All categories

3 years ago

True or false if an object has energy it must be moving

2 answers:

8 0

False. Non moving objects can have energy in the form of gravitational potential energy, elastic potential energy, thermal energy, internal energy, nuclear energy, or chemical energy.

Drupady [299]3 years ago

3 0

Answer:

Explanation:

If an object is moving, it is said to have kinetic energy (KE). Potential energy (PE) is energy that is "stored" because of the position and/or arrangement of the object. The classic example of potential energy is to pick up a brick. When it's on the ground, the brick had a certain amount of energy.

You might be interested in

Describe two ways unbalanced forces help you in your day to<br> day life.

Alexxandr [17]

Answer:

we need unbalance force to lift objects
we need unbalance force to drag objects

6 0

3 years ago

A 2,000 g quantity of C-14 is left to undergo radioactive decay. The half-life of C-14 is approximately 5,700 years. After going

Katyanochek1 [597]

Start with 2,000 grams.
After 1 half-life, 1,000 grams are left.
After another half-life, 500 are left.
After another half-life, 250 are left.
After another half-life, 125 are left.

That was FOUR half-lifes.

X = 4 .

8 0

4 years ago

Read 2 more answers

A certain quantity of steam has a temperature of 100.0 oC. To convert this steam into ice at 0.0 oC, energy in the form of heat

KonstantinChe [14]

Answer:

2452.79432 m/s

Explanation:

m = Mass of ice

$L_s$ = Latent heat of steam

$s_w$ = Specific heat of water

$L_i$ = Latent heat of ice

v = Velocity of ice

$\Delta T$ = Change in temperature

Amount of heat required for steam

$Q_1=mL_s\\\Rightarrow Q_1=m(2.256\times 10^6)$

Heat released from water at 100 °C

$Q_2=ms_w\Delta T\\\Rightarrow Q_2=m4186\times (100-0)\\\Rightarrow Q_2=m0.4186\times 10^6$

Heat released from water at 0 °C

$Q_3=mL_i\\\Rightarrow Q_3=m(333.5\times 10^3)\\\Rightarrow Q_3=m(0.3335\times 10^6)$

Total heat released is

$Q=Q_1+Q_2+Q_3\\\Rightarrow Q=m(2.256\times 10^6)+m0.4186\times 10^6+m(0.3335\times 10^6)\\\Rightarrow Q=3008100m$

The kinetic energy of the bullet will balance the heat

$K=Q\\\Rightarrow \frac{1}{2}mv^2=3008100m\\\Rightarrow v=\sqrt{2\times 3008100}\\\Rightarrow v=2452.79432\ m/s$

The velocity of the ice would be 2452.79432 m/s

6 0

4 years ago

I need help on number 5. We need to use one of the four kinematics equations.

Gnom [1K]

ANSWER

$35.02m$

EXPLANATION

Parameters given:

Initial velocity, u = 26.2 m/s

When the vase reaches its maximum height, its velocity becomes 0 m/s. That is the final velocity.

We can now apply one of Newton's equations of motion to find the height:

$v^2=u^2-2as$

where a = g = acceleration due to gravity = 9.8 m/s²

Therefore, we have that:

$\begin{gathered} 0=26.2^2-2(9.8)s \\ \Rightarrow19.6s=686.44 \\ s=\frac{686.44}{19.6} \\ s=35.02m \end{gathered}$

That is the height that the vase will reach.

5 0

1 year ago

1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n

sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

$(1).\: x =v_0t$

$(2).\: y= 15m-\dfrac{1}{2}gt^2$

where $v_0$ is the initial velocity.

(a).

When the projectile hits the 50m mark, $y=0$ ; therefore,

$0=15-\dfrac{1}{2}gt^2$

solving for $t$ we get:

$t= 1.75s.$

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

$50m = v_0(1.75s)$

which gives

$\boxed{v_0 = 28.58m/s.}$

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

$v_x = 28.58m/s.$

the vertical component of the velocity is

$v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.$

which gives a speed $v$ of

$v = \sqrt{v_x^2+v_y^2}$

$\boxed{v =33.3m/s.}$

4 0

3 years ago