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attashe74 [19]
3 years ago
10

A ship travels 84 km on a bearing of 17°, and then travels on a bearing of 107° for 135 km. Find the distance of the end of the

trip from the starting point, to the nearest kilometer.

Physics
1 answer:
Neporo4naja [7]3 years ago
6 0

Answer:

159km

Explanation:

Please view the diagram below;

The distance x of the ship from the staring point could be simply obtained by Pythagoras's theorem as follows;

x^2=135^2+84^2\\x^2=18225+7056\\x^2=25281\\x=\sqrt{25281}\\ x=159km

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nirvana33 [79]

Answer:

A

Explanation:

The answer would be A because when feet push down on anything, the force will push down the skateboard

5 0
3 years ago
21. Calculate the acceleration of the bus from point D to E. Show your work.
Marat540 [252]

21) Acceleration from D to E: 1 m/s^2

22) The acceleration of the bus from D to E is 1 m/s^2

Explanation:

21)

The acceleration of an object is equal to the rate of change of velocity of the object. Mathematically:

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

where

u is the initial velocity

v is the final velocity

t is the time elapsed

In this problem, we want to measure the acceleration of the bus from point D to point E. We have:

- Initial velocity at point D: u = 0

- Final velocity at point E: v = 5 m/s

- Time elapsed from D to E: t = 21 - 16 = 5 s

Therefore, the acceleration between D and E is

a=\frac{5-0}{5}=1 m/s^2

22) This question is the same as 21), so the result is the same.

Learn more about acceleration:

brainly.com/question/9527152

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#LearnwithBrainly

4 0
3 years ago
A block of mass 57.1 kg rests on a slope having an angle of elevation of 28.3°. If pushing downhill on the block with a force ju
stira [4]

Answer:

The coefficient is 0.90

Explanation:

Drawing a diagram makes thing easier, we will assume that the acceleration tends to zero because it start barely moving.

-F_s+mg*sin(\theta)+F=0\\F_s=57.1kg*9.8m/s^2*sin(28.3)+177N\\F_s=442N\\F_s=\µ*N\\N=m*g*cos(\theta)\\N=57.1*9.8*cos(28.3)=493N\\\\\µ=\frac{442N}{493N}=0.90

3 0
3 years ago
A 70 kg student stands on top of a 5.0 m platform diving board . how much gravitational potential energy does the student have?
Marianna [84]

Answer:

a. P.E = 3430Joules.

b. Workdone = 3430Nm

Explanation:

<u>Given the following data;</u>

Mass = 70kg

Distance = 5m

We know that acceleration due to gravity is equal to 9.8m/s²

To find the potential energy;

Potential energy = mgh

P.E = 70*9.8*5

<em>P.E = 3430J</em>

b. To find the workdone;

Workdone = force * distance

But force = mass * acceleration

Force = 70*9.8

Force = 686 Newton.

Workdone = 686 * 5

<em>Workdone = 3430Nm</em>

6 0
2 years ago
Five identical quintuplets leave earth when they reach the age of 21, in the year 2121. Each quintuplet goes on a spaceship jour
Elena-2011 [213]

Answer:

Explanation:

This is a problem based on time dilation , a theory given by Albert Einstein .

The formula of time dilation is as follows .

t₁ = \frac{t}{\sqrt{1-\frac{v^2}{c^2} } }

t is time measured on the earth and t₁ is time measured by man on ship .

A ) Given t = 20 years , t₁ = ? v = .4c

\frac{20}{\sqrt{1-\frac{.16c^2}{c^2} } }

=1.09 x 20

t₁= 21.82 years

B ) Given t = 5 years , t₁ = ? v = .2c

\frac{5}{\sqrt{1-\frac{.04c^2}{c^2} } }

=1.02 x 5

t₁= 5.1 years

C ) Given t = 10 years , t₁ = ? v = .8c

\frac{10}{\sqrt{1-\frac{.64c^2}{c^2} } }

=1.67 x 10

t₁= 16.7  years

D ) Given t = 10 years , t₁ = ? v = .4c

\frac{10}{\sqrt{1-\frac{.16c^2}{c^2} } }

=1.09 x 10

t₁= 10.9  years

E ) Given t = 20 years , t₁ = ? v = .8c

\frac{20}{\sqrt{1-\frac{.64c^2}{c^2} } }

=1.67 x 20

t₁= 33.4   years

7 0
2 years ago
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