The speed after 10 s is 20 m/s
Given that we have the following;
force = 2800 N
Mass = 1400 kg
time = 10 s
Recall that;
F = ma
F = force
m = mass
a = acceleration
a = F/m
a = 2800 N/1400 kg
a = 2 m/s^2
Now;
v = u + at
v = ??
u = 0 m/s
a= 2 m/s^2
t = 10 s
v = at
v = 2 * 10
v = 20 m/s
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Given below the arrangement of loading on the larger boat by two tug boats.
F₁ = 5 N
F₂ = 4 N
Angle between them θ = 90⁰
Resultant between two vectors,
Substituting
So magnitude of the net force on the block = 6.403 N
Answer:
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Explanation:
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