Help me with this please

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

Yuliya22 [10]

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.

7 0

2 years ago

Determine the angle of an incline that would yield a constant velocity, given the coefficient of kinetic friction is 0.10.

azamat

Answer:

$\theta=5.71^{o}$

Explanation:

In order to solve this problem, we mus start by drawing a free body diagram of the given situation (See attached picture).

From the free body diagram we can now do a sum of forces in the x and y direction. Let's start with the y-direction:

$\sum F_{y}=0$

$-W_{y}+N=0$

$N=W_{y}$

so:

$N=mgcos(\theta)$

now we can go ahead and do a sum of forces in the x-direction:

$\sum F_{x}=0$

the sum of forces in x is 0 because it's moving at a constant speed.

$-f+W_{x}=0$

$-\mu_{k}N+mg sen(\theta)=0$

$-\mu_{k}mg cos(\theta)+mg sen(\theta)=0$

so now we solve for theta. We can start by factoring mg so we get:

$mg(-\mu_{k} cos(\theta)+sen(\theta))=0$

we can divide both sides into mg so we get:

$-\mu_{k} cos(\theta)+sen(\theta)=0$

this tells us that the problem is independent of the mass of the object.

$\mu_{k} cos(\theta)=sen(\theta)$

we now divide both sides of the equation into $cos(\theta)$ so we get:

$\mu_{k}=\frac{sen(\theta)}{cos(\theta)}$

$\mu_{k}=tan(\theta)$

so we now take the inverse function of tan to get:

$\theta=tan^{-1}(\mu_{k})$

so now we can find our angle:

$\theta=tan^{-1}(0.10)$

so

$\theta=5.71^{o}$

8 0

2 years ago

6. A warehouse employee is pushing a 15.0 kg desk across a floor at a constant speed of 0.50 m/s. How much work must the employe

MariettaO [177]

Answer:

7.5 J

Explanation:

To answer the question given above, we need to determine the energy that will bring about the speed of 1 m/s. This can be obtained as follow:

Mass (m) = 15 Kg

Velocity (v) = 1 m/s

Energy (E) =?

E = ½mv²

E = ½ × 15 × 1²

E = ½ × 15 × 1

E = ½ × 15

E = 7.5 J

Therefore, to change the speed to 1 m/s, the employee must do a work of 7.5 J.

3 0

2 years ago

Our data for how much force is needed can be qualitative. This means we don’t have to know exact values for the force in Newtons

kirill [66]

Answer:

For macroscopic objects, we use units such as grams and kilograms to state their masses, but. There will not be much error if you estimate the mass of an atom by simply. to what we did for our hypothetical example, but the percentages are different: Add two atomic masses of carbon and six atomic masses of hydrogen:

Explanation:

8 0

2 years ago

Read 2 more answers

A vw bug is driving north on a striaght two lane road at a constant 38mph and a chevy impala is driving south at 29mph on the sa

DedPeter [7]

The Answer is B. 67mph, north

5 0

3 years ago