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2 years ago

. In any energy transformation, energy is _____. destroyed created conserved

Physics

2 answers:

zubka84 [21]2 years ago

8 0

Answer: i believe it is conserved or stays the same.

Explanation: energy cant be destroyed no matter what and no energy is being created

I hope this helps a thank and a brainlist would be greatly appreciated

Mrac [35]2 years ago

6 0

Answer: Nor

Explanation: In any energy transformation, energy is nor created destroyed or conserved

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A projectile is fired vertically from Earth's surface with an initial speed v0. Neglecting air drag, how far above the surface o

lidiya [134]

This may helpv^2=u^2+2as. v=0 at top of flight. a=acceleration of gravity(vo^2)/2a=s.

3 0

3 years ago

An object is pushed from rest across a sheet of ice, accelerating at 8.0 m/s^2 [E] over a displacement of 1.05 m [E]. The object

tatiyna

Answer:

$D_T=18.567m$

Explanation:

From the question we are told that:

Acceleration $a=8.0 m/s^2$

Displacement $d=1.05 m$

Initial time $t_1=6.0s$

Final Time $t_2=2.5s$

Generally the equation for Velocity of 1.05 travel is mathematically given by

Using Newton's Law of Motion

$V^2=2as$

$V=\sqrt{2*6*1.05}$

$V=4.1m/s$

Generally the equation for Distance traveled before stop is mathematically given by

$d_2=v*t_1$

$d_2=3.098*4$

$d_2=12.392$

Generally the equation for Distance to stop is mathematically given by

Since For this Final section

Final velocity $v_3=0 m/s$

Initial velocity $u_3=4.1 m/s$

Therefore

Using Newton's Law of Motion

$-a_3=(4.1)/(2.5)$

$-a_3=1.64m/s^2$

Giving

$v_3^2=u^2-2ad_3$

Therefore

$d_3=\frac{u_3^2}{2ad_3}$

$d_3=\frac{4.1^2}{2*1.64}$

$d_3=5.125m$

Generally the Total Distance Traveled is mathematically given by

$D_T=d_1+d_2+d_3$

$D_T=5.125m+12.392+1.05 m$

$D_T=18.567m$

6 0

2 years ago

30 PTS

tigry1 [53]

Spreads out in the medium, first choice

7 0

3 years ago

Two cars are heading towards one another. Car A is moving with an acceleration of aA = 4 m/s2. Car B is moving with an accelerat

Paladinen [302]

Answer:

Car B reaches car A in 19.7 s.

Explanation:

Hi there!

The equation of the position of an object moving in a straight line at constant acceleration is as follows:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the object at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration

When both cars meet, their positions are the same. At the meeting point:

position of car A = position of car B

xA = xB

x0A + v0A · t + 1/2 · aA · t² = x0B + v0B · t + 1/2 · aB · t²

Let´s place the origin of the frame of reference at the point where A is located. In that case x0A = 0 and x0B = 2900 m. Since both cars are initially at rest, v0A and v0B = 0. So, the equation gets reduced to this:

1/2 · aA · t² = x0B + 1/2 · aB · t²

If we replace with the data we have and solve for t:

1/2 · 4 m/s² · t² = 2900 m - 1/2 · 11 m/s² · t²

2 m/s² · t² = 2900 m - 5.5 m/s² · t²

5.5 m/s² · t² + 2 m/s² · t² = 2900 m

7.5 m/s² · t² = 2900 m

t² = 2900 m / 7.5 m/s²

t = 19.7 s

Car B reaches car A in 19.7 s.

4 0

3 years ago

if a person with a mass of 70kg is standing on scale in an elevator when the cable snaps, what will the reading on the scale be

dybincka [34]

Answer:

The reading will be the same.

Explanation:

Mass does not depend upon anything and it remains the same anywhere. What changes is the weight of the body because it depends upon gravity and is different at different places.

Giving me the brainest will be helpful.

3 0

2 years ago