Answer:
Explanation:
Given parameters:
Weight of object = 49N
Force applied = 12N
Unknown:
Acceleration of object = ?
Solution:
The acceleration of the object is found by dividing the force by the weight;
Acceleration = = 0.25m/s²
A gas expands from an initial volume of 0.040 m^3 and an initial pressure of 210 kPa to a final volume of 0.065 m^3 while its te
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Answer:
(a) The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced
(b) it’s possible for a diver to enter the water with the velocity of 25 m/s if he has initial velocity of 14.4 m/s. The upward initial velocity can’t be physically attained
Explanation:
(a)
To find the final velocity for an object traveling distance h taking the initial vertical component of velocity as
the kinematics equation is written as
where a is acceleration
Substituting g for a where g is gravitational force value taken as 9.81
Since the initial velocity is zero, we can solve for final velocity by substituting figures, note that 70 ft is 21.3 m for h
= 20.44275
Therefore, the divers enter with a speed of 20.4 m/s
The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced
(b)
The divers can enter water with a velocity of 25 m/s only if they have some initial velocity. Using the kinematic equation
Since we have final velocity of 25 m/s
= 14.390761 m/s
Therefore, it’s possible for a diver to enter the water with the velocity of 25 m/5 if he has initial velocity of 14.4 m/s
In conclusion, the upward initial velocity can’t be physically attained
Answer:
The answer to the question is
The distance d, which locates the point where the light strikes the bottom is 29.345 m from the spotlight.
Explanation:
To solve the question we note that Snell's law states that
The product of the incident index and the sine of the angle of incident is equal to the product of the refractive index and the sine of the angle of refraction
n₁sinθ₁ = n₂sinθ₂
y = 2.2 m and strikes at x = 8.5 m, therefore tanθ₁ = 2.2/8.5 = 0.259 and
θ₁ = 14.511 °
n₁ = 1.0003 = refractive index of air
n₂ = 1.33 = refractive index of water
Therefore sinθ₂ = =
= 0.1885 and θ₂ = 10.86 °
Since the water depth is 4.0 m we have tanθ₂ = or x₂ =
=
= 20.845 m
d = x₂ + 8.5 = 20.845 m + 8.5 m = 29.345 m.