Answer:
oo.p i wish I could answer that
Explanation:
The net force on the barge is 8000 N
Explanation:
In order to find the net force on the badge, we have to use the rules of vector addition, since force is a vector quantity.
In this problem, we have two forces:
- The force of tugboat A,
, acting in a certain direction - The force of tugboat B,
, also acting in the same direction
Since the two forces act in the same direction, this means that we can simply add their magnitudes to find the net combined force on the barge. Therefore, we get
and the direction is the same as the direction of the two forces.
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Answer:
T = 3.23 s
Explanation:
In the simple harmonic movement of a spring with a mass the angular velocity is given by
w = √ K / m
With the initial data let's look for the ratio k / m
The angular velocity is related to the frequency and period
w = 2π f = 2π / T
2π / T = √ k / m
k₀ / m₀ = (2π / T)²
k₀ / m₀ = (2π / 3.0)²
k₀ / m₀ = 4.3865
The period on the new planet is
2π / T = √ k / m
T = 2π √ m / k
In this case the amounts are
m = 6 m₀
k = 10 k₀
We replace
T = 2π√6m₀ / 10k₀
T = 2π √6/10 √m₀ / k₀
T = 2π √ 0.6 √1 / 4.3865
T = 3.23 s
Explanation:
- In chemical reactions, chemical changes occur.
- Atoms are simply rearranged and new bonds are formed.
- Chemical reactions are driven by a need for atoms to attain stability in their structure.
- In all chemical reactions, a reactant or reactants gives new product i.e new substances are formed.
- Most these reactions are not easily reversible.
- They are usually accompanied by the release of energy.
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Answer:
0° C
Explanation:
Given that
Mass of ice, m = 50g
Mass of water, m(w) = 50g
Temperature of ice, T(i) = 0° C
Temperature of water, T(w) = 80° C
Also, it is known that
Specific heat of water, c = 1 cal/g/°C
Latent heat of ice, L(w) = 89 cal/g
Let us assume T to be the final temperature of mixture.
This makes the energy balance equation:
Heat gained by ice to change itself into water + heat gained by melted ice(water) to raise its temperature at T° C = heat lost by water to reach at T° C
m(i).L(i) + m(i).c(w)[T - 0] = m(w).c(w)[80 - T], on substituting, we have
50 * 80 + 50 * 1(T - 0) = 50 * 1(80 - T)
4000 + 50T = 4000 - 50T
0 = 100 T
T = 0° C
Thus, the final temperature is 0° C
