Answer:
Incomplete question... We are suppose to be given the force equation and i think it is Fvec = (cx - 3.00x^2)i
Explanation:
Check attachment for solution
Five
The diagram that represents rays from the sun is given below.
The answer from the diagram is parallel lines.
Six
The formula for this is n_i * Sin(i) = n_r * sin(r)
n1 = 1.333 water
n2 = 1.309 ice
i (angle) = 40o
r (angle) = ?
Solution
1.333 * sin(40) = 1.309*sin(r)
1.333 * sin(40)/1.309 = sin(r)
0.6546 = sin(r)
sin-1(0.6546) = 40.88 degrees.
I guess my closest answer would be A. I'm not rounding at all and that's the answer I get.
Answer:
Explanation:
Initial kinetic energy = 29 J
work done against gravity = mgsin33 x d , m is mass of the block
= 1.2 x 9.8 sin 33 x .9
= 5.76 J
potential energy stored in compressed spring
= 1/2 k x², k is spring constant and x is compression
= .5 x 200 x .3²
= 9
energy left = 29 - ( 5.76 + 9 )
= 14.24 J
b )
energy stored in spring when compression is .4 m
= 1/2 x 200 x .4²
= 16 J
required kinetic energy = 16 + 5.76
= 21.76 J
Block must be projected with energy of 21.76 J .
The amount of work which Sam does need to stop the boat is 784Joule if boat mass is 800kg and boat drift in 1.4m/sec.
We know that according to work-energy theorem, change in kinetic energy of the body from one speed to another is equivalent to amount of work done by all forces acting on the body.
So, here we can see that final speed of boat is =0m/sec since Sam need to stop it.
Initial speed of boat is 1.4m/sec.
Also, we know that kinetic energy is given by the below formula:
Kinetic energy=(1/2)mv²
where m is the mass of the body
and v is the speed of the body
Now, from work energy theorem, we get
=>W=(1/2)m
- (1/2)m
=>W=0-(1/2)×800×(1.4)²
=>W= - 400× 1.4 × 1.4
=>W = -400×1.96
=>W= -784Joule
Here negative sign denotes work has to done against the original motion of boat.
Hence, amount of work needs to be done by Sam is 784Joule.
To know more about work,visit here:
brainly.com/question/18094932
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Answer: I observed that the electric field strength is the same at all points between the plates. The value of the field is 70 volts per meter. This is exactly 100 times the applied voltage. The electric field lines point from the positive plate to the negative plate, as the downward arrow on the detector shows.
Explanation:
the sample answer, don't directly copy it!