The thinnest layer of the earth is the:

A 242-g block is pressed against a spring of force constant 1.62 kN/m until the block compresses the spring 10.0 cm. The spring

hammer [34]

Answer:

a) = 3.94 m

b) = 3.15 m

Explanation:

Given

Mass of the block, m = 242 g

Force constant, k = 1.62 kN/m

Compression of the spring, x = 10 cm

Angle of inclination = 60°

a) if we equate the energy at the bottom of the ramp to the energy at a distance d up the ramp, we have

1/2kx² = mgh where, h = dsinΦ

1/2kx² = mgdsinΦ

1/2 * 1.62*10^3 * 0.1² = 0.242 * 9.8 * dsin 60

1/2 * 16.2 = 2.3716 * d sin 60

d sin 60 = 8.1 / 2.3716

0.866 d = 3.415

d = 3.415 / 0.866

d = 3.94 m

b) net force on the block = mgd sin 60 + µ mgd cos 60

8.1 = d[mg sin 60 + µ mg cos 60]

8.1 = d [0.242 * 9.8 * 0.866 + 0.44 * 0.242 * 9.8 * 0.5]

8.1 = d (2.05 + 0.52)

8.1 = 2.57 d

d = 8.1 / 2.57

d = 3.15 m

3 0

3 years ago

Read 2 more answers

One method of determining the location of the center of gravity of a person is to weigh the person as he/she lies on a board of

andrey2020 [161]

Answer:

x = 1.018 m

Explanation:

given,

height of man = 190 cm

= 1.9 m

scale reading on left = 450 N

scale reading on the right = 390 N

Let center of gravity of man be x distance from feet, feet is on right side.

For system to be in equilibrium moment about center should be equal to zero.

∑M = 0

now,

450(1.9 - x ) - 390 × x = 0

450(1.9 - x ) = 390 × x

855 - 450 x = 390 x

840 x = 855

$x = \dfrac{855}{840}$

x = 1.018 m

hence, point of center of gravity from feet is equal to x = 1.018 m

7 0

4 years ago

#26: Why is spear fishing much more successful for scuba divers than someone spear fishing from above the water? Explain your re

frozen [14]

The water has an irregular surface so if the light rays fall on top then it will reflect oppositely or in some cases may refract, so the fish will seem a bit smaller or further or closer in anyway, example if u put a straw in water it looks bent but is it ? nope. So the main reason is that it refracts which makes the accuracy lesser which make it difficult.

6 0

3 years ago

A cannonball is catapulted toward a castle. The cannonball's velocity when it leaves the catapult is 40 m/s at an angle of 37° w

sleet_krkn [62]

Answer:

a) Maximum height = 36.6 m

b) Horizontal distance at which the ball lands = 166.1 m

c) x-component = 32 m/s. y-component = - 27 m/s

Explanation:

Please, see the attached figure for a description of the problem.

The velocity vector "v" of the cannonball has two components, a horizontal component, "vx", and a vertical component "vy". Notice that at the maximum height, the vertical component "vy" of the velocity vector is 0.

In the same way, the position vector "r" is composed by "rx", its horizontal component, and "ry", the vertical component.

The velocity vector "v" ad the position vector "r" at time "t" are given by the following equations:

v = (v0 * cos α, v0 * sin α + g * t)

r = (x0 + v0 * t * cos α, y0 + v0 * t * sin α + 1/2 * g * t²)

Where

v0 = magnitude of the initial velocity vector

α = launching angle

g = gravity acceleration (-9.8 m/s², because the y-axis points up)

t = time

x0 = initial horizontal position

y0 = initial vertical position

If we consider the origin of the system of reference as the point at which the cannonball leaves tha catapult, then, x0 and y0 = 0

a) We know that at maximum height, the vertical component of the vector "v" is 0, because the ball does not move up nor down at that moment (see figure). Then:

0 = v0 * sin α + g * t

-v0 * sin α / g = t

-40 m/s * sin 37° / -9.8 m/s² = t

t = 2.5 s

We can now calculate the position of the cannonball at time t=2.5 s to obtain the maximum height:

r = (x0 + v0 * t cos α, y0 + v0 * t * sin α + 1/2 * g * t²)

The max height is the magnitude of the vector ry max (see figure). The vector ry max is:

ry = (0, y0 + v0 t sin α + 1/2 g * t²)

magnitude of ry = $|ry|= \sqrt{(0m)^{2} + (y0 + v0* t*sin \alpha+ 1/2*g*t^{2})^{2}}= y0 + v0*t*sin \alpha + 1/2*g*t^{2}$ )

Then:

max height = y0 + v0 * t * sin α + 1/2 * g * t²

max height = 0 m + 40 m/s * 2.5 s * sin 37° - 1/2* 9.8 m/s² * (2.5 s)² = 29.6 m

Since the ball leaves the catapult 7 m above the ground, the max height above the ground will be 29.6 m + 7 m = 36.6m

max height = 36.6 m

b) When the ball hits the ground, the position is given by the vector "r final" (see figure). The magnitude of "rx", the horizontal component of "r final", is the horizontal distance between the catapult and the wall.

r final = ( x0 + v0 * t * cos α, y0 + v0 * t * sin α + 1/2 * g * t²)

We know that the vertical component of "r final" is -7 (see figure).

Then, we can obtain the time when the the ball hits the ground:

y0 + v0 * t * sin α + 1/2 * g * t² = -7 m

0 m + 40 m/s * t * sin 37° + 1/2 g * t² = -7 m

7 m + 40 m/s * t * sin 37° + 1/2 (-9.8 m/s²) * t² = 0

7 m + 24.1 m/s * t - 4.9 m/s² * t² = 0

solving the quadratic equation:

t = 5.2 s (The negative solution is discarded).

With this time, we can calculate the value of the horizontal component of "r final"

Distance to the wall = |rx| = x0 + v0 t cos α

|rx| = 0m + 40 m/s * 5.2 s * cos 37° = 166.1 m

c) With the final time obtained in b) we can calculate the velocity of the ball:

v = (v0 * cos α, v0 * sin α + g * t)

v =(40 m/s * cos 37°, 40 m/s * sin 37° -9.8 m/s² * 5.2 s)

v =(32 m/s, -27 m)

x-component = 32 m/s

y-component = - 27 m/s

7 0

4 years ago

Small blocks A and B are held at rest on a smooth plane inclined at 30° to the horizontal. Each is held in equilibrium by a forc

Olin [163]

<h3>Answer :</h3>

First of all, See the attachment for better understanding $.$

_________________________________

❂ Weight of block A :

➝ mg sin30° = 18

➝ W (1/2) = 18

➝ W = 18×2

➝ W = 36N 

❂ Weight of block B :

➝ N sin30° = 18

➝ (Mg cos30°) sin30° = 18

➝ W' (√3/2)(1/2) = 18

➝ W' (√3/4) = 18

➝ W' = 72/√3

➝ W' = 41.61N

5 0

3 years ago