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sukhopar [10]
3 years ago
15

At last, it's time to go diving! You load your stuff into the boat and head to an offshore dive site.

Physics
1 answer:
Ulleksa [173]3 years ago
4 0
A. The boat uses an engine which provides it the kinetic energy to move. The engine uses gasoline, or manual paddles, push pole or propeller.  
B. In terms of thermodynamics law- for every action, there is an equal and opposite reaction. In this case, the thrust causes movement in case of the boat.
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A hockey puck on a frozen pond is given an initial speed of 20.0 m/s. If the puck always remains on the ice and slides 115 m bef
bekas [8.4K]

Answer:

μ_k = 0.1773

Explanation:

We are given;

Initial velocity;u = 20 m/s

Final velocity;v = 0 m/s (since it comes to rest)

Distance before coming to rest;s = 115 m

Let's find the acceleration using Newton's second law of motion;

v² = u² + 2as

Making a the subject, we have;

a = (v² - u²)/2s

Plugging relevant values;

a = (0² - 20²)/(2 × 115)

a = -400/230

a = -1.739 m/s²

From the question, the only force acting on the puck in the x direction is the force of friction. Since friction always opposes motion, we see that:

F_k = −ma - - - (1)

We also know that F_k is defined by;

F_k = μ_k•N

Where;

μ_k is coefficient of kinetic friction

N is normal force which is (mg)

Since gravity acts in the negative direction, the normal force will be positive.

Thus;

F_k = μ_k•mg - - - (2)

where g is acceleration due to gravity.

Thus,equating equation 1 and 2,we have;

−ma = μ_k•mg

m will cancel out to give;

-a = μ_k•g

μ_k = -a/g

g has a constant value of 9.81 m/s², so;

μ_k = - (-1.739/9.81)

μ_k = 0.1773

3 0
3 years ago
Four aqueous solutions and their concentrations are shown in the above illustration. which of the solutions is most likely to be
Elza [17]
The solution that would most likely be a strongest conductor of electricity is the solution that is most saturated or concentrated. This is because the atoms that are found within the aqueous solutions have become positively charged resulting to the attraction of negatively charged ions that are found in electricity. On the other hand, the least conductive from the aqueous solutions would be the most unsaturated one because of less conductive ions present.
3 0
2 years ago
A force of 5N and a force of 8N act to the same point and are inclined at 45degree to each other. Find the magnitude and directi
Alex_Xolod [135]
  • Magnitude: 12.1 N.
  • Direction: 17.0° to the 8 N force.
<h3>Explanation</h3>

Refer to the diagram attached (created with GeoGebra). Consider the 5 N force in two directions: parallel to the 8 N force and normal to the 8 N force.

  • \displaystyle F_{\text{1, Parallel}} = F_1 \cdot \cos{45^\textdegree} = \dfrac{5\sqrt{2}}{2}\;\text{N}.
  • \displaystyle F_{\text{1, Normal}} = F_1 \cdot \sin{45^\textdegree} = \dfrac{5\sqrt{2}}{2}\;\text{N}.

The sum of forces on each direction will be the resultant force on that direction:

  • Resultant force parallel to the 8 N force: (8 + \dfrac{5\sqrt{2}}{2})\;\text{N}.
  • Resultant force normal to the 8 N force: \dfrac{5\sqrt{2}}{2}\;\text{N}.

Apply the Pythagorean Theorem to find the magnitude of the resultant force.

\displaystyle \Sigma F = \sqrt{{(8 + \dfrac{5\sqrt{2}}{2})}^2 + {(\dfrac{5\sqrt{2}}{2})}^2} = 12.1\;\text{N} (3 sig. fig.).

The size of the angle between the resultant force and the 8 N force can be found from the tangent value of the angle. Tangent of the angle:

\displaystyle \dfrac{\Sigma F_\text{Normal}}{\Sigma F_\text{Parallel}} = \dfrac{8 + \dfrac{5\sqrt{2}}{2}}{\dfrac{5\sqrt{2}}{2}} \approx 0.306491.

Find the size of the angle using inverse tangent:

\displaystyle \arctan{ \dfrac{\Sigma F_\text{Normal}}{\Sigma F_\text{Parallel}}} = \arctan{0.306491} = 17.0\textdegree.

In other words, the resultant force is 17.0° relative to the 8 N force.

4 0
3 years ago
If the sprinter accelerates at that rate for a distance of 15 m, and then maintains the velocity he has at that point for the re
ahrayia [7]

Answer:

The time for the entire race is 11.39 sec.

Explanation:

Given that,

Distance = 15 m

Remainder distance = 100 m

Suppose A sprinter begins a race with an acceleration of 3.4 m/s².

We need to calculate the time

Using equation of motion

s=ut+\dfrac{1}{2}at^2

Put the value in the equation

15=0+\dfrac{1}{2}\times3.4\times t^2

t^2=\dfrac{30}{3.4}

t=2.97\ sec

We need to calculate the final velocity of sprinter

Using equation of motion again

v=u+at

Put the value into the formula

v=0+3.4\times2.97

v=10.09\ m/s

We need to calculate the distance covers by sprinter

d=100-15=85\ m

The sprinter need to covers only 85 m.

We need to calculate the time

Using formula of time

t=\dfrac{d}{v}

Put the value into the formula

t'=\dfrac{85}{10.09}

t'=8.42\ sec

We need to calculate the time for the entire race

t''=t+t'

Put the value into the formula

t''=2.97+8.42

t''=11.39\ sec

Hence, The time for the entire race is 11.39 sec.

4 0
3 years ago
Please help me i’ll give you brainlist
Diano4ka-milaya [45]

Answer:

True

Explanation:

6 0
2 years ago
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