Answer:
50.3N
Explanation:
Work done = force x distance
422J. = force x 8.39m
÷8.39 both side to get force
Force is 50.3N to 1 d.p.
Check:
50.3 x 8.39=422.017J
Same as 422J to 1 d.p
Answer:
Water.
Explanation:
This means:
1) For the temperature of water to raise at any point to the next degree by 1°C, will require a specific heat capacity of 4.184 J/Kg°C
2) For the temperature of wood to raise at any point to the next degree by 1°C, will require a specific heat capacity of 1.760 J/Kg°C
Note that: specific heat is directly proportional to energy, therefore the higher the heat capacity, the higher the energy.
4.184 J/Kg°C is higher than 1.760 J/Kg°C, hence WATER needs more energy.
The answer is A: Gorilla.
In the animal kingdom, all organisms are multicelluar.
The gravitational force of the shell exerts is 4.25m x 10¯¹² N.
We need to know about gravitational force to solve this problem. The gravitational force is the force caused by two masses of objects. The magnitude of gravitational force can be determined as
F = G.m1.m2 / R²
where F is the gravitational force, G is the gravitational constant (6.674 × 10¯¹¹ Nm²/kg²), m1 and m2 are the mass of the object and R is the radius.
From the question above, we know that
m1 = 1.6 kg
m2 = m
R = 5.01 m
By substituting the following parameters, we get
F = G.m1.m2 / R²
F = 6.674 × 10¯¹¹ . 1.6 . m / 5.01²
F = 4.25m x 10¯¹² N
where m is the mass of the shell
For more on gravitational force at: brainly.com/question/19050897
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Answer:
The time taken for the commercial Jet liner to reach the end of its runway is 10.18 s.
Explanation:
Given;
average acceleration of the commercial Jet liner, a = 3g = 3 x 9.8 m/s² = 29.4 m/s²
distance traveled by the commercial Jet liner, s = 1542 m
The time taken for the commercial Jet liner to reach the end of its runway is calculated as follows;
s = ut + ¹/₂at²
where;
u is the initial velocity of the commercial Jet liner = 0
s = 0 + ¹/₂at²
s = ¹/₂at²
2s = at²
Therefore, the time taken for the commercial Jet liner to reach the end of its runway is 10.18 s.
