Se deja caer una caja de madera de 4.5kg de masa desde una altura de 2.25metros .

A spiral spring has a length of 14 cm when a force of 4 N is hung on it. A force of 6 N extends

stira [4]

Answer:

bussy

Explanation:

shart

5 0

2 years ago

Mike is standing on the roof of a building looking at the roof of the neighboring building that is 15 meters away and 10 meters

amid [387]

Answer:

Part a)

$t = 1.65 s$

Part b)

$x = 40.4 m$

Since the distance of other building is 15 m so YES it can make it to other building

Part c)

$v = 27.3 m/s$

direction of velocity is given as

$[tex]\theta = 26.35 degree$

Explanation:

Part a)

acceleration due to gravity on this planet is 3/4 times the gravity on earth

So the acceleration due to gravity on this new planet is given as

$a = \frac{3}{4}(9.81)$

$a = 7.36 m/s^2$

now the vertical displacement covered by the canister is given as

$y = 10 m$

now by kinematics we have

$y = \frac{1}{2}gt^2$

$10 = \frac{1}{2}(7.36)t^2$

$t = 1.65 s$

Part b)

Horizontal speed of the canister is given as

$v_x = 24.5 m/s$

now the distance moved by it

$x = v_x t$

$x = 24.5 (1.65)$

$x = 40.4 m$

Since the distance of other building is 15 m so YES it can make it to other building

Part c)

Final velocity in X direction will remains the same

$v_x = 24.5 m/s$

final velocity in Y direction

$v_y = v_i + at$

$v_y = 0 + (7.36)(1.65)$

$v_y = 12.14 m/s$

now magnitude of velocity is given as

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

$v = \sqrt{24.5^2 + 12.14^2}$

$v = 27.3 m/s$

direction of velocity is given as

$\theta = tan^{-1}\frac{v_y}{v_x}$

$\theta = tan^{-1}\frac{12.14}{24.5}$

$[tex]\theta = 26.35 degree$

6 0

3 years ago

A ball of moist clay falls 17.3 m to the ground. It is in contact with the ground for 24.0 ms before stopping. (a) What is the a

gizmo_the_mogwai [7]

Answer:

Acceleration, $767.08\ m/s^2$

Explanation:

Given that,

Height from a ball falls the ground, h = 17.3 m

It is in contact with the ground for 24.0 ms before stopping.

We need to find the average acceleration the ball during the time it is in contact with the ground.

Firstly, find the velocity when it reached the ground. So,

$v^2=u^2+2ah$

u = initial velocity=0 m/s

a = acceleration=g

$v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 17.3} \\\\v=18.41\ m/s$

It is in negative direction, u = -18.41 m/s

Let a is average acceleration of the ball. Consider, v = and u = -18.41 m/s.

$a=\dfrac{v-u}{t}\\\\a=\dfrac{0-18.41}{24\times 10^{-3}}\\\\a=767.08\ m/s^2$

So, the average acceleration of the ball during the time it is in contact is $767.08\ m/s^2$ .

4 0

3 years ago

The solution to a disjunction is the what of the two solutions

astra-53 [7]

Answer:

The solution set of a disjunction is the union of the solution sets of the individual inequalities. A convenient way to graph a disjunction is to graph each individual inequality above the number line, then move them both onto the actual number line

Explanation:

6 0

3 years ago

kristine speeds past a parked police car at 32 m/s. The police car starts from rest with a uniform acceleration of 2.5 m/s^2. Ho

Digiron [165]

$32 = 0 \times t + \frac{1}{2} \times 2.5 \times t^{2} \\ 32 = 0 + 1.25 \times t {}^{2} \\ 32 = 1.25t {}^{2} \\ \frac{32}{1.25} = \frac{1.25t {}^{2} }{1.25} \\ t {}^{2} = 25.6 \\ \sqrt{t {}^{2} } = \sqrt{25.6} \\ t = 5.1seconds \\$

7 0

3 years ago