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2 years ago

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

Physics

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 baseball approaches home plate at a speed of 44.0 m/s, moving horizontally just before being hit by a bat. The batter hits a p

Luda [366]

Explanation:

It is given that,

Speed of the baseball, u = 44 m/s

Speed of the baseball, v = 53 m/s

Mass of the ball, m = 145 g = 0.145 kg

Time of contact between the ball and the bat, t = 2.2 ms = 0.0022 s

$F=ma$

$F=\dfrac{mv}{t}$

$F_1=\dfrac{0.145\ kg\times 44\ m/s}{0.0022\ s}$

F₁ = 2900 N...........(1)

$F=ma$

$F=\dfrac{mv}{t}$

$F_2=\dfrac{0.145\ kg\times 53\ m/s}{0.0022\ s}$

F₂ = 3493.18 N.........(2)

In average vector form force is given by :

$F=F_1+F_2$

$F=(2900i+(-3493.18)\ N$

$F=(2900i-3493.18j)\ N$

Hence, this is the required solution.

6 0

3 years ago

Before the cells in a person's body can use the food that the person eats, the food must be A. chemically broken down into sugar

larisa86 [58]

A chemically broken down into sugar

6 0

3 years ago

Read 2 more answers

Your starship, the Aimless Wanderer,lands on the mysterious planet Mongo. As chief scientist-engineer,you make the following mea

Setler [38]

Answer:

a) M = 4,997 10²⁰ kg , b) T = 1.43 10³ s

Explanation:

a) This exercise should be solved in several parts, let's start by calculating the acceleration of gravity of this planet from kinematics

v = v₀ - a t

As it indicates that there is no atmosphere, the friction force is zero and the initial and final velocity have the same module, but the opposite direction

a = (v₀ - v) / t

a = (15 - (-15)) /9.00 = 30/9

a = 3.33 m / s²

Now we use Newton's second law where force is the force of universal attraction

F = m a

G m M / r² = m a

M = a r² / G

Let's calculate

M = 3.33 (1.00 10⁵)² / 6.67 10⁻¹¹

M = 4,997 10²⁰ kg

b) The period of the ship's orbit

In this case we have a centripetal acceleration

The radius of the orbit is the radius of the plant plus the height of the ship from the surface

R = $R_{m}$ + h

R = 1 10⁵ + 2.00 10⁴

R = 12 10⁴ m

F = m a

G m M / R² = m a

Centripetal acceleration is

a = v² / R

The orbit is circular therefore the velocity module is constant, so we can use the equation of uniform motion, where the distance is the length of the orbit, for a circle

d = 2π R

v = d / t

v = 2π R / T

Let's replace

G m M / R² = m (2π R / T)² / R

G M = R³ 4π² / T²

T² = 4π² R³ / G M

T² = (4π² (12 10⁴)³ / (6.67 10⁻¹¹ 4,997 10²⁰)

T² = 6.82 10¹⁶ / 3.33 10¹⁰

T = √ (2,048 10⁶)

T = 1.43 10³ s

3 0