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Masja
6 days ago
10

A marble rolls with a velocity of 10 mm/s [E] on a game board that is being pulled [60o N of E] at 40.0 mm/s. What is the veloci

ty of the marble relative to the floor?

Physics
1 answer:
ValentinkaMS [1.1K]6 days ago
7 0

Answer:

R = 45.82[mm/s], with an angle of 48.95°

Explanation:

To solve this, the two vectors will be illustrated in a diagram, allowing someone viewing from outside the game board to see that the collective motion comes from both the marble's trajectory and the board's movement.

The image attached displays this vector addition. By utilizing the parallelogram law, we can derive the resultant vector, shown in red.

Next, we will break down the velocity vector of 40 [mm/s] into its x and y components.

x = 40*cos(60) = 20 [mm/s]

y = 40*sin(60) = 34.64 [mm/s]

<pConsequently, the resultant vector is:

Rx = 20 + 10 = 30 [mm/s]

Ry = 34.64 [mm/s]

<pApplying the Pythagorean theorem, we can ascertain the vector’s magnitude

R = \sqrt{(30)^{2} +(34.64)^{2} }\\ R = \sqrt{2099.92}\\R=45.82[mm/s]

Next, we can calculate the angle relative to the horizontal using the previously obtained components.

tan α = 34.46 / 30

α = 48.95°

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Yuliya22 [1153]

Explanation:

We are given that,

Mass of the rocket, m=250\times 10^3\ slugs

(a) The standard unit of mass is kilogram (kg). The conversion between slugs and kilograms is as follows:

1 slug = 14.59 kg

Thus, 250\times 10^3\ slugs=250\times 10^3\times 14.59

Mass of the rocket, m = 3647500 kg

(b) The weight of the rocket can be expressed as:

W = m g

W=3647500\times 9.8=35745500\ N

or

W=3.5\times 10^7\ N

(c) If the rocket were on the moon, the gravitational acceleration on the moon is given as a=5.3\ ft/s^2=1.61\ m/s^2

Mass refers to the quantity of matter present in an object. Therefore, the mass of the rocket remains constant at 3647500 kg

The weight of the rocket on the moon would be, W=mg

W=3647500\times 1.61

W = 5872475 N

or

W=5.8\times 10^6\ N

Thus, this is the final answer required.

8 0
3 days ago
For a group class project, students are building model roller coasters. Each roller coaster needs to begin at the top of the fir
Keith_Richards [1034]

Case A:

A.75 kg 65 N/m 1.2 m

m = weight of the car = 0.75 kg

k = spring's stiffness = 65 N/m

h = elevation of the hill = 1.2 m

x = spring's compression = 0.25 m

Applying the principle of energy conservation from the Top of the hill to the Bottom of the hill

Energy at the Top of the hill equals Energy at the Bottom of the hill

spring energy + gravitational potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (65) (0.25)² + (0.75 x 9.8 x 1.2) = (0.5) (0.75) v²

v = 5.4 m/s



Case B:

B.60 kg 35 N/m.9 m

m = weight of the car = 0.60 kg

k = spring's stiffness = 35 N/m

h = height of the hill = 0.9 m

x = spring's compression = 0.25 m

Using energy conservation from the Top of the hill to the Bottom of the hill

Top hill energy = Bottom hill energy

spring energy + gravitational potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (35) (0.25)² + (0.60 x 9.8 x 0.9) = (0.5) (0.60) v²

v = 4.6 m/s




Case C:

C.55 kg 40 N/m 1.1 m

m = weight of the car = 0.55 kg

k = spring's stiffness = 40 N/m

h = height of the hill = 1.1 m

x = spring's compression = 0.25 m

Using conservation of energy from the Top of the hill to the Bottom of the hill

Energy at the Top of the hill = Energy at the Bottom of the hill

spring energy + gravitational potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (40) (0.25)² + (0.55 x 9.8 x 1.1) = (0.5) (0.55) v²

v = 5.1 m/s




Case D:

D.84 kg 32 N/m.95 m

m = weight of the car = 0.84 kg

k = spring's stiffness = 32 N/m

h = height of the hill = 0.95 m

x = spring's compression = 0.25 m

Using energy conservation from the Top of the hill to the Bottom of the hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + gravitational potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (32) (0.25)² + (0.84 x 9.8 x 0.95) = (0.5) (0.84) v²

v = 4.6 m/s


thus, the closest result is from case C at 5.1 m/s




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Respuesta:

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Explicación:

Se nos proporciona;

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Así, se puede expresar como;

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Ahora, la presión se daría como;

P = ρgh

Y así,

ρgh_1 + P_(bomba) = ρgh_2

<ppor lo="" tanto="">

P_(bomba) = ρg(h_2 - h_1)

<pal sustituir="" los="" valores="" pertinentes="" obtenemos="">

P_(bomba) = 1000•9.8(30 - 20)

P_(bomba) = 98,000 Pa

</pal></ppor>
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2 days ago
A student performs an experiment that involves the motion of a pendulum. The student attaches one end of a string to an object o
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Answer:

-utilize precisely the same apparatus

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Explanation:

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