Answer:0.302
Explanation:
Given
mass of crate m=27 kg
Force required to set crate in motion is 80 N
Once the crate is set in motion 56 N is require to move it with constant velocity
i.e. 80 N is the amount of force needed to just overcome static friction and 56 is the kinetic friction force
thus

where
is the coefficient of static friction and N is Normal reaction





Uranium-238 decays<span> by alpha emission </span>into<span> thorium-234, which itself </span>decays<span> by beta emission to protactinium-234, which </span>decays<span> by beta emission to </span>uranium<span>-234, and so on. The various </span>decay<span> products, (sometimes referred to as “progeny” or “daughters”) form a series starting at </span>uranium-238<span>.</span>
Answer:


Explanation:
From the question we are told that
Initial velocity of 60 m/s
Wind speed 
Generally Resolving vector mathematically

Generally the equation Pythagoras theorem is given mathematically by



Therefore Resultant velocity (m/s)

b)Resultant direction
Generally the equation for solving Resultant direction

Therefore


From the calculation, the gravitational force of attraction is 1.33 * 10^-14 N.
<h3>What is the gravitational force?</h3>
The gravitational force is an attractive force that acts between any two masses.
It is given by;
F = Gm1m2/r^2
F = 6.67 * × 10−11 * 2.5 * 5/(250)^2
F = 83.4 × 10−11 /62500
F= 1.33 * 10^-14 N
Learn more about gravitational force:brainly.com/question/12528243
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