Mework
1 answer:
The force applied by the student is 75 N
Explanation:
We can solve this problem by using Newton's second law, which states that the force applied on an object is equal to the product between the mass of the object and its acceleration:
where
F is the force applied
m is the mass of the object
a is the acceleration
In this problem,
m = 50 kg
Substituting, we find the force applied by the student:
Learn more about Newton's second law of motion:
#LearnwithBrainly
You might be interested in
The average speed of the given car is 2.22 s and 3.13 s for 0.25 m and 0.50 m distance respectively.
<h3>How to calculate the Average speed?</h3>The average speed can be calculated by adding the speed of each trial divided by the number of trials,
For 0.25 m the average speed will be:
For the 0.50 m, the average speed will:
Therefore, the average speed of the given car is 2.22 s and 3.13 s for 0.25 m and 0.50 m distance respectively.
Learn more about Average speed:
Missing question in the text of the exercise. Found on internet:
"What is the acceleration due to gravity on the surface of the planet?"
Solution:
The gravitational acceleration at Earth's surface is given by:
where
G is the gravitational constant
M is the Earth's mass
r is the Earth's radius
The Earth's mass can be rewritten also as the product between the Earth's density,
, and its volume (the volume of a sphere of radius r):
Now let's call M' the mass of the new planet, r' its radius and d' its density. The acceleration due to gravity on the surface of the new planet is
(1)
so we need to find M' and r'.
The problem says the radius of the new planet is twice the Earth's radius:
(2)
and that its density is 2/3 of Earth's density:
so the mass M' of the new planet is, with respect to the Earth's mass:
(3)
And if we substitute (2) and (3) into (1), we find the gravitational acceleration on the surface of the new planet:
And since
, we find
"What is the acceleration due to gravity on the surface of the planet?"
Solution:
The gravitational acceleration at Earth's surface is given by:
where
G is the gravitational constant
M is the Earth's mass
r is the Earth's radius
The Earth's mass can be rewritten also as the product between the Earth's density,
Now let's call M' the mass of the new planet, r' its radius and d' its density. The acceleration due to gravity on the surface of the new planet is
so we need to find M' and r'.
The problem says the radius of the new planet is twice the Earth's radius:
and that its density is 2/3 of Earth's density:
so the mass M' of the new planet is, with respect to the Earth's mass:
And if we substitute (2) and (3) into (1), we find the gravitational acceleration on the surface of the new planet:
And since
Answer:
D = 2.828 m
Explanation:
given,
distance of source of light = 1.24 m below surface of pool
refractive index of the water = n₁ = 1.33
refractive index of air = n₂ = 1
refraction angle be = 90°
let C be the critical angle
Radius = d tan C
d is the depth of the source
Using Snell's law
n₁ sin C = n₂ sin R
1.33 x sin C = 1 x sin 90°
C = 48.75°
hence,
R = 1.24 x tan 48.75°
R = 1.414 m
Diameter = 2 x R
D = 2 x 1.414
D = 2.828 m