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denis-greek [22]

2 years ago

8

A crane moves a 250 kg scoreboard from the ground to the height of 100 m. What is the work done on the scoreboard?

2 answers:

Mazyrski [523]2 years ago

7 0

If you’re doing potential and kinetic energy then the answer is potential.

Sveta_85 [38]2 years ago

5 0

Answer:

You will have to do the last PE - the initial PE

Explanation:

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A person gives a shopping cart an initial push along a horizontal floor to get it moving, and then lets go. The cart travels for

mash [69]

Answer:

Only a backward force is acting, no forward force.

Explanation:

Once released from the initial push, in absence of friction, the shopping cart would continue moving forward at a constant speed forever.
As it would move at a constant speed, no net force would be acting on it.
So, if it is gradually slowing, there must be a net force producing an acceleration in a direction opposite to the movement.
This force is the kinetic friction force, and is the only force acting on the cart in the horizontal direction.
As any friction force, opposes to the relative movement between the cart and the horizontal floor, which means that is directed backward.
This is consistent with the direction of the acceleration of the cart.

8 0

3 years ago

1 A 75-g ball is projected from a height of 1.6 m with a horizontal velocity of 2 m/s and bounces from a 400-g smooth plate supp

Tanzania [10]

Answer with explanation:

We are given that

Mass of ball, $m_1=$ 75 g= $\frac{75}{1000}=0075kg$

1 kg=1000 g

Height, $h_1=1.6 m$

$h_2=0.6 m$

Horizontal velocity, $v_x=2 m/s$

Mass of plate $m_2=400 g=\frac{400}{1000}=0.4 kg$

a.Initial velocity of plate, $u_2=0$

Velocity before impact= $u_1=\sqrt{2gh_1}=\sqrt{2\times 9.8\times 1.6}=5.6m/s$

Where $g=9.8 m/s^2$

Velocity after impact, $v_1=\sqrt{2gh_2}=\sqrt{2\times 9.8\times 0.6}=3.4m/s$

According to law of conservation of momentum

$m_1u_1+m_2u_1=-m_1v_1+m_2v_2$

Substitute the values

$0.075\times 5.6+0=-0.075\times 3.4+0.4v_2$

$0.4v_2=0.075\times 5.6+0.075\times 3.4$

$v_2=\frac{0.075\times 5.6+0.075\times 3.4}{0.4}=1.69 m/s$

Velocity of plate=1.69 m/s

b.Initial energy= $\frac{1}{2}m_1v^2_x+m_1gh_1=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 1.6=1.326 J$

Final energy= $\frac{1}{2}m_1v^2_x+m_1gh_2+\frac{1}{2}m_2v^2_2$

Final energy= $\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 0.6+\frac{1}{2}(0.4)(1.69)^2=1.162 J$

Energy lost due to compact=Initial energy-final energy=1.326-1.162=0.164 J

6 0

2 years ago

What are some types of landforms on Earth’s surface?<br><br><br><br> PLS ANSWER QUICK 11 POINTS

Brrunno [24]

Answer:

plateau, mountains, hills, plains

5 0

2 years ago

Read 2 more answers

Three science questions help please!

Nimfa-mama [501]

6. D
7. D
8. B
let me know if you need clarification

4 0

3 years ago

Enunciado: Una bola se lanza verticalmente de la parte superior de un edificio con una velocidad inicial de 25 m/s. La bola impa

bearhunter [10]

Responder:

Explicación:

Usaremos la ecuación de movimiento para determinar la altura de la bola medida desde la parte superior del edificio.

Usando la ecuación para obtener la altura de caída

S = ut + 1 / 2gt²

u es la velocidad inicial = 25 m / s

g es la aceleración debida a la gravedad = 9,81 m / s²

t es el tiempo = 7 segundos

S es la altura de la caída

S = 25 (7) +1/2 (9,81) × 7²

S = 175 + 4,905 (49)

S = 175 + 240,345

S = 415,35 m

Esto significa que la pelota se elevó a 415,35 m de altura

7 0

3 years ago