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2 years ago

In which situation can you be at rest and moving at the same time

1 answer:

5 0

Answer:Being at rest is a relative term. Relative to earth we are at rest when we are sitting down. However, you are moving because the earth is moving its orbit around the sun.

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Use this formula to solve this problem:

Maslowich

Well, you gave us the formula to calculate power from work and time,
but you didn't give us the formula for work. We have to know that.

   Work = (force) x (distance)

The work to raise Sara to the top of the hill is

   Work = (300 N) x (15 meters)

   = 4,500 newton-meters = 4,500 joules .

Now we're ready to use the formula that you gave us. (Thank you.)

Power = (work) / (time)

= (4,500 joules) / (10 seconds)

450 joules/second = 450 watts.

6 0

2 years ago

Read 2 more answers

A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s

aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is $U_{s}$ = 0

And, initial potential gravitational potential energy of the rod is $U_{g} = \frac{mgL}{2}$ .

It is given that,

mass of the bar = 0.795 kg

g = 9.8 $m/s^{2}$

L = length of the rod = 0.2 m

Initial total energy T = $\frac{mgL}{2}$

Now, when the rod is in horizontal position then final total energy will be as follows.

T = $\frac{1}{2}kx^{2} + I \omega^{2}$

where, I = moment of inertia of the rod about the end = $\frac{mL^{2}}{3}$

Also, $\omega = \frac{\nu}{L}$

where, $\nu$ = speed of the tip of the rod

x = spring extension

The initial unstrained length is $x_{o}$ = 0.1 m

Therefore, final length will be calculated as follows.

x' = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m

Then, x = $x' - x_{o}$

x = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m - 0.1 m

= 0.1236 m

k = 25 N/m

So, according to the law of conservation of energy

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}$

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

Putting the given values into the above formula as follows.

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

$\frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}$

v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

7 0

2 years ago

Why is this website not work how i want it this is so annoying

Mice21 [21]

It depends how you want it to work do you want to take a picture oh yea and to do that you must create a account btw

5 0

2 years ago

Read 2 more answers

Water has the ability to dissolve polar covalent molecules like sucrose by

Alla [95]

Explanation:

Sucrose is a disaccharide which is composed of fructose and glucose. Sucrose molecule has oxygen atoms bonded to hydrogen atoms (O-H bonds - Polar groups) on all ends of its double 6-Carbon ring. The areas near the oxygen atoms are slightly negative, and the areas near the hydrogen atoms are slightly positive that is, the O-H bonds are polar. They bond with the neighbouring Oxygen and Hydrogen atoms because of their

dipole - dipole attractions and hence hydrogen bonds are formed.

However, the covalent bonds within the molecule aren't broken. But rather, the hydrogen bonds holding the sucrose molecules in the crystalline lattice.

5 0

2 years ago

When grocery shopping, the mass of the cart changes as you start to fill up your cart. How does the change in mass of your cart

Tasya [4]

Answer:

Explanation:

4 0

2 years ago