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Sphinxa [80]
3 years ago
14

Samuel is running in a park’s nature trail. At the first checkpoint, he ran 925 m in 10 minutes. At the second checkpoint, he sl

owed down and only ran 673 m in 13 minutes. What was the average velocity of Samuel’s entire trip (in m/s)?
2.38 m/s

1.16 m/s

3.18 m/s

1.89 m/s
Physics
1 answer:
svetoff [14.1K]3 years ago
8 0

Answer:

Average Velocity= 1.16m/s

Explanation:

Distance initially run= d1= 925m

distance finally run= d2= 673m

Time= t1= 10m

time= t2= 13m

Average Velocity= Total distance covered/ total time

                           =d1+d2/t1+t2

                  = 925+673/ 10(60) + 13(60)  {converting mins to secs)

                = 1598/1380= 1.157≅ 1.16m/s

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Let r denote the distance between the center of the earth and the center of the moon. What is the magnitude of the acceleration
stepan [7]

Answer:

The magnitude of the acceleration ae of the earth due to the gravitational pull of the moon is \mathbf{3.3187\times10^{-5}}\frac{m}{s^{2}}

Explanation:

By Newton's gravitational law, the magnitude of the gravitational force between two objects is:

F=G\frac{Mm}{r^{2}}(1)

With G the gravitational constant, M the mass of earth, m the mass of the moon and r the distance between the moon and the earth, a quick search on physics books or internet websites give us the values:

M=5.972\times10^{24}\,kg

m=7.34767309\times10^{22}\,kg

r=384400\,km

G=6.674\times10^{-11}\,\frac{N\,m^{3}}{kg^{2}}

Using those values on (1)

F=(6.674\times10^{-11})*\frac{(5.972\times10^{24})(7.34767309\times10^{22})}{(384400\times10^{3})^{2}}

F\approx1.98193\times10^{20}N

Now, by Newton's second Law we can find the acceleration of earth ae due moon's pull:

F=M*ae\Longrightarrow ae=\frac{F}{M}=\frac{1.98193\times10^{20}}{5.972\times10^{24}}\approx\mathbf{3.3187\times10^{-5}}

6 0
4 years ago
A 2-m long string is stretched between two supports with a tension that produces a wave speed equal to vw=50.00m/s. What are the
svetoff [14.1K]

Answer

given,

Length of the string, L = 2 m

speed of the wave , v = 50 m/s

string is stretched between two string

For the waves the nodes must be between the strings

the wavelength  is given by

           \lambda = \dfrac{2L}{n}

where n is the number of antinodes; n = 1,2,3,...

the frequency expression is given by

            f = n\dfrac{v}{2L}

now, wavelength calculation

      n = 1

           \lambda_1 = \dfrac{2\times 2}{1}

                    λ₁ = 4 m

      n = 2

           \lambda_2 = \dfrac{2\times 2}{2}

                   λ₂ = 2 m

      n =3

           \lambda_3 = \dfrac{2\times 2}{3}

                    λ₃ = 1.333 m

now, frequency calculation

      n = 1

            f = n\dfrac{v}{2L}

            f_1 =1\times \dfrac{50}{2\times 2}

                    f₁ = 12.5 Hz

      n = 2

            f = n\dfrac{v}{2L}

            f_2 =2\times \dfrac{50}{2\times 2}

                    f₂= 25 Hz

      n = 3

            f = n\dfrac{v}{2L}

            f_3 =3\times \dfrac{50}{2\times 2}

                    f₃ = 37.5 Hz

8 0
3 years ago
What's √ 25what's √ 25 ​
malfutka [58]

Answer:

the square root of 25 is 5

Explanation:

\sqrt{25}=5

5 0
3 years ago
Two blocks of masses 6 kg and 5.5 kg are
sp2606 [1]

When you squish the spring, you put some energy into it, and after the cord
burns and they go boing in opposite directions, that energy that you stored
in the spring is what gives the blocks their kinetic energy.

But linear momentum still has to be conserved.  It was zero while they were
tied together and nothing was moving, so it has to be zero after they both
take off.

Momentum = (mass) x (velocity)

After the launch, the 5.5-kg moves to the right at 6.8 m/s,
so its momentum is
                               (5.5 x 6.8) = 37.4 kg-m/s to the right.

In order for the total momentum to be zero, the other block has to
carry the same amount of momentum in the opposite direction.

               M x V = (6 x speed) = 37.4 kg-m/s to the left.

Divide each side by  6 :      Speed = 37.4 / 6 =<em>  6.2333... m/s left</em>

(That number is  (6 and 7/30) m/s .)
5 0
3 years ago
Describe how the particles of a substance behave at its boiling point.
fgiga [73]
Particles begin to move out of their ordered arrangement due to energy. They are unable to escape because of external pressure pushing down on the liquid.
4 0
3 years ago
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