A change in the size or shape of a substance is a ___________ change. physical, chemical, or nuclear?

What electrons are in “motion”, what do you have?

Sergeeva-Olga [200]

Answer:

When a positive charged object is placed near a conductor electrons are attracted the the object. ... When electric voltage is applied, an electric field within the metal triggers the movement of the electrons, making them shift from one end to another end of the conductor. Electrons will move toward the positive side. As you know, electrons are always moving. They spin very quickly around the nucleus of an atom. As the electrons zip around, they can move in any direction, as long as they stay in their shell.

7 0

4 years ago

You throw a ball upward from ground level with initial upward speed v0. What is the max height of the trajectory?

Inga [223]

Answer:

The max height of the ball is y = -1/2 (v0²/g).

It takes the ball t = -2 · v0/g to hit the ground.

The speed of the ball when it hits the ground is v = -v0.

Explanation:

The height and velocity of the ball is given by the following equations:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the ball at time t

y0 = initial height

v0 = initial velocity

t = time

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity at time t

When the ball is at max height, the velocity is 0. So, let´s find the time at which the velocity of the ball is 0.

v = v0 + g · t

0 = v0 + g · t

t = -v0/g

Now, replacing t = -v0/g in the equation of height, we will obtain the maximum height:

y = y0 + v0 · t + 1/2 · g · t² (y0 = 0 because the origin of the frame of reference is located on the ground)

y = v0 · t + 1/2 · g · t²

Replacing t:

y = v0 · (-v0/g) + 1/2 · g · (-v0/g)²

y = -(v0²/g) + 1/2 · (v0²/g)

y = -1/2 (v0²/g)

The max height of the ball is y = -1/2 (v0²/g). Remember that g is negative.

Since the acceleration of the ball is always the same, the time it takes the ball to impact the ground will be twice the time it takes to reach its max height, t = -2 v0/g.

However, let´s calculate that time knowing that at that time the height is 0:

y = y0 + v0 · t + 1/2 · g · t²

0 = v0 · t + 1/2 · g · t²

0 = t · ( v0 + 1/2 · g · t)

0 = v0 + 1/2 · g · t

-2 · v0/g = t

It takes the ball t = -2 · v0/g to hit the ground.

Let´s use the equation of velocity at final time (t = -2 · v0/g):

v = v0 + g · t

v = v0 + g · ( -2 · v0/g)

v = v0 - 2· v0

v = -v0

The speed of the ball when it hits the ground is v = -v0.

7 0

4 years ago

Can carrying an object be said to be doing work on that object?

Schach [20]

Answer:

Only if that object is being moved while carried

Explanation:

Work = Force • distance travelled

While holding an object would need force equal to that objects mass multiplied by 9.8, if it hasn't moved, then no work is done.

if you were to carry the object and have it travel some distance, then work would be done on that object.

7 0

3 years ago

Read 2 more answers

If mass 1 is 2.0 m to the right of the fulcrum and weighs 4.0 kg, and mass 2 weighs 6.0 kg, what distance will mass 2 need to be

STALIN [3.7K]

Explanation:

C 1.33 m to the left

.........................

8 0

3 years ago

PLS I NEED HELP ASAAP

Lerok [7]

Answer:

Option A

Explanation:

The Equation represents the displacement of the object which is represented by x

$x=t-t^2$

so, $x_0$ means when time is zero so we replace t with zero in the equation,

$x_0=(0)-(0)^2\\x_0=0$

now for v which is velocity we need to differentiate the function as the formula for velocity is rate of change of displacement over time so we derivate the equation once and get,

$v=1-2t\\$

now for $v_0$ we insert t = 0 and get

$v_0=1-2(0)\\v_0=1$

now for a which is acceleration the formula of acceleration is rate of change of velocity over time, so we differentiate the the equation of v(velocity) once or the equation of x(displacement) twice so now we get,

$a=-2$

so Option A is your answer.

Remember derivative of a constant is always zero because a constant value has no rate of change has its a constant hence the derivative is 0

5 0

3 years ago