The distance traveled by the particle at the given time interval is 0.28 m.
<h3>
Position of the particle at time, t = 0</h3>
The position of the particle at the given time is calculated as follows;
x = 2 sin2(t)
y = 2 cos2(t)
x(0) = 2 sin2(0) = 0
y(0) = 2 cos2(0) = 2(1) = 2
<h3>
Position of the particle at time, t = 4</h3>
x = 2 sin2(t)
y = 2 cos2(t)
x(4) = 2 sin2(4) = 0.28
y(4) = 2 cos2(4) = 2(1) = 1.98
<h3>Distance traveled by the particle at the given time interval</h3>
d = √[(x₄ - x₀)² + (y₄ - y₀)²]
d = √[(0.28 - 0)² + (1.98 - 2)²]
d = 0.28 m
Thus, the distance traveled by the particle at the given time interval is 0.28 m.
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Answer:
a)V= 0.0827 m³
b)P=181.11 x 10² N/m²
Explanation:
Given that
m = 81.5 kg
Density ,ρ = 985 kg/m³
As we know that
Mass = Volume x Density
81.5 = V x 985
V= 0.0827 m³
The force exerted by weight = m g
F= m g= 81.5 x 10 = 815 N ( Take ,g= 10 m/s²)
Area ,A= 4.5 x 10⁻² m²
The Pressure P
P=181.11 x 10² N/m²
Answer:
v = 1.30 m/s
Explanation:
given,
mass hung = 0.35 Kg
spring stretched when load is hanged (x)= 0.13 m
now,
weight of the mass attached = Kx
m g = k x
0.35 x 9.8 = k x 0.13
k = 26.38 N/m
now, using conservation of energy
v = 1.30 m/s
Answer:
Explanation:
It is given that,
Speed of the projectile is 0.5 v. Let h is the height above the ground. Using the first equation of motion to find it.
Initial speed of the projectile is v and final speed is 0.5 v.
g is the acceleration due to gravity
Let h is the height above the ground. Using the second equation of motion as :
So, the height of the projectile above the ground is . Hence, this is the required solution.
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