The net force acting on the crate is determined as 176 N to the left.
<h3>Net force acting on the crate</h3>
The net force acting on the crate is calculated as follows;
∑F = F1 + F2 + F3 + F4
F(net) = -440y + 176x + 440y - 352x
F(net) = -176 x
The resultant force is pointing in negative x direction.
Thus, the net force acting on the crate is determined as 176 N to the left.
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Answer:
C.) The amount of mass used up in holding a nucleus together.
Explanation:
The mass defect of a nucleus represents the mass of the energy binding the nucleus. It is the difference between the mass of the nucleus and the sum of the masses of the nucleons of which it is composed.
Regards!
Answer:
a) truc is C, b) correct result is the B
Explanation:
As the speed of the competition is very high, for the judges the speed is
v = d / t
v = 3 109 m / 20
v = 1.5 108 m / s
This is half the speed of light. For these high speeds we must use the relations of special relativity.
For the time t = to γ
For distance L = Lo / γ
γ = √ (1-v2 / c2)
Own time and distance (to and Lo) corresponds to the observer who is not moving the judges in this case
Let's look for the range value
γ = 1 / √ (1 - (1.5 / 3) 2) = 1 / 0.866 = 1.15
The time t = 20 1.15 = 23 s
The distance L = 3 10 9 /1.15 = 2.60 109 m
From these results we see that time increases and the distance is shorter.
Let's review the claims
A) False. It's the opposite
B) False
C) True. It is according to the result found
D) False.
In the nuclear fusion process, we will also use the special relativity that has a relationship between energy and mass
ΔE = c² Δm
As in the process energy is released, for the law of conservation of the mass of energy to be fulfilled, the total mass of the products, He atom, must be reduced.
Therefore the correct result is the B
I think the correct answer is
D) Ted associated being asked a question with embarrassment.
Glad I could help, and good luck!
AnonymousGiantsFan
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>