Answer:

Explanation:
When a spring is compressed, the force exerted by the spring is given by:

where
k is the spring constant
x is the compression of the spring
In this problem we have:
k = 52 N/m is the spring constant
x = 43 cm = 0.43 m is the compression
Therefore, the force exerted by the spring on the dart is

Now we can apply Newton' second law of motion to calculate the acceleration of the dart:

where
F = 22.4 N is the force exerted on the dart by the spring
m = 75 g = 0.075 kg is the mass of the dart
a is its acceleration
Solving for a,

Answer:
Explanation:
Give that,
Spring constant (k)=40N/m
Force applied =75N
Since the force is applied to the right, we don't know if it is compressing or stretching the spring
So let assume it compress
Using hooke's law
F=-ke
e=-F/k
Then, e=-75/40
e=-1.875m
The deformation is 1.875m.
Let assume it stretch
Using hooke's law
-F=-ke
e=F/k
Then, e=75/40
e=1.875m
The elongation is 1.875m
Answer:
d.none
Explanation:
because shape size and physical actually do are dependant
Answer:
Yes the body will receive a dangerous shock in both cases.
Explanation:
Different parts of the body has different resistance. skin has the high resistance as compared to other organs of the body.
Dry skin has high resistance than wet skin this is because water is relatively good conductor of electricity, it adds parallel path to the current flow and hence reduces skin resistance.
Dry hands body has approximately 500 kΩ resistance and if 120 V electricity supply current received will be:
I = V/R= 120/ 500*10^3
I= 0.24 mA
Even the current seems is much lower than the safe zone but this is the case in case of DC voltage in case of AC voltage the body will receive a shock this is because the skin pass more current when the voltage is changing i.e. AC.
Similarly for wet hands body resistance is 1 kΩ. so the current through the body seems to be:
I = 120 / 1000
I = 12 mA
The current is higher than safe zone so the body will receive a dangerous shock.