Automobile traveling at 65 mph constant on the road described below. Find rate at which radar must rotate when theta = 15 deg. A
1 answer:
Answer:
The rate at which radar must rotate is 0.335 rad/s.
Explanation:
Given that,
Velocity = 65 m/h = 29.0576 m/s
Angle = 15°
Suppose, the radius given by
We need to calculate the rate at which radar must rotate
Using formula of linear velocity
Where, v = velocity
r = radius
Put the value into the formula
Hence, The rate at which radar must rotate is 0.335 rad/s.
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Answer:
The maximum radius the asteroid can have for her to be able to leave it entirely simply by jumping straight up is approximately 1782.45 meters
Explanation:
Whereby the height the astronaut can jump on Earth = 0.500 m, we have the following kinematic equation;
v² = u² - 2·g·h
Where;
v = The final velocity
u = The initial velocity
g = The acceleration due to gravity ≈ 9.8 m/s²
h = The height she jumps
At the maximum height, = 0.500 m, she jumps, v = 0, therefore, we have;
0² = u² - 2·g·
u² = 2 × 9.8 × 0.5 = 9.8
u = √9.8 ≈ 3.13
u = 3.13 m/s
Her initial jumping velocity ≈ 3.13 m/s
Escape velocity,
Where;
M = The mass of the asteroid
G = The Universal gravitational constant = 6.67408 × 10⁻¹¹ m³/(kg·s²)
r = The radius of the asteroid
The average density of the Earth = 5515 kg/m³
The mass of the asteroid, M = Density × Volume = 5515 kg/m³× 4/3 × π × r³
The escape velocity, she has, ≈ 3.13 m/s is therefore;
Therefore, the maximum radius of the asteroid can have for her jumping velocity to be equal to the escape velocity for her to be able to leave it entirely simply by jumping straight up = r ≈ 1782.45 meters.
Answer:
0.82 MPa
Explanation:
the change in pressure 'σ'=160kPa
K= σ/∈ => σ/3∈
K= 160/(3 x 0.065)
K=820 kPA=0.82 MPa
Thus,the bulk modulus of the tissue 'K' is 0.82 MPa