When you ask for "joules per second", you're asking for "watts".
The rate of energy "transfer" is 'power'. In this case, the light bulb
transfers energy out of the electrical circuit and into the space
around it, in the form of light and heat radiation.
Electrical power = (voltage) x (current) =
(6 volts) x (0.5 ampere) =
3 watts = 3 joules per second.
Answer: yes a quantity have different dimensions in different system of units . No,because in different system of units doesn't change the quantity but it only changes the numerical.
Good luck !
Hy tikki! I've asked some questions, so of you find the questions as easy, then answer it. I'll surely mark you as brainliest :)
<h2>The distance between students is 2.46 m</h2>
Explanation:
The force of attraction due to Newton's gravitation law is
F =
Here G is the gravitational constant
m₁ is the mass of one student
m₂ is the mass of second student .
and r is the distance between them
Thus r =
If we substitute the values in the above equation
r =
= 2.46 m
Answer:
The launching point is at a distance D = 962.2m and H = 39.2m
Explanation:
It would have been easier with the drawing. This problem is a projectile launching exercise, as they give us data after the window passes and the wall collides, let's calculate with this data the speeds at the point of contact with the window.
X axis
x = Vox t
t = x / vox
t = 7.1 / 340
t = 2.09 10-2 s
In this same time the height of the window fell
Y = Voy t - ½ g t²
Let's calculate the initial vertical speed, this speed is in the window
Voy = (Y + ½ g t²) / t
Voy = [0.6 + ½ 9.8 (2.09 10⁻²)²] /2.09 10⁻² = 0.579 / 0.0209
Voy = 27.7 m / s
We already have the speed at the point of contact with the window. Now let's calculate the distance (D) and height (H) to the launch point, for this we calculate the time it takes to get from the launch point to the window; at this point the vertical speed is Vy2 = 27.7 m / s
Vy = Voy - gt₂
Vy = 0 -g t₂
t₂ = Vy / g
t₂ = 27.7 / 9.8
t₂ = 2.83 s
This is the time it also takes to travel the horizontal and vertical distance
X = Vox t₂
D = 340 2.83
D = 962.2 m
Y = Voy₂– ½ g t₂²
Y = 0 - ½ g t2
H = Y = - ½ 9.8 2.83 2
H = 39.2 m
The launching point is at a distance D = 962.2m and H = 39.2m