In which scenario is the least amount of work done on a crate?

a volleyball is hit upward with an initial velocity of 7.5 m/s. calculate the displacement of the volleyball when its final velo

Luden [163]

Answer:

The displacement of the volleyball is 2.62 m

Explanation:

Given;

initial velocity of the volleyball, u = 7.5 m/s

final velocity of the volleyball, v = 2.2 m/s

displacement of the volleyball, d = ?

Apply the following kinematic equation;

v² = u² - 2gd

2gd = u² - v²

$d = \frac{u^{2}-v^{2} }{2g}\\\\d = \frac{7.5^{2}-2.2^{2} }{2*9.8}\\\\d = 2.62 \ m$

Therefore, the displacement of the volleyball is 2.62 m

7 0

3 years ago

A house brick has a volume of 1900 cm³ and a weight in air of 80N.What is its apparent weight in water?The density of water is 1

Shtirlitz [24]

Answer:

61 N

Explanation:

We'll begin by calculating the mass of the brick when placed in water. This can be obtained as follow:

Volume of brick = 1900 cm³

Density of water = 1 g/cm³

Mass of brick in water =…?

Density = mass / volume

1 = mass of brick in water / 1900

Cross multiply

Mass of brick in water = 1 × 1900

Mass of brick in water = 1900 g

Next, we shall convert 1900 g to Kg.

1000 g = 1 Kg

Therefore,

1900 g = 1900 g × 1 Kg / 1000 g

1900 g = 1.9 Kg

Next, we shall determine the weight in water. This can be obtained as follow:

Mass (m) = 1.9 Kg

Acceleration due to gravity (g) = 10 m/s²

Weight (W) =?

W = m × g

W = 1.9 × 10

W = 19 N

Thus, the weight of the brick in water is 19 N.

Finally, we shall determine the apparent weight of the brick in water. This can be obtained as follow:

Weight in air = 80 N

Weight in water = 19 N

Apparent weight =?

Apparent weight = weight in air – weight in water

Apparent weight = 80 – 19

Apparent weight = 61 N

3 0

3 years ago

Assume the Earth is a ball of perimeter 40, 000 kilometers. There is a building 20 meters tall at point a. A robot with a camera

torisob [31]

Answer:

Approximately 21 km.

Explanation:

Refer to the not-to-scale diagram attached. The circle is the cross-section of the sphere that goes through the center C. Draw a line that connects the top of the building (point B) and the camera on the robot (point D.) Consider: at how many points might the line intersects the outer rim of this circle? There are three possible cases:

No intersection: There's nothing that blocks the camera's view of the top of the building.
Two intersections: The planet blocks the camera's view of the top of the building.
One intersection: The point at which the top of the building appears or disappears.

There's only one such line that goes through the top of the building and intersects the outer rim of the circle only once. That line is a tangent to this circle. In other words, it is perpendicular to the radius of the circle at the point A where it touches the circle.

The camera needs to be on this tangent line when the building starts to disappear. To find the length of the arc that the robot has travelled, start by finding the angle $\angle \mathrm{B\hat{C}D}$ which corresponds to this minor arc.

This angle comes can be split into two parts:

$\angle \mathrm{B\hat{C}D} = \angle \mathrm{B\hat{C}A} + \angle \mathrm{A\hat{C}D}$ .

Also,

$\angle \mathrm{B\hat{A}C} = \angle \mathrm{D\hat{A}C} = 90^{\circ}$ .

The radius of this circle is:

$\displaystyle r = \frac{c}{2\pi} = \rm \frac{4\times 10^{7}\; m}{2\pi}$ .

The lengths of segment DC, AC, BC can all be found:

$\rm DC = \rm \left(1.75 \displaystyle + \frac{4\times 10^{7}\; m}{2\pi}\right)\; m$ ;
$\rm AC = \rm \displaystyle \frac{4\times 10^{7}}{2\pi}\; m$ ;
$\rm BC = \rm \left(20\; m\displaystyle +\frac{4\times 10^{7}}{2\pi} \right)\; m$ .

In the two right triangles $\triangle\mathrm{DAC}$ and $\triangle \rm BAC$ , the value of $\angle \mathrm{B\hat{C}A}$ and $\angle \mathrm{A\hat{C}D}$ can be found using the inverse cosine function:

$\displaystyle \angle \mathrm{B\hat{C}A} = \cos^{-1}{\rm \frac{AC}{BC}}$

$\displaystyle \angle \mathrm{D\hat{C}A} = \cos^{-1}{\rm \frac{AC}{DC}}$

$\displaystyle \angle \mathrm{B\hat{C}D} = \cos^{-1}{\rm \frac{AC}{BC}} + \cos^{-1}{\rm \frac{AC}{DC}}$ .

The length of the minor arc will be:

$\displaystyle r \theta = \frac{4\times 10^{7}\; \rm m}{2\pi} \cdot (\cos^{-1}{\rm \frac{AC}{BC}} + \cos^{-1}{\rm \frac{AC}{DC}}) \approx 20667 \; m \approx 21 \; km$ .