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Pepsi [2]
3 years ago
6

A ship leaves a harbor and sails at 23.5 ∘ to the north of due west. After traveling 575 km, how far west is the ship from the h

arbor?
Physics
1 answer:
solniwko [45]3 years ago
6 0

Answer:

D = 527.31 Km

Explanation:

given,

angle of ship, θ = 23.5° N of W

distance travel in the direction = 575 Km

Distance of ship in west from harbor = ?

now,

Distance of the ship in the west direction

D = d cos θ

d = 575 Km

θ = 23.5°

inserting all the values

D = 575 x cos 23.5°

D = 575 x 0.91706

D = 527.31 Km

Hence, the distance travel by the ship in west from harbor is equal to D = 527.31 Km  

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Arunner starts from rest and accelerates uniformly to a speed of 8.0 meters per
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Answer:

2.0 m/s/s

Explanation:

The acceleration of an object is the rate of change of velocity of the object.

Mathematically, it is given by:

a=\frac{v-u}{t}

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u is the initial velocity

v is the final velocity

t is the time taken for the velocity to change from u to v

Acceleration is a vector, so it has both a magnitude and a direction.

For the runner in this problem, we have:

u = 0 is the initial velocity (he starts from rest)

v = 8.0 m/s is the final velocity

t = 4.0 s is the time taken

Substituting, we find

a=\frac{8.0-0}{4.0}=2.0 m/s^2

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3 years ago
How is a projectile different from an object in free fall
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Explanation:

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Bodies A and B have equal mass. Body B is initially at rest. Body A collides with body B in a one-dimensional elastic collision.
jek_recluse [69]
According to the statement " Collision <span>between two bodies in which the total kinetic energy of the two bodies after the collision is equal to their total kinetic energy before the collision."
The best answer is :
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What is unique about the movement of uranus compared to the other outer planets
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where σ(t) and σ(0) represents the time-dependent and initial (i.e., time =0) stresses, respectively, and t and τ denote elapsed
lesya [120]

Answer:

E_r(6)=4.35614\ MPa

Explanation:

\epsilon = Strain = 0.49

\sigma _0 = 3.1 MPa

At t = Time = 32 s \sigma = 0.41 MPa

\tau = Time-independent constant

Stress relation with time

\sigma=\sigma _0exp\left(-\frac{t}{\tau}\right)

at t = 32 s

0.41=3.1exp\left(-\frac{32}{\tau}\right)\\\Rightarrow exp\left(-\frac{32}{\tau}\right)=\frac{0.41}{3}\\\Rightarrow -\frac{32}{\tau}=ln\frac{0.41}{3}\\\Rightarrow \tau=-\frac{32}{ln\frac{0.41}{3}}\\\Rightarrow \tau=16.0787\ s

The time independent constant is 16.0787 s

E_{r}(t)=\frac{\sigma(t)}{\epsilon_0}

At t = 6

\\\Rightarrow E_{r}(6)=\frac{\sigma(6)}{\epsilon_0}

From the first equation

\sigma(t)=\sigma _0exp\left(-\frac{t}{\tau}\right)\\\Rightarrow \sigma(6)=3.1exp\left(-\frac{6}{16.0787}\right)\\\Rightarrow \sigma(6)=2.13451

E_r(6)=\frac{2.13451}{0.49}\\\Rightarrow E_r(6)=4.35614\ MPa

E_r(6)=4.35614\ MPa

6 0
3 years ago
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