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lara [203]
3 years ago
13

What are all of the si units

Physics
1 answer:
mina [271]3 years ago
5 0

Here you go

There are seven basic units in the SI system: the meter (m), the kilogram (kg), the second (s), the kelvin (K), the ampere (A), the mole (mol), and the candela (cd).


Hope it helps you

Please mark me as Brainsliest

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I want ti know how to study​
arlik [135]

Answer:

Make sure everything is organized have a planner it can help

Get rid of all distractions

Listen to music if it helps you concentrate

Have your notes

Being willing to stay focus on what you are doing

Understand what you are doing

And most off all Be Happy and Remain Calm : )

3 0
3 years ago
Read 2 more answers
A speed-time graph is shown below:
VladimirAG [237]

Answer: 0.5 m/s^{2}

Explanation:

Average acceleration a_{ave} is the variation of velocity \Delta V over a specified period of time \Delta t:

a_{ave}=\frac{\Delta V}{\Delta t}}

Where:

\Delta V=V_{f}-V_{o} being V_{o}=0 cm/s the initial velocity and V_{f}=4 cm/s the final velocity  (according to the information given from the described graph)

\Delta t=8 s

Then:

a_{ave}=\frac{4 cm/s -0 cm/s}{8 s}}

a_{ave}=0.5 m/s^{2}

5 0
2 years ago
Two lightbulbs both operate on 120V . One has a power of 25W and the other 100W. (ii) Which lightbulb carries more current? Choo
Vikki [24]

The lightbulb that carries more current will be the 25W bulb.

<h3>How to explain the information?</h3>

It should be noted that an electric current simply means the stream of charged particles that move through an electrical conductor or space.

In this case, it should be noted that the power is the same for both bulbs. Therefore, the 25W bulb will have the higher resistance so that it will have lower power.

Therefore, the lightbulb that carries more current will be the 25W bulb.

Learn more about current on:

brainly.com/question/1100341

#SPJ4

7 0
1 year ago
An object moving on a horizontal, frictionless surface makes a glancing collision with another object initially at rest on the s
cluponka [151]

Answer:

Momentum is always conserved, and kinetic energy may be conserved.

Explanation:

For an object moving on a horizontal, frictionless surface which makes a glancing collision with another object initially at rest on the surface, the type of collision experienced by this objects can either be elastic or an inelastic collision depending on whether the object sticks together after collision or separates and move with a common velocity after collision.

If the body separates and move with a common velocity after collision, the collision is elastic but if they sticks together after collision, the collision is inelastic.

Either ways the momentum of the bodies are always conserved since they will always move with a common velocity after collision but their kinetic energy may or may not be conserved after collision, it all depends whether they separates or stick together after collision and since we are not told in question whether or not they separate, we can conclude that their kinetic energy "may" be conserved.

6 0
2 years ago
The experimental apparatus shown in the figure above contains a pendulum consisting of a 0.66 kg ball attached to a string of le
lara31 [8.8K]

The problem is solved and the questions are answered below.

Explanation:

a. To calculate the speed of the 0.66 kg ball just before the collision

V₀ + K₀ = V₁ + K₁

= mgh₀ = 1/2 mv₁²

where, h= r - r cosθ

V = \sqrt{2gh}

 V = 2.42 m/s

b. Calculate the speed of the 0.22 kg ball immediately after the collision

y = y₀ + Vy₀t - 1/2 gt²

0 = 1.2 - 1/2 gt²

t = 0.495 s

x = x₀ + Vx₀t

1.4 = 0 + vx₀ (0.495)

Vx₀ = 2.83 m/s

C. To Calculate the speed of the 0.66 kg ball immediately after the collision

m₁ v₁ = m₁ v₃ + m₂ v₄

(0.66)(2.42) = (0.66) v₃ + (0.22)(2.83)

V₃ = 1.48 m/s

D. To Indicate the direction of motion of the 0.66 kg ball immediately after the collision is to the right.

E. To Calculate the height to which the 0.66 kg ball rises after the collision

V₀ + k₀ = V₁ + k₁

1/2 mv₀² = mgh₁

h₁ = v₀²/2 g

  = 0.112 m

F. Based on your data, No the collision is not elastic.

Δk = 1/2 m₁v₃² =1/2 m₂v₄² - 1/2 m₁v₁²

     = 1/2 (0.66)(1.48)² + 1/2 (0.22)(2.83)² - 1/2 (0.66)(2.42)²

    = - 0.329 J

Hence, kinetic energy is not conserved.

8 0
2 years ago
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