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2 years ago

You skip north for 12 minutes to your best friend's house that is 1.5 kilometers away. What is your average velocity?

Physics

1 answer:

Svetllana [295]2 years ago

5 0

Answer:

The average velocity is 7.5 km/h

Explanation:

Were Gonna convert minutes to hours

12 minutes = 12/60 hours = 0.2 hours

Now we estimate the average velocity of the distance traveled over this time

1.5 / 0.2 km/h = 7.5 km/h

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Which of the following is not a galilean moon?

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The answer is B) titan

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3 years ago

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If the mass of a material is 50 grams and the volume of the material is 12 cm^3, what would the density of the material be?

AlekseyPX

The density of the material would be
25/6 grams per cm^3.
to obtain the result above this is what we do:
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3 years ago

A satellite is put into an orbit at a distance from the center of the Earth equal to twice the distance from the center of the E

Sveta_85 [38]

Answer:

995 N

Explanation:

Weight of surface, w= 4000N

Gravitational constant, g, is taken as 9.81 hence mass, m of surface is W/g where W is weight of surface

m= 4000/9.81= 407.7472

Using radius of orbit of 6371km

The force of gravity of satellite in its orbit, $F=\frac {GMm}{(2r)^{2}}=\frac {GMm}{4(r)^{2}}$

Where $G=6.67*10^{-11}$ and $M=5.94*10^{24}$

$F=\frac {(6.67*10^{-11}*5.94*10^{24}*407.7472)}{4*({6.371*10^{6}m)}^{2}}$

F= 995.01142 then rounded off

F=995N

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3 years ago

A barge floating in fresh water (rho = 1000 kg/m3) is shaped like a hollow rectangular prism with base area A = 550 m2 and heigh

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3 years ago

A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long

Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

$W=-\mu mg\times l$

Put the value into the formula

$W=-0.30\times62\times9.8\times3.50$

$W=-637.98\ J$

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

$K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}$

$\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}$

Final potential energy is zero

So, $\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2$

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl$

Put the value into the formula

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50$

$v_{2}=\sqrt{2\times55.915}$

$v_{2}=10.57\ m/s$

The speed of skier is 10.57 m/s.

Hence, (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0

3 years ago