1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Natali [406]
3 years ago
11

Convert 580 g to kg.

Physics
2 answers:
MArishka [77]3 years ago
7 0

The Answer Is - 0.58 Kilogram's

Hope This Help's

irina1246 [14]3 years ago
4 0

• Answer:

0.58 kg

• Explanation:

1 kg = 1000 g

580 g = 0.58 kg

You might be interested in
Electricity flows from what to what
DerKrebs [107]
Electricity flows from positive to negative
5 0
2 years ago
Read 2 more answers
How much energy is stored in the capacitor when it is aa fully charged
lakkis [162]

Answer:

0.5*10uF * 16*16 =0.0128

Explanation:

I have no explanation just like my soul.

6 0
3 years ago
A spring is stretched from x=0 to x=d, where x=0 is the equilibrium position of the spring. It is then compressed from x=0 to x=
VARVARA [1.3K]

Answer:

The same amount of energy is required to either stretch or compress the spring.

Explanation:

The amount of energy required to stretch or compress a spring is equal to the elastic potential energy stored by the spring:

U=\frac{1}{2}k (\Delta x)^2

where

k is the spring constant

\Delta x is the stretch/compression of the spring

In the first case, the spring is stretched from x=0 to x=d, so

\Delta x = d-0=d

and the amount of energy required is

U=\frac{1}{2}k d^2

In the second case, the spring is compressed from x=0 to x=-d, so

\Delta x = -d -0 = -d

and the amount of energy required is

U=\frac{1}{2}k (-d)^2= \frac{1}{2}kd^2

so we see that the amount of energy required is the same.

7 0
4 years ago
A boy standing on the bridge kicks a stone into the water below. He kicks the stone with a horizontal velocity of 8.06 m/s. It l
Ronch [10]

Time taken to reach water :

t = \dfrac{D}{v}\\\\t=\dfrac{6.78}{8.06}\ s\\\\t=0.84\ s

Now, initial vertical speed , u = 0 m/s.

By equation of motion :

h = ut +\dfrac{at^2}{2}

Here, a = g = acceleration due to gravity = 9.8 m/s².

So,

h = 0(t) +\dfrac{9.8\times 0.84^2}{2}\\\\h=3.46\ m

Therefore, the height of the bridge is 3.46 m.

Hence, this is the required solution.

6 0
3 years ago
The classic Millikan oil-drop experiment was the first to obtain an accurate measurement of the charge on an electron. In it, oi
Sati [7]

Answer:

1.56\times 10^5 N/C        

Explanation:

The electric field that will balance the weight of the oil drop can be calculated using the following:

electric force, F = e E ( where, e is the charge of an electron and E is the electric field)

weight, W = 2.5 ×10⁻¹⁴ N

e E = W

E =\frac{W}{e}

Substitute the values:

E =\frac{ 2.5 \times 10^{-14} N}{1.6\times10^{-19}C}= 1.56\times 10^5 N/C

6 0
3 years ago
Read 2 more answers
Other questions:
  • What are gamma rays used for
    15·2 answers
  • How does the ringing sound of a telephone travels from the phone to your ear
    13·2 answers
  • g An object with mass 1kg travels at 3 m/s and collides with a stationary object whose mass is 0.5kg. The two objects stick toge
    12·1 answer
  • Electromagnetic induction means that moving a magnet through a loop of wire creates an electric current.
    8·2 answers
  • What is indicated by the slope of an acceleration vs. time graph?
    7·1 answer
  • What is the full name of emf​
    5·1 answer
  • uniform mementum mas is 90g is pivoted at 40cm mark if the metra rula ia in equilibrium with an mark unknown mass m place at the
    11·1 answer
  • A speedboat is moving at a rate of 45 km per hour travels a distance of 27 km. How long did it take to go to 27 km.
    15·1 answer
  • An object of mass 100kg is moving with a velocity of 5m/s. Calculate the kinetic energy of that object
    13·2 answers
  • A spring with k = 136 N/m is compressed by 12.5 cm. How much elastic potential energy does the
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!