1) the weight of an object at Earth's surface is given by

, where m is the mass of the object and

is the gravitational acceleration at Earth's surface. The book in this problem has a mass of m=2.2 kg, therefore its weight is

2) On Mars, the value of the gravitational acceleration is different:

. The formula to calculate the weight of the object on Mars is still the same, but we have to use this value of g instead of the one on Earth:

3) The weight of the textbook on Venus is F=19.6 N. We already know its mass (m=2.2 kg), therefore by re-arranging the usual equation F=mg, we can find the value of the gravitational acceleration g on Venus:

4) The mass of the pair of running shoes is m=0.5 kg. Their weight is F=11.55 N, therefore we can find the value of the gravitational acceleration g on Jupiter by re-arranging the usual equation F=mg:

5) The weight of the pair of shoes of m=0.5 kg on Pluto is F=0.3 N. As in the previous step, we can calculate the strength of the gravity g on Pluto as

<span>6) On Earth, the gravity acceleration is </span>

<span>. The mass of the pair of shoes is m=0.5 kg, therefore their weight on Earth is
</span>

<span>
</span>
Answer:
remains the same
Explanation:
Momentum refers to the quantity of motion of a body. When any body of mass moves, it possess momentum. Numerically,
Momentum = mass x velocity
i.e. momentum is the product of the mass x velocity
Momentum of a body is always conserved.
In the context, the skateboard has certain momentum before Freddy lands on it. After Freddy lands, the momentum of skateboard remains the same, there is no change in the momentum.
This is because, here the momentum is conserved. After Freddy lands on the skateboard, the total mass on the skateboard increases and so the velocity decreases making the momentum same before the landing.
Answer:
Explanation:
This is a circular motion questions
Where the oscillation is 27.3days
Given radius (r)=3.84×10^8m
Circular motion formulas
V=wr
a=v^2/r
w=θ/t
Now, the moon makes one complete oscillation for 27.3days
Then, one complete oscillation is 2πrad
Therefore, θ=2πrad
Then 27.3 days to secs
1day=24hrs
1hrs=3600sec
Therefore, 1day=24×3600secs
Now, 27.3days= 27.3×24×3600=2358720secs
t=2358720secs
Now,
w=θ/t
w=2π/2358720 rad/secs
Now,
V=wr
V=2π/2358720 ×3.84×10^8
V=1022.9m/s
Then,
a=v^2/r
a=1022.9^2/×3.84×10^8
a=0.0027m/s^2
The similarities and the differences between gravitational and electric force are listed below
Explanation:
- The magnitude of the gravitational force between two objects is given by Newton's law of gravitation:
where
is the gravitational constant
are the masses of the two objects
r is the separation between them
- Coloumb's law gives instead the strength of the electrostatic force between two charged objects, which is
where:
is the Coulomb's constant
are the two charges
r is the separation between the two charges
By comparing the two equations, we find the following similarities:
- Both the forces are inversely proportional to the square of the distance between the two objects,

- Both the forces are proportional to the product between the "main quantity" of each force, which is the mass for the gravitational force (
) and the charge for the electric force (
Instead, we have the following differences:
- The gravitational force is always attractive, since the sign of
is always positive, while the electric force can be either attractive or repulsive, since the sign of
can be either positive or negative - The value of the gravitational costant G is much smaller than the value of the Coulomb's constant, so the gravitational force is much weaker than the electric force
Learn more about gravitational force and electric force:
brainly.com/question/1724648
brainly.com/question/12785992
brainly.com/question/8960054
brainly.com/question/4273177
#LearnwithBrainly
Answer:
The force required to begin to lift the pole from the end 'A' is 240 N
Explanation:
The given parameters for the pole AB are;
The length of the pole, l = 10.0 m
The weight of the pole, W = 600 N ↓
The distance of the center of gravity of the pole from the side 'A' = 4.0 m
Let '
' represent the force required to begin to lift the pole from the end 'A' and let a force applied in the upwards direction be positive
For equilibrium, the sum of moment about the point 'B' = 0, therefore, taking moment about 'B', we have
× 10.0 m - W × 4.0 m = 0
∴
× 10.0 m = W × 4.0 m = 600 N × 4.0 m
× 10.0 m = 600 N × 4.0 m
∴
= 600 N × 4.0 m/(10.0 m) = 240 N
The force required to begin to lift the pole from the end 'A',
= 240 N.