Answer:
the intensity will be 4 times that of the earth.
Explanation:
let us assume the following:
intensity of light on earth =J
distance of earth from sun = d
intensity of light on other planet = K
distance of other planet from sun = 
(from the question, the planet is half as far from the sun as earth)
from the question the intensity is inversely proportional to the square of the distance, hence
- intensity on earth : J =
J
= 1 ... equation 1
- intensity on other planet : K =
(the planet is half as far from the sun as earth)
K
= 1 ....equation 2
- equating both equation 1 and 2 we have
J = K
J = K
J = 
K = 4J
intensity of light on other planet (K) = 4 times intensity of light on earth (J)
For this case, the first thing you should do is write the kinematic motion equation of the block.
We have then:
vf = vo + a * t
Where,
vf: Final speed.
vo: Initial speed.
a: acceleration.
t: time.
Substituting the values:
(16) = (0) + a * (16)
Clearing the acceleration:
a = 16/16 = 1m / s ^ 2
Note: the other data for this case are not used in this problem.
answer:
The acceleration of the box is 1m / s ^ 2
7500J
Explanation:
Given parameters:
Weight of child = 150N
Height of lifting = 50m
Distance = 50m
Unknown:
Work done by the woman = ?
Solution:
Work done by a body is the force applied to move a body through a given distance.
Work done = Force x distance
Weight is a force impacted by the mass of a body and the gravity on it.
input the parameters:
Work done = 150 x 50 = 7500J
learn more:
Work done brainly.com/question/9100769
#learnwithBrainly
<span>95 km/h = 26.39 m/s (95000m/3600 secs)
55 km/h = 15.28 m/s (55000m/3600 secs)
75 revolutions = 75 x 2pi = 471.23 radians
radius = 0.80/2 = 0.40m
v/r = omega (rad/s)
26.39/0.40 = 65.97 rad/s
15.28/0.40 = 38.20 rad/s
s/((vi + vf)/2) = t
471.23 /((65.97 + 38.20)/2) = 9.04 secs
(vf - vi)/t = a
(38.20 - 65.97)/9.04 = -3.0719
The angular acceleration of the tires = -3.0719 rad/s^2
Time is required for it to stop
(0 - 38.20)/ -3.0719 = 12.43 secs
How far does it go?
65.97 - 38.20 = 27.77 M</span>
