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4 years ago

11

Suppose a pulley with an IMA of 2 was used with an input force of 50 N to lift a box 45 cm that weighs 100 N. What distance was

the rope pulled?

1 answer:

IRINA_888 [86]4 years ago

6 0

The ideal mechanical advantage (IMA) is the number of times in which the input force is multiplied under ideal conditions. If the real force was only 50N, the distance at which the rope was pulled will be twice the distance given in this item. The answer is 90 cm.

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3. A sprinter leaves the starting blocks with an acceleration of 4.5 m/s2. What is the

UkoKoshka [18]

Hi there! :)

$\large\boxed{v_{f} = 18 m/s}$

Use the following kinematic equation to solve for the final velocity:

$v_{f} = v_{i} + at$

In this instance, the runner started from rest, so the initial velocity is 0 m/s. We can rewrite the equation as:

$v_{f} = at$

Plug in the given acceleration and time:

$v_{f} = 4.5 * 4 = 18 m/s$

5 0

3 years ago

A 2.93 kg particle has a velocity of (2.98 i hat - 3.98 j) m/s.

cupoosta [38]

Answer:

a) The x and y components of the momentum are $8.731\,\frac{kg\cdot m}{s}$ and $-11.661\,\frac{kg\cdot m}{s}$ , respectively.

b) The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

Explanation:

a) The vectorial equation of momentum is represented by the following expression:

$\vec p = m\cdot \vec v$ (1)

Where:

$\vec p$ - Vector momentum, measured in kilogram-meters per second.

$m$ - Mass of the particle, measured in kilograms.

$\vec v$ - Vector velocity, measured in meters per second.

If we know that $m = 2.93\,kg$ and $\vec v = 2.98\,\hat{i}-3.98\,\hat{j}\,\,\,\left[\frac{m}{s} \right]$ , then the momentum is:

$\vec p = (2.93)\cdot (2.98\,\hat{i}-3.98\,\hat{j})\,\,\,\left[\frac{kg\cdot m}{s} \right]$

$\vec p = 8.731\,\hat{i}-11.661\,\hat{j}\,\,\,\left[\frac{kg\cdot m}{s} \right]$

The x and y components of the momentum are $8.731\,\frac{kg\cdot m}{s}$ and $-11.661\,\frac{kg\cdot m}{s}$ , respectively.

b) The magnitude and direction of momentum are represented by the following expressions:

$\|\vec p \| = \sqrt{p_{x}^{2}+p_{y}^{2}}$ (2)

$\theta = \tan^{-1}\left(\frac{p_{y}}{p_{x}} \right)$ (3)

Where:

$\|\vec p\|$ - Magnitude of momentum, measured in kilogram-meters per second.

$\theta$ - Direction of momentum, measured in sexagesimal degrees.

If we know that $p_{x} = 8.731\,\frac{kg\cdot m}{s}$ and $p_{y} = -11.661\,\frac{kg\cdot m}{s}$ , then the magnitude and direction of momentum are, respectively:

$\|\vec p\| = \sqrt{\left(8.731\,\frac{kg\cdot m}{s} \right)^{2}+\left(-11.661\,\frac{kg\cdot m}{s} \right)^{2}}$

$\|\vec p\| \approx 14.567\,\frac{kg\cdot m}{s}$

$\theta =\tan^{-1}\left(\frac{-11.661\,\frac{kg\cdot m}{s} }{8.731\,\frac{kg\cdot m}{s} } \right)$

$\theta \approx 306.823^{\circ}$

The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

6 0

3 years ago

In general, if the temperature of a chemical reaction is increased, the reaction rateA. IncreasesB. decreasesC. remains the same

In-s [12.5K]

A. Increases

I would assume this to be the answer because heat is another form of energy. If there is more energy the molecules will become more active. This makes A the most logical answer.

6 0

3 years ago

As you found out in the previous part, bernoulli's equation tells us that a fluid element that flows through a flow tube with de

8090 [49]

This causes the fluid to increase its speed. Bernoulli's principle tells us that an increase in the speed of a fluid happens at the same time with a reduction in pressure or a reduction in the fluid's potential energy. This necessitates that the amount of kinetic energy, potential energy and internal energy stays persistent.

8 0

3 years ago

What is the kinetic energy of a 0.25kg ball rolling at a speed of 2.5m/s

Lisa [10]

Answer:

Explanation:

Kinetic Energy formula:

KE = $\frac{1}{2}$ mv²

m=mass

v=speed

Given:

m=0.25kg

v=2.5m/s

Plug the values in:

KE = 1/2(0.25kg)(2.5m/s)²

KE = 0.78125 J (Joules)

4 0

3 years ago