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Damm [24]
3 years ago
15

A cruise started its trip 8:00. Its average speed was 100 km/h.

Physics
2 answers:
Amanda [17]3 years ago
7 0

Answer: 120

Explanation:

Aneli [31]3 years ago
5 0

Answer:

120 k/m

Explanation:

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Initially, a 2.00-kg mass is whirling at the end of a string (in a circular path of radius 0.750 m) on a horizontal frictionless
drek231 [11]

Answer:

v_f = 15 \frac{m}{s}

Explanation:

We can solve this problem using conservation of angular momentum.

The angular momentum \vec{L} is

\vec{L}  = \vec{r} \times \vec{p}

where \vec{r} is the position and \vec{p} the linear momentum.

We also know that the torque is

\vec{\tau} = \frac{d\vec{L}}{dt}  = \frac{d}{dt} ( \vec{r} \times \vec{p} )

\vec{\tau} =  \frac{d}{dt}  \vec{r} \times \vec{p} +   \vec{r} \times \frac{d}{dt} \vec{p}

\vec{\tau} =  \vec{v} \times \vec{p} +   \vec{r} \times \vec{F}

but, as the linear momentum is \vec{p} = m \vec{v} this means that is parallel to the velocity, and the first term must equal zero

\vec{v} \times \vec{p}=0

so

\vec{\tau} =   \vec{r} \times \vec{F}

But, as the only horizontal force is the tension of the string, the force must be parallel to the vector position measured from the vertical rod, so

\vec{\tau}_{rod} =   0

this means, for the angular momentum measure from the rod:

\frac{d\vec{L}_{rod}}{dt} =   0

that means :

\vec{L}_{rod} = constant

So, the magnitude of initial angular momentum is :

| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i| cos(\theta)

but the angle is 90°, so:

| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i|

| \vec{L}_{rod_i} | = r_i * m * v_i

We know that the distance to the rod is 0.750 m, the mass 2.00 kg and the speed 5 m/s, so:

| \vec{L}_{rod_i} | = 0.750 \ m \ 2.00 \ kg \ 5 \ \frac{m}{s}

| \vec{L}_{rod_i} | = 7.5 \frac{kg m^2}{s}

For our final angular momentum we have:

| \vec{L}_{rod_f} | = r_f * m * v_f

and the radius is 0.250 m and the mass is 2.00 kg

| \vec{L}_{rod_f} | = 0.250 m * 2.00 kg * v_f

but, as the angular momentum is constant, this must be equal to the initial angular momentum

7.5 \frac{kg m^2}{s} = 0.250 m * 2.00 kg * v_f

v_f = \frac{7.5 \frac{kg m^2}{s}}{ 0.250 m * 2.00 kg}

v_f = 15 \frac{m}{s}

8 0
3 years ago
A person's center of mass is easily found by having the person lie on a reaction board. A horizontal, 2.1-m-long, 6.6 kg reactio
MrRissso [65]
Answer is x hopefully this helps your stupid head
5 0
3 years ago
Which step of mitosis involves the chromosomes lining up in the center of the cell?
gizmo_the_mogwai [7]

Answer:

Metaphase

Explanation:

During metaphase,chromosome line up in the middle of the cell.

5 0
3 years ago
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Which law of motion accounts for the following statement?
julia-pushkina [17]
That would be the second law
5 0
3 years ago
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A bee flies at a constant rate of 5 m s until it notices a flower. It then comes to rest in 5 seconds. Find the acceleration of
FrozenT [24]

Answer:

B) -1m/s^2

Explanation:

Final speed = 0 m/s

Initial speed = 5m/s

Time taken for it to come to rest(0m/s) = 5

then use the formula;

[v = u + at],where v is the final speed..u is the initial speed..t is time taken for it to come to rest and a is the acceleration

; 0 = 5 + 5a

; -5 = 5a

;Acceleration = -1 m/s^2

5 0
3 years ago
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