A force is a push or pull upon an object resulting from the object's interaction with another object.
Answer : The correct option is, (a) No. Potatoes will be warmer.
Explanation :
Formula used :
where,
Q = heat taken
m = mass of substance
c = specific heat of substance = ?
= change in temperature
As we are given that the potatoes have a specific heat of 
and raisins have a specific heat of 
. That means higher the value of specific heat the more heat taken by the substance (more warmer will be the substance). So, the potatoes will be more warmer as compared to raisins.
Hence, the correct option is, (a) No. Potatoes will be warmer.
The final velocity of the projectile when it strikes the ground below is 198.51 m/s.
<h3>
Time of motion of the projectile</h3>
The time taken for the projectile to fall to the ground is calculated as follows;
h = vt + ¹/₂gt²
where;
- h is height of the cliff
- v is velocity
- t is time of motion
265 = (185 x sin45)t + (0.5)(9.8)t²
265 = 130.8t + 4.9t²
4.9t² + 130.8t - 265 = 0
solve the quadratic equation using formula method,
t = 1.89 s
<h3>Final velocity of the projectile</h3>
vyf = vyi + gt
where;
- vyf is the final vertical velocity
- vyi is initial vertical velocity
vyf = (185 x sin45) + (9.8 x 1.89)
vyf = 149.322 m/s
vxf = vxi
where;
- vxf is the final horizontal velocity
- vxi is the initial horizontal velocity
vxf = 185 x cos(45)
vxf = 130.8 m/s
vf = √(vyf² + vxf²)
where;
- vf is the speed of the projectile when it strikes the ground below
vf = √(149.322² + 130.8²)
vf = 198.51 m/s
Learn more about final velocity here: brainly.com/question/6504879
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Answer:
your answer should be astronomy
Explanation:
Answer: A symbolic expression for the net force on a third point charge +Q located along the y axis
![F_N=k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}](https://tex.z-dn.net/?f=F_N%3Dk_e%5Cfrac%7BQ%5E2%7D%7Bd%5E2%7D%5Ctimes%20%5Csqrt%7B%5B4%2B%5Cfrac%7B1%7D%7B4%7D-%5Csqrt%7B2%7D%5D%7D)
Explanation:
Let the force on +Q charge y-axis due to +2Q charge be 
and force on +Q charge y axis due to -Q charge on x-axis be 
.
Distance between the +2Q charge and +Q charge = d units
Distance between the -Q charge and +Q charge = 
units
= Coulomb constant
Net force on +Q charge on y-axis is:
![|F_N|=|k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}|](https://tex.z-dn.net/?f=%7CF_N%7C%3D%7Ck_e%5Cfrac%7BQ%5E2%7D%7Bd%5E2%7D%5Ctimes%20%5Csqrt%7B%5B4%2B%5Cfrac%7B1%7D%7B4%7D-%5Csqrt%7B2%7D%5D%7D%7C)
The net froce on the +Q charge on y-axis is
