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goblinko [34]
3 years ago
9

two ships leave a port at the same time. The first ship sails on a bearing of 40 degrees at 18 knots and the second at a bearing

of 130 degrees at 26 knots. How far apart are they after 1.5 hours?

Physics
1 answer:
Yuliya22 [10]3 years ago
7 0

Answer:

The distance between the ships is 87.84 km.

Explanation:

Given that,

Angle of first ship= 40°

Speed of first ship = 18 knots

Angle of second ship= 130°

Speed of second ship = 26 knots

We need to calculate the resultant velocity

Using cosine rule

v=\sqrt{v_{1}^2+v_{2}^2-2v_{1}v_{2}\cos\theta}

Put the value into the formula

v=\sqrt{18^2+26^2-2\times18\times26\times\cos90}

v=\sqrt{18^2+26^2}

v=\sqrt{324+676}

v=10\sqrt{10}

We need to calculate the distance between the ships

d =v\times t

Put the value into the formula

d=10\sqrt{10}\times1.5\times1.852

d=87.84\ km

Hence, The distance between the ships is 87.84 km.

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2 years ago
At what point in its motion is the KE of a pendulum bob a maximum? 1. The KE does not change. 2. at the lowest point 3. at the h
Andrew [12]

Answer:

2. at the lowest point

Explanation:

The motion of the pendulum is a continuous conversion between kinetic energy (KE) and gravitational potential energy (GPE). This is because the mechanical energy of the pendulum, which is sum of KE and GPE, is constant:

E = KE + GPE = const.

Therefore, when KE is maximum, GPE is minimum, and viceversa.

So, the point of the motion where the KE is maximum is where the GPE is minimum: and since the GPE is directly proportional to the heigth of the bob:

GPE=mgh

we see that GPE is minimum when the bob is at the lowest point,so the correct answer is

2. at the lowest point

3 0
3 years ago
During a race, a runner runs at a speed of 6 m/s. 2 seconds later, she is running at a speed of 10 m/s. What is the runner’s acc
Lapatulllka [165]
Let's calculate the average acceleration. It is the rate of changing speeds. Hence, we need to calculate the difference of speeds. 10-6=4 m/s. The rate is now \frac{4m/s}{2s} =2m/s^2.
In general, the formula for the mean acceleration between two times 1 and 2 is given by:
\frac{u_2-u_1}{T} where v1 and v2 are the speeds at the respective points and T is the time interval between them.
5 0
3 years ago
A 2000-kg car moving with a speed of 20 m/s collides with and sticks to a 1500-kg car at rest at a stop sign. Show that because
amid [387]

Answer:

13.33m/s

Explanation:

Given data

m1= 2000kg

u1= 20m/s

m2= 1500kg

u2= 0m/s

v1= 10m/s

Required

The speed of the sticks

We know that  from the expression for the conservation of momentum

m1u1+m2u2= m1v1+m2v2

2000*20+1500*0=2000*10+1500*v2

40000=20000+1500v2

collect like terms

40000-20000= 1500v2

20000= 1500v2

v2= 20000/1500

v2= 13.33 m/s

Hence the velocity of the sticks is 13.33m/s

8 0
3 years ago
Consider a highway composed of concrete. If the road is 4.5 km long, 30 m wide, and 0.70 m thick, what is its mass?
Aloiza [94]

Answer:

Explanation:

We need to assume that the density of the concrete is about 2350 Kg/m^3. And using the dimensions of the highway we can calculate the volume of the highway.

vh=4500 * 30 * 0,70\\vh=94500 m^3\\den=m/v\\m=den*v\\m=2350*94500\\\\m=222075 ton

5 0
3 years ago
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