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2 years ago

13. If you have a 12 kg object, how do you find the weight?

1 answer:

5 0

The weight is the force of gravity acting on the object. So assuming it’s on earth, 12(9.81) also, F(weight in Newtons)=mass( in kg)acceleration (due to gravity)

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Jan ran 4 miles north in 28 minutes. What was Jan's average velocity?

fenix001 [56]

Answer:

3.83 m/s

Explanation:

Given that,

Distance covered by Jan, d = 4 miles

1 mile = 1609.34 m

4 miles = 6437.38 m

Time, t = 28 minutes = 1680 s

Jan's average speed,

v = d/t

$v=\dfrac{6437.38\ \text{m}}{1680\ \text{s}}\\\\v=3.83\ \text{m/s}$

Hence, the average velocity of Jan is 3.83 m/s.

6 0

3 years ago

(PLEASE HELP FAST 20 POINTS)

Lera25 [3.4K]

Everything we see or do in everyday life that involves electricity in any way is the result of electrons moving from one place to another, or from one object to another. <em> (last choice)</em>

4 0

2 years ago

A 3-m-high, 7-m-wide rectangular gate is hinged at the top edge and is restrained by a fixed ridge. Determine the hydrostatic fo

Shalnov [3]

Answer:

The Hydrostatic force is $F = 137.2 kN$

The location of pressure center is $Z = 1.333 \ m$

Explanation:

From the question we are told that

The height of the gate is $h = 3 \ m$

The weight of the gate is $w = 7 \ m$

The height of the water is $h_w = 2 \ m$

The density of water is $\rho_w = 1000 \ kg/m^3$

Note used $h_w$ for height of water and height of gate immersed by water since both have the same value

The area of the gate immersed in water is mathematically represented as

$A = h_w * w$

substituting values

$A = 2* 7$

$A = 14 \ m^2$

The hydrostatic force is mathematically represented as

$F = \rho_w * g * h_f * A$

Where

$h_f =h- h_w$

$h_f =3 -2$

$h_f = 1\ m$

$F = 1000 * 9.8 * 1 * 14$

$F = 137.2 kN$

The center of pressure is mathematically represented as

$Z = h_f + \frac{I_g}{h_f * A}$

Where $I_g$ is the moment of inertia of the gate which mathematically represented as

$I_g = \frac{w * h_w^2}{12}$

The $h_w$ is the height of gate immersed in water

$I_g = \frac{7 * 2^2 }{12}$

$I_g = 4.667\ kg m^2$

Thus

$Z = 1 + \frac{4.66}{1 * 14}$

$Z = 1.333 \ m$

3 0

2 years ago

A sound source is moving at 80 m/s toward a stationary listener that is standing in still air (a) Find the wavelength of the sou

Setler [38]

Answer:

a. wavelength of the sound, $\vartheta = 1.315\vartheta_{o}$

b. observed frequecy, $\lambda = 0.7604\lambda_{o}$

Given:

speed of sound source, $v_{s}$ = 80 m/s

speed of sound in air or vacuum, $v_{a}$ = 343 m/s

speed of sound observed, $v_{o}$ = 0 m/s

Solution:

From the relation:

v = $\vartheta \lambda$ (1)

where

v = velocity of sound

$\vartheta$ = observed frequency of sound

$\lambda$ = wavelength

(a) The wavelength of the sound between source and the listener is given by:

$\lambda = \frac{v_{a}}{\vartheta }$ (2)

(b) The observed frequency is given by:

$\vartheta = \frac{v_{a}}{v_{a} - v_{s}}\vartheta_{o}$

$\vartheta = \frac{334}{334 - 80}\vartheta_{o}$

$\vartheta = 1.315\vartheta_{o}$ (3)

Using eqn (2) and (3):

$\lambda = \frac{334}{1.315} = \frac{1}{1.315}\frac{v_{a}}{\vartheta_{o}}$

$\lambda = 0.7604\lambda_{o}$

4 0

2 years ago

A steel ball rolls with a constant velocity on a tabletop 0.950 m high it rolls off and hit the ground 0.352 m from the edge of

sp2606 [1]

Answer:

0.799 m/s if air resistance is negligible.

Explanation:

For how long is the ball in the air?

Acceleration is constant. The change in the ball's height $\Delta h$ depends on the square of the time:

$\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t$ ,

where

$\Delta h$ is the change in the ball's height.
$g$ is the acceleration due to gravity.
$t$ is the time for which the ball is in the air.
$v_0$ is the initial vertical velocity of the ball.

The height of the ball decreases, so this value should be the opposite of the height of the table relative to the ground. $\Delta h = -0.950\;\text{m}$ .
Gravity pulls objects toward the earth, so $g$ is also negative. $g \approx -9.81\;\text{m}\cdot\text{s}^{-2}$ near the surface of the earth.
Assume that the table is flat. The vertical velocity of the ball will be zero until it falls off the edge. As a result, $v_0 = 0$ .

Solve for $t$ .

$\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t$ ;

$\displaystyle -0.950 = \frac{1}{2} \times (-9.81) \cdot t^{2}$ ;

$\displaystyle t^{2} =\frac{-0.950}{1/2 \times (-9.81)}$ ;

$t \approx 0.440315\;\text{s}$ .

What's the initial horizontal velocity of the ball?

Horizontal displacement of the ball: $\Delta x = 0.352\;\text{m}$ ;
Time taken: $\Delta t = 0.440315\;\text{s}$

Assume that air resistance is negligible. Only gravity is acting on the ball when it falls from the tabletop. The horizontal velocity of the ball will not change while the ball is in the air. In other words, the ball will move away from the table at the same speed at which it rolls towards the edge.

$\begin{aligned}\text{Rolling Velocity}&=\text{Horizontal Velocity} \\&= \text{Average Horizontal Velocity}\\ &=\frac{\Delta x}{\Delta t}=\frac{0.352\;\text{m}}{0.440315\;\text{s}}=0.0799\;\text{m}\cdot\text{s}^{-1}\end{aligned}$ .

Both values from the question come with 3 significant figures. Keep more significant figures than that during the calculation and round the final result to the same number of significant figures.

3 0

2 years ago