What is 1 abiotic factors shown in this diagram?

A 0.245 kg ball is thrown straight up from 2.07 m above the ground. Its initial vertical speed is 8.00 m/s. A short time later,

iris [78.8K]

Answer:

The work done by gravity is $4.975 \: Joules$

Explanation:

The data given in the question is :

Mass is $0.245 kg$

Height from ground is $2.07 m$

As we know , the work done is state function , it depends on initial and final position not on the path followed.

So, work done by gravity = change in potential energy

Work done = Initial potential energy - final potential energy

Insert values from question

Work done = $mass \times gravity \times (change \: in \: height)$

Work done = $0.245 kg \times 9.81 m/s^{2} \times 2.07 m$

So, work done = $4.975 Joules$

Hence the work done by gravity is $4.975 \: Joules$

3 0

3 years ago

B.A certain piece of copper wire is determined to have a mass of 2.00 g per meter.

masha68 [24]

Answer:

14cm

Explanation:

Mass per gram of the piece of wire;

2g of the wire is found in 1m

Since

100cm = 1m;

So;

100cm of the wire contains 2g of the wire

To provide 0.28g

Since;

2g of wire is made up of 100cm

0.28g of wire will be contained in $\frac{100 x 0.28}{2}$ = 14cm

14cm of the wire will contain 0.28g

8 0

2 years ago

A cube, whose edges are aligned with the , and axes, has a side length . The field is immersed in an electric field aligned with

Ivenika [448]

Complete Question

Complete Question is attached below

Answer:

$q=1.558*10^{-9}c$

Explanation:

From the question we are told that:

Side length s=1.13m

Left field strength $E_l=784.75N/m$

Right field strength $E_r=776.38 N/m$

Front field strength $E_f=725.5 N/m$

Back field strength $E_b=749.54 N/m$

Top field strength $E_t=944.95 N/m$

Bottom field strength $E_{bo}=1082.58 N/m$

Generally, the equation for Charge flux is mathematically given by

$\phi=EAcos\theta$

Where

Theta for Right,Left,Front and Back are at an angle 90

$cos 90=0$

Therefore

$\phi =0$ with respect to Right,Left,Front and Back

Generally, the equation for Charge Flux is mathematically also given by

$\phi=\frac{q}{e_o}$

Where

$Area =L*B\\\\A=1.13*1.13\\\\A=1.2769m^2$

Therefore

$Q_{net}=E_{bo}Acos\theta_{bo} +E_tAcos\theta_t$

$Q_{net}=1082.85*1.2769*cos0=944.95*1.2769cos (180)$

$Q_{net}=176N/C m^2$

Giving

$q=\phi*e_0$

$q=176N/C m^2*1.558*10^{-12}c$

$q=1.558*10^{-9}c$

4 0

2 years ago

1.

8_murik_8 [283]

Answer: 2.61 s

Explanation:

We are given the following data:

$s=5 ft$ is the initial height of the object

$v=40 ft/s$ is the initial height of the object

In addition, the motion of the object is given by:

$h=-16t^{2}+ vt +s$ (1)

Where:

$h=0 ft$ is the final height of the object

$t$ is the time the object is in the air before hitting the ground

Rewritting (1) with the given data:

$0=-16t^{2}+ 40t +5$ (2)

Solving the quadratic equation with the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ , where $a=-16$ , $b=40$ , $c=5$ and choosing the positive value of time:

$t=\frac{-40\pm\sqrt{(-40)^{2}-4(-16)(5)}}{2(-16)}$ (3)

$t=2.61 s$ (4) This is the time

5 0

3 years ago

g A solid disk rotates in the horizontal plane at an angular velocity of rad/s with respect to an axis perpendicular to the disk

maksim [4K]

Complete question:

A solid disk rotates in the horizontal plane at an angular velocity of 0.067 rad/s with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is 0.18 kg.m2. From above, sand is dropped straight down onto this rotating disk, so that a thin uniform ring of sand is formed at a distance of 0.40 m from the axis. The sand in the ring has a mass of 0.50 kg. After all the sand is in place, what is the angular velocity of the disk?

Answer:

The angular velocity of the disk is 0.0464 rad/s

Explanation:

Given;

initial angular velocity of disk, ωi = 0.067 rad/s

initial moment of inertia of the disk, I₁ = 0.18 kg.m²

radius of sand on the disk, R = 0.40 m

mass of sand, m = 0.50 kg

Initial angular momentum = Final angular momentum

$I_i \omega_i = I_f \omega_f\\\\I_i \omega_i = (I_{sand} + I_{disk})\omega _f$

Moment of inertia of sand ring = MR²

$\omega_f = \frac{I_i\omega_i}{I_{sand} +I_{disk}} = \frac{0.18*0.067}{MR^2 +0.18} \\\\\omega_f = \frac{0.18*0.067}{0.5*0.4^2 +0.18}= 0.0464 \ rad/s$

Therefore, the angular velocity of the disk is 0.0464 rad/s

8 0

3 years ago