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2 years ago

14

Landon attends an early childhood program that is located at a community center which also runs an adult care program.

1 answer:

Ira Lisetskai [31]2 years ago

3 0

Good for Landon. What’s the question?

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How do paved parking lots and roads with concrete or asphalt affect water flow on the land?

harina [27]

On the regular ground, water usually seeps in to the soil, which has a number of benefits for the wildlife in that area. However, with asphalt and concrete, there is no soft and absorbent soil to take in the water, so it just keeps flowing down to the lowest part of land it can. Additionally, some water runoff can carry fertilizer and other harmful chemicals with it in to the oceans and lakes it’s dumped in, which harms the ecosystem in them as well.

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3 years ago

Bill Nye help.. I didn’t get link to vid, hoping y’all have seen it. 20 POINTS!

Lyrx [107]

Answer:

Potential, Kinetic and Chemical energy.

Explanation:

btw, congratulations for turning into an expert.

6 0

3 years ago

I need help on 2/3 please asap thx ♥️

ch4aika [34]

<u>questions 2</u>

F=m

therefore

a=F/m

a=408/68

=6m/s^2.

5 0

3 years ago

A 2.1 ✕ 103-kg car starts from rest at the top of a 5.9-m-long driveway that is inclined at 19° with the horizontal. If an avera

Lina20 [59]

Answer:

3.9 m/s

Explanation:

We are given that

Mass of car,m= $2.1\times 10^3 kg$

Initial velocity,u=0

Distance,s=5.9 m

$\theta=19^{\circ}$

Average friction force,f= $4.0\times 10^3 N$

We have to find the speed of the car at the bottom of the driveway.

Net force, $F_{net}=mgsin\theta-f=2.1\times 10^3\times 9.8sin19-4.0\times 10^3$

Where $g=9.8 m/s^2$

Acceleration, $a=\frac{F_{net}}{m}=\frac{2.1\times 10^3\times 9.8sin19-4.0\times 10^3}{2.1\times 10^3}$

$v=\sqrt{2as}$

$v=\sqrt{2\times \frac{2.1\times 10^3\times 9.8sin19-4.0\times 10^3}{2.1\times 10^3}\times 5.9}$

v=3.9 m/s

7 0

3 years ago

Particle 1 and particle 2 have masses of m1 = 2.2 × 10-8 kg and m2 = 4.8 × 10-8 kg, but they carry the same charge q. The two pa

Lorico [155]

Answer: $r_2=11.81 cm$

Explanation:

Given

$m_1=2.2\times 10^{-8} kg$

$m_2=4.8\times 10^{-8} kg$

same charge on both masses

potential Energy due to Magnetic Field =Kinetic Energy of Particle

$qV=\frac{mv^2}{2}$

$v=\sqrt{\frac{2qV}{m}}$

and we know

Force due to magnetic field will Provide centripetal Force

$qvB=\frac{mv^2}{r}$

$B=\frac{\sqrt{\frac{2Vm}{q}}}{r}$

and B is equal for both particles

thus $\frac{m}{r^2}=constant$

$\frac{m_1}{r_1^2}=\frac{m_2}{r_2^2}$

$r_2^2=\frac{4.8}{2.2}\times 8^2$

$r_2=11.81 cm$

4 0

3 years ago