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sashaice [31]
3 years ago
15

What is a watt a unit of? light time distance power

Physics
2 answers:
weeeeeb [17]3 years ago
6 0

Answer:

power...............m....

Len [333]3 years ago
5 0

Answer:

The dimension of power is energy divided by time. In the International System of Units (SI), the unit of power is the watt (W), which is equal to one joule per second.

SI unit: watt (W)

In SI base units: kg⋅m2⋅s−3

Explanation:

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Early humans used rocks as tools to make other things as well as to construct buildings. Why was it better for them to use rocks
babymother [125]
Wood isn't as tough as rock. Wood also breaks down in weather and cracks under pressure. Plus rocks were more accessible.
6 0
3 years ago
Read 2 more answers
A space station, in the form of a wheel 140 m in diameter, rotates to provide an "artificial gravity" of 3.90 m/s2 for persons w
Zigmanuir [339]
Radial acceleration is given by

a_{rad}= \frac{v^2}{r}
where 

v=r \omega
then

a_{rad}= \frac{r^2 \omega^2}{r}=r\omega^2

Now

70\omega^2=3.90 \frac{m}{s^2}  \\  \\ \omega= \sqrt{ \frac{3.9}{70} }

Using the relation

\omega=2 \pi f

2 \pi f= \sqrt{ \frac{3.9}{70} }\\  \\ f= \frac{1}{2 \pi}\sqrt{ \frac{3.9}{70} }Hz

Putting into rpm

\frac{60}{2 \pi}\sqrt{ \frac{3.9}{70}} =2.254rpm

8 0
3 years ago
A baseball rolls off a 1.20m high desk and strikes the floor 0.50m away from the base of the desk . How fast was it rolling?
noname [10]

The initial velocity of the ball is 1.01 m/s

Explanation:

The motion of the ball rolling off the desk is a projectile motion, which consists of two independent motions:

- A uniform horizontal motion with constant horizontal velocity

- A vertical accelerated motion with constant acceleration (g=9.8 m/s^2, acceleration due to gravity)

We start by analyzing the vertical motion: we can find the time of flight of the ball by using the following suvat equation

s=ut+\frac{1}{2}gt^2

where

s = 1.20 m is the vertical displacement (the height of the desk)

u = 0 is the initial vertical velocity

g=9.8 m/s^2

t is the time of flight

Solving for t,

t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(1.20)}{9.8}}=0.495 s

Now we analyze the horizontal motion. We know that the ball covers a horizontal distance of

d = 0.50 m

in a time

t = 0.495 s

Therefore, since the horizontal velocity is constant, we can calculate it as

v_x = \frac{d}{t}=\frac{0.50}{0.495}=1.01 m/s

So, the ball rolls off the table at 1.01 m/s.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

4 0
3 years ago
TRUE OR FALSE: The following drops were most likely dropped from a 90 degree angle.
horsena [70]

Answer:

True

Explanation:

If it weren't from a 90 degree angle then the circle would be a bit more oval shaped

7 0
3 years ago
Read 2 more answers
Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a
34kurt

Answer:

E=\frac{KQ}{2\sqrt 2a^2}

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}

Substitute x=a and R=a

Then, we get

E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}

E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}

E=\frac{KQa}{2\sqrt 2a^3}

E=\frac{KQ}{2\sqrt 2a^2}

Where K=9\times 10^9 Nm^2/C^2

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=\frac{KQ}{2\sqrt 2a^2}

4 0
3 years ago
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