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3 years ago

How can I act more like Runa from Kakegurui?

2 answers:

8 0

This isn't to hard just put your fingers up ball them in a fist put them next to your shoulders and start running. Also absolutely love that show. Great taste.

igomit [66]3 years ago

4 0

Answer:

LOVE HER, and just study her, and sometimes just pause and try and act like that, or watch the backstory and go through everything they went through.

Explanation:

Hope this helps

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A hand pump is being used to inflate a bicycle tire that has a gauge pressure of 59.0 lb/in2. If the pump is a cylinder of lengt

bagirrra123 [75]

Answer:

∴The air cannot be made to flow in with the given pump at the given conditions.

Explanation:

Given:

gauge pressure of bicycle tyre, $P_g=59\ lb.in^{-2}$

length of cylinder of the pump, $l=17.4\ in$

area of the the cylinder of the pump, $a= 3\ in^2$

we have the density of air at STP, $\rho=4.4256\times 10^{-5}\ lb.in^{-2}$

The piston must be pushed more than the pressure inside the tyre:

$P_g=\rho\times V\div a$

$59=4.4256\times 10^{-5}\times a\times h\div a$

$h=13.33\times 10^5\ in$

∴The air cannot be made to flow in with the given pump at the given conditions.

4 0

3 years ago

A car travels due east with a speed of 38.0 km/h. Raindrops are falling at a constant speed vertically with respect to the Earth

DiKsa [7]

Answer: 116.926 km/h

Explanation:

To solve this we need to analise the relation between the car and the Raindrops. The cars moves on the horizontal plane with a constant velocity.

Car's Velocity (Vc) = 38 km/h

The rain is falling perpedincular to the horizontal on the Y-axis. We dont know the velocity.

However, the rain's traces on the side windows makes an angle of 72.0° degrees. ∅ = 72°

There is a relation between this angle and the two velocities. If the car was on rest, we will see that the angle is equal to 90° because the rain is falling perpendicular. In the other end, a static object next to a moving car shows a horizontal trace, so we can use a trigonometric relation on this case.

The following equation can be use to relate the angle and the two vectors.

Tangent (∅) = Opposite (o) / adjacent (a)

Where the Opposite will be the Rain's Vector that define its velocity and the adjacent will be the Car's Velocity Vector.

Tan(72°) = Rain's Velocity / Car's Velocity

We can searching for the Rain's Velocity

Tan(72°) * Vc = Rain's Velocity

Rain's Velocity = 116.926 km/h

3 0

4 years ago

5) A 20.0 kg cart with no friction wheels sits on a table. A light string is attached to it and runs over a low friction pulley

ella [17]

Answer:

1) Please find attached, created with Microsoft Visio

2) The acceleration of the masses connected by the light string is 0.00735 m/s²

3) The tension in the cord is 0.147 N

4) The time it would take the block to go 1.2 m to the edge of the table is approximately 18.07 s

5) The velocity of the cart as soon as it gets to the edge of the table is 0.042 m/s

Explanation:

1) Please find attached, the required free body diagram, showing the tension, weight and frictional (zero friction) forces acting on the cart and the mass created with Microsoft Visio

2) The acceleration of the masses connected by the light string is given as follows;

F = Mass, m × Acceleration, a

The mass of the truck, M = 20.0 kg

The mass attached to the string, hanging rom the pulley, m = 0.0150 kg

The force, F acting on the system = The pulling force on the cart = The tension on the cable = The weight of the hanging mass = 0.0150 × 9.8 = 0.147 N

The pulling force acting on the cart, F = M × a

∴ F = 0.147 N = 20.0 kg × a

a = 0.147 N/(20.0 kg) = 0.00735 m/s²

The acceleration of the truck = a = 0.00735 m/s²

3) The tension in the cord = F = 0.147 N

4) The time, t, it would take the block to go 1.2 m to the edge of the table is given by the kinematic equation, s = u·t + 1/2·a·t²

Where;

s = The distance to the edge of the table = 1.2 m

u = The initial velocity = 0 m/s (The cart is assumed to be initially at rest)

a = The acceleration of the cart = 0.00735 m/s²

t = The time taken

Substituting the known values, gives;

s = u·t + 1/2·a·t²

1.2 = 0 × t + 1/2 ×0.00735 × t²

1.2 = 1/2 ×0.00735 × t²

t² = 1.2/(1/2 ×0.00735) ≈ 326.5306

t = √(1.2/(1/2 ×0.00735)) ≈ 18.07

The time it would take the block to go 1.2 m to the edge of the table = t ≈ 18.07 s

5) The velocity, v, of the cart as soon as it gets to the edge of the table is given by the kinematic equation, v² = u² + 2·a·s as follows;

v² = u² + 2·a·s

u = 0 m/s

v² = 0² + 2 × 0.00735 × 1.2 = 0.001764

v = √(0.001764) = 0.042

The velocity of the cart as soon as it gets to the edge of the table = v = 0.042 m/s.

7 0

3 years ago

Read 2 more answers

A descent vehicle landing on the moon has a vertical velocity toward the surface of the moon of 29.6 m/s. At the same time, it h

inessss [21]

If the vertical component is 29.6 m/s down, and the horizontal component
is 54.8 m/s parallel to the surface, then the magnitude of the slanty vector is

√(29.6² + 54.8²) = √(876.16 + 3003.04) = √3879.2 = 62.28 m/s .

That's 139 mph ! Wow !

6 0

3 years ago

006 (part 1 of 2) 10.0 points

Wittaler [7]

Snow balls can be fun. In the winter time, and in the Turkey season

8 0

3 years ago