Which of these is the same as a newton [n]? a. m/s b. m2/s c. m/s2 d. kg-m/s2

If it takes you 10 seconds to move a chair 5 meters across the floor, using a force of 2 Newtons, how much power did you put out

Maru [420]

Answer:

power=work done÷time taken

2×5=10

10÷10=1

ans 1J per second

5 0

2 years ago

Question 7 of 10

Dahasolnce [82]

Kinetic friction (also referred to as dynamic friction) is the force that resists the relative movement of the surfaces once they're in motion.
https://www.khanacademy.org › stat...
Static and kinetic friction example (video) | Khan Academy

Answer a would be static friction
Answer b is fluid friction
(Air resistance is fluid friction. Fluid friction is the friction experienced by objects which are moving in a fluid and the air is a fluid.)
Answer c is static friction
ANSWER D IS KINETIC FRICTION

Hope this helps :D

4 0

3 years ago

A simple pendulum consists of a mass M attached to a string oflength L andnegligible mass. For this system, when undergoing smal

PilotLPTM [1.2K]

The frequency of the pendulum is independent of the mass on the end. (c)

This means that it doesn't matter if you hang a piece of spaghetti or a school bus from the bottom end. If there is no air resistance, and no friction at the top end, and the string has no mass, then the time it takes the pendulum to swing from one side to the other only depends on the length of the string.

8 0

3 years ago

A nonconducting container filled with 25 kg of water at 23°C is fitted with a stirrer, which is made to turn by gravity acting o

Paul [167]

Explanation:

Given that,

Weight of water = 25 kg

Temperature = 23°C

Weight of mass = 32 kg

Distance = 5 m

(a). We need to calculate the amount of work done on the water

Using formula of work done

$W=mgh$

$W=32\times9.8\times5$

$W=1568\ J$

The amount of work done on the water is 1568 J.

(b). We need to calculate the internal-energy change of the water

Using formula of internal energy

The change in internal energy of the water equal to the amount of the work done on the water.

$\Delta U=W$

$\Delta U=1568\ J$

The change in internal energy is 1568 J.

(c). We need to calculate the final temperature of the water

Using formula of the change internal energy

$\Delta U=mc_{p}\Delta T$

$\Delta U=mc_{p}(T_{2}-T_{1})$

$T_{2}=T_{1}+\dfrac{\Delta U}{mc_{p}}$

$T_{2}=23+\dfrac{1568}{25\times4.18\times10^{3}}$

$T_{2}=23.01^{\circ}\ C$

The final temperature of the water is 23.01°C.

(d). The amount of heat removed from the water to return it to it initial temperature is the change in internal energy.

The amount of heat is 1568 J.

Hence, This is the required solution.

6 0

3 years ago

How to calculate the angle of a resultant

stiv31 [10]

Answer:

The direction angle θ of the resultant in the Polar (positive) specification is then θ = α + 60°. The Law of Cosines is used to calculate the magnitude (r) and the Law of Sines is used to calculate the angle (α).

3 0

3 years ago