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Vladimir79 [104]

2 years ago

14

Calculate the velocity of the object in the graph during the following time interval 0-6 seconds. Also, during which time interv

al is the object accelerating and how can you tell?

1 answer:

cestrela7 [59]2 years ago

3 0

Its accelarating from 1-10 i know that because the line is going up. i hope i helped

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If the net torque is zero, what does this imply about the clockwise and counterclockwise torques.

lapo4ka [179]

Answer:

Explanation:

When two forces acting on a line of action and they are equal in magnitude but opposite it direction, it forms a couple.

Torque is defined as the product of either force and the perpendicular distance between the two forces.

It is a vector quantity.

The net torque is zero, it means the anticlockwise torque is equal to the clockwise torque.

It means they balances each other.

6 0

3 years ago

Is it accurate to say that galaxies take up most of the volume of space? Why or why not?

RUDIKE [14]

Answer:

ntergalactic space takes up most of the volume of the universe, but even galaxies and star systems consist almost entirely of empty space.

Explanation:

so to be exact galixes do make up most of the volume of space because galaxies are what make up space!

4 0

1 year ago

Look at figure D . Explain what will happen if a larger mass it put on the force meter

aev [14]

it will experience great force

5 0

3 years ago

Playing near a road construction site, a child falls over a barrier and down onto a dirt slope that is angled downward at 33° to

Debora [2.8K]

Answer:

$\mu_{k} \approx 0.719$

Explanation:

The equations of equilibrium for the child are: (x' in the direction parallel to slope, y' in the direction perpendicular to slope)

$\Sigma F_{x'} = m\cdot g \cdot \sin \theta - \mu_{k}\cdot N = m\cdot a\\\Sigma F_{y'} = N - m\cdot g\cdot \cos \theta = 0$

After some algebraic manipulation, an expression for the coefficient of kinetic friction is obtained:

$m\cdot g\cdot \sin \theta - \mu_{k}\cdot m \cdot g \cos \theta = m \cdot a$

$g \cdot (\sin \theta - \mu_{k}\cdot \cos \theta) = a$

$\mu_{k}\cdot \cos \theta = \sin \theta - \frac{a}{g}$

$\mu_{k} = \frac{1}{\cos \theta}\cdot (\sin \theta - \frac{a}{g} )$

$\mu_{k} = \frac{1}{\cos 33^{\textdegree}}\cdot \left(\sin 33^{\textdegree}-\frac{(-0.57\,\frac{m}{s^{2}}) }{9.807\,\frac{m}{s^{2}} } \right)$

$\mu_{k} \approx 0.719$

5 0

3 years ago

Which instrument would play lower tones?

Elanso [62]

An upright base / 6ft long

8 0

3 years ago