Answer: Heat energy is transferred from warmer objects to cooler objects.
Explanation:
The spring constant is 4 N/m
Explanation:
When a spring is stretched/compressed by the application of a force, the relationship between the magnitude of the force applied and the elongation of the spring is given by Hooke's law:
where
F is the magnitude of the spring applied
k is the spring constant
x is the elongation of the spring, relative to its equilibrium position
For the spring in this problem, we have:
F = 0.12 N (force applied)
x = 3 cm = 0.03 m (elongation of the spring)
Therefore, we can solve the formula for k to find the spring constant:
Learn more about forces:
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A salt is dissolved in water which has a freezing point of 0 degrees celsius. the freezing point of the solution would be dependent on the concentration of the salt in the solution. It is explained by the colligative properties. These <span>are </span>properties<span> that depend upon the concentration of solute molecules or ions, but not upon the identity of the solute. Hope this answers the question.</span>
1750 meters.
First, determine how long it takes for the kit to hit the ground. Distance over constant acceleration is:
d = 1/2 A T^2
where
d = distance
A = acceleration
T = time
Solving for T, gives
d = 1/2 A T^2
2d = A T^2
2d/A = T^2
sqrt(2d/A) = T
Substitute the known values and calculate.
sqrt(2d/A) = T
sqrt(2* 1500m / 9.8 m/s^2) = T
sqrt(3000m / 9.8 m/s^2) = T
sqrt(306.122449 s^2) = T
17.49635531 s = T
Rounding to 4 significant figures gives 17.50 seconds. Since it will take
17.50 seconds for the kit to hit the ground, the kit needs to be dropped 17.50
seconds before the plane goes overhead. So just simply multiply by the velocity.
17.50 s * 100 m/s = 1750 m
Answer:
Explanation:
1) The time of flight equation for projectile motion can be used here to find total time in air.
t = 2vsin∅ / g
where v is speed, Ф is launch angle
t = 2×4×sin 60 / 9.8
t = 0.71 seconds
2) Distance where it hit the ground is called as range and has the following standard equation
D = v² sin2Ф/g
D = 4²sin 2×60 / 9.8
D = 1.41m
3) Maximum elevation is maximum time reached
h = v² sin²Ф / 2g
h = 4²sin² 60 / 2*9.8
h = 0.61 m
