A 4 kg car with frictionless wheels is on a ramp that makes a 25° angle with the horizontal. The car is connected to a 1 kg mass
by a string that passes over a frictionless pulley.
What is the acceleration of the car?
What is the acceleration of the car?
1 answer:
Refer to the diagram shown below.
W₁ = (4 kg)*(9.8 m/s²) = 39.2 N
W₂ = (1 kg)*(9.8 m/s²) = 9.8 N
The normal reaction on the 4-kg mass is
N = (39.2 N)*cos(25°) = 35.5273 N
The force acting down the inclined plane due to the weight is
F = (39.2 N)*sin(25°) = 16.5666 N
The net force that accelerates the 4-kg mass at a m/²s down the plane is
F - W₂ = (4 kg)*(a m/s²)
4a = 16.5666 - 9.8
a = 1.6917 m/s²
Answer: 1.69 m/s² (nearest hundredth)
W₁ = (4 kg)*(9.8 m/s²) = 39.2 N
W₂ = (1 kg)*(9.8 m/s²) = 9.8 N
The normal reaction on the 4-kg mass is
N = (39.2 N)*cos(25°) = 35.5273 N
The force acting down the inclined plane due to the weight is
F = (39.2 N)*sin(25°) = 16.5666 N
The net force that accelerates the 4-kg mass at a m/²s down the plane is
F - W₂ = (4 kg)*(a m/s²)
4a = 16.5666 - 9.8
a = 1.6917 m/s²
Answer: 1.69 m/s² (nearest hundredth)
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Explanation:
It is known that relation between pressure and density is as follows.
P =
where, P = pressure
= density
g = acceleration due to gravity
h = height
Putting the given values into the above formula as follows.
P =
=
= 110495000 Pa
Now, relation between pressure and force is as follows.
P =
or, F = PA
F =
=
Thus, we can conclude that a force of can be experienced at such depth.
Answer:
16.9000000000000001 J
Explanation:
From the given information:
Let the initial kinetic energy from point A be = 1.9000000000000001 J
and the final kinetic energy from point B be = ???
The charge particle Q = 6 mC = 6 × 10⁻³ C
The change in the electric potential from point B to A;
i.e. V_B - V_A = -2.5 × 10³ V
According to the work-energy theorem:
-Q × ΔV = ΔK