F = 2820.1 N
Explanation:
Let the (+)x-axis be up along the slope. The component of the weight of the crate along the slope is -mgsin15° (pointing down the slope). The force that keeps the crate from sliding is F. Therefore, we can write Newton's 2nd law along the x-axis as
Fnet = ma = 0 (a = 0 no sliding)
= F - mgsin15°
= 0
or
F = mgsin15°
= (120 kg)(9.8 m/s^2)sin15°
= 2820.1 N
The Earth's core is estimated to be bigger than both Mars and Mercury. However, I would say that Mercury is the safest choice. Mercury is also known to be smaller than Jupiter's Moon!
<h2>Answer: The second Statement
</h2>
<h2>
The algebraic sum of the currents flowing through each of the three resistors is equal to the current through the battery. </h2><h2 />
In a series circuit, the value of the equivalent resistance 
is equal to the sum of the values of each of them:
Where:
<h2>The equivalent resistance of the combination of resistors is greater than the resistance of any one of three resistors. </h2>
In this case the current 
flowing through the resistors is the same in each one. This is because the current flowing through the circuit only has one way to go, so the current intensity is the same throughout the circuit.
Therefore:
<h2>The current flowing through each of the resistors is the same and is equal to the current through the battery. </h2><h2>The algebraic sum of the voltages across the three resistors is equal to the voltage across the battery. </h2>
The battery provides a voltage 
that is the sum of the different voltages at the ends of the resistors:
Where the Voltage, according to Ohm's law is:
Hence, the second statement of this question is <u>True
</u>
After one day, the rate of increase in Delta Cephei's brightness is;0.46
We are informed that the function has been used to model the brightness of the star known as Delta Cephei at time t, where t is expressed in days;
B(t)=4.0+3.5 sin(2πt/5.4)
Simply said, in order to determine the rate of increase, we must determine the derivative of the function that provides
B'(t)=(2π/5.4)×0.35 cos(2πt/5.4)
Currently, at t = 1, we have;
B'(1)=(2π/5.4)×0.35 cos(2π*1/5.4)
Now that the angle in the bracket is expressed in radians, we can use a radians calculator to determine its cosine, giving us the following results:
B'(1)=(2π/5.4)×0.3961
B'(1)≈0.46
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