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3 years ago

Describe your acceleration as you skate down your neighborhood sidewalk.

Physics

1 answer:

Paraphin [41]3 years ago

6 0

Answer: when you increase or decrease your speed.

Explanation:

Moving a skate at rest, you need to apply force in order to cause acceleration.

F = ma Where

F = force applied

m = mass of the skate

a = acceleration

The initial velocity u will be equal to zero and the skate will acceleration to a certain velocity.

as you skate down your neighborhood sidewalk, you will accelerate when you increase your speed. Because

Acceleration is the rate of change of velocity. That is,

Acceleration = change in velocity/ time.

And also, you will decelerate when you reduce the speed or velocity down your neighborhood sidewalk.

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A 9.5 V battery supplies a 3.5 mA current to a circuit for 7.0 h. Part A How much charge has been transferred from the negative

pychu [463]

Answer:

88.2 C

Explanation:

The current can be defined as the rate of flow of charge in a conductor.

The relation between charge current and time is given as

I = Q/T

I = current, Q= charge and T = time

that is ampere = coulomb / second

The amount of charge passed is from the negative to the positive terminal

shall be given by:

Q = I * t = 3.5mA * 7h * 3600s/h = 88.2 C

Note: take care of the units.

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3 years ago

When a solid compound dissolves in water

VladimirAG [237]

Answer:

The solid separates and disperse uniformly throughout the solution.

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3 years ago

What countries could benefit most from solar energy

Alecsey [184]

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Hoped this helped

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3 years ago

Una flecha tiene una rapidez de lanzamiento inicial de 18 m/s. Si debe dar en un blanco a 31 m de distancia, que está a la misma

maks197457 [2]

Answer:

34.8 and 55.2º

Explanation:

This is a projectile launching exercise, as we are told that the range of the arrow must be equal to its range and = 31 m let's use the equation

The scope equation is

R = v₀² sin 2θ /g

sin 2 θ = R g / v₀²

sin 2 θ = 31 9.8 / 18²

2 θ = sin⁻¹ 0.93765

θ = 34.8º

At the launch of projectiles we have two complementary angles with the same range in this case 34.8 and (90-34.8) = 55.2º

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3 years ago

Joey, whose mass is m = 36 kg, stands at rest at the outer edge of a frictionless merry-go-round with the mass M = 300 kg and th

Evgen [1.6K]

Answer:

$\omega=0.24\ rad.s^{-1}$

Explanation:

Given:

mass of person, $m=36\ kg$

mass of merry go-round, $M=300\ kg$

radius of merry go-round, $R=2\ m$

velocity of the person running, $v=4\ m.s^{-1}$

We consider merry go-round as a ring:

Now the moment of inertial of the ring is given as,

$I=M.R^2$

$I=300\times 2^2$

$I=1200\ kg.m^{-2}$

Moment of inertia of the person considering as a point mass:

$I_p=m.R^2$

$I_p=36\times 2^2$

$I_p=144\ kg.m^2$

Now according to the conservation of angular momentum:

$I.\omega=I_p.\omega_p$

where:

$\omega =$ angular velocity of the merry-go-round

$\omega_p=$ angular velocity of the person running

$1200\times \omega=144\times \frac{v}{R}$

$\omega=\frac{144}{1200} \times \frac{4}{2}$

$\omega=0.24\ rad.s^{-1}$

4 0

3 years ago

Read 2 more answers