length of compression in the spring

We know kinetic energy is given by:

$KE=\frac{1}{2} m.v^2$

Hence the kinetic energy of both the blocks is equal when they are released to move free.

8 0

3 years ago