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4 years ago

Does anyone wanna be my friend????

Physics

2 answers:

Alla [95]4 years ago

7 0

// Okay, I would love to be your friend :D //

Marrrta [24]4 years ago

4 0

Answer:

Yes I will be your friend no problem bro

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Difference between liquid solid and gas in the arrangement of molecules

marin [14]

Answer:

Note that:

Particles in a:

gas are well separated with no regular arrangement.

liquid are close together with no regular arrangement.

solid are tightly packed, usually in a regular pattern.

Explanation:

3 0

3 years ago

Read 2 more answers

A marble resting near the edge of .90 m high table is given an initial horizontal speed of 1.24 m/s. What will be it’s horizonta

Maru [420]

Answer:

0.53 m

Explanation:

First of all, we have to consider the vertical motion of the ball, in order to find the time it takes for the marble to reach the ground. The initial height is $h=0.90 m$ , the initial vertical velocity is zero, while the acceleration is $g=-9.81 m/s^2$ , so the vertical position at time t is given by

$y(t)=h-\frac{1}{2}gt^2$

By demanding y(t)=0, we find the time t at which the ball reaches the ground:

$0=h-\frac{1}{2}gt^2$

$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(0.9 m)}{9.81 m/s^2}}=0.43 s$

Now we can find the horizontal range of the marble: we know the initial horizontal speed (v=1.24 m/s), we know the total time of the motion (t=0.43 s), and since the horizontal speed is constant, the total distance traveled on the horizontal direction is

$x=vt=(1.24 m/s)(0.43 s)=0.53 m$

8 0

4 years ago

What is fundamental questioning came now<br>cfy_tuhq_wjt

Nadya [2.5K]

Answer:

juz taking points nvm no sry for u cos u too wasted point

8 0

3 years ago

Decide which of the following day-to-day activities might be considered aerobic exercise?

erastova [34]

Answer:

3. Mowing the lawn

Explanation:

Aerobic exercise is any type of cardiovascular conditioning.

Mowing the lawn is the only thing that is actively moving.

3 0

3 years ago

Read 2 more answers

A uniformly charged conducting sphere of 1.1 m diameter has a surface charge density of 6.2 µC/m2. (a) Find the net charge on th

ira [324]

Answer:

(a) q = 2.357 x 10⁻⁵ C

(b) Φ = 2.66 x 10⁶ N.m²/C

Explanation:

Given;

diameter of the sphere, d = 1.1 m

radius of the sphere, r = 1.1 / 2 = 0.55 m

surface charge density, σ = 6.2 µC/m²

(a) Net charge on the sphere

q = 4πr²σ

where;

4πr² is surface area of the sphere

q is the net charge on the sphere

σ is the surface charge density

q = 4π(0.55)²(6.2 x 10⁻⁶)

q = 2.357 x 10⁻⁵ C

(b) the total electric flux leaving the surface of the sphere

Φ = q / ε

where;

Φ is the total electric flux leaving the surface of the sphere

ε is the permittivity of free space

Φ = (2.357 x 10⁻⁵) / (8.85 x 10⁻¹²)

Φ = 2.66 x 10⁶ N.m²/C

8 0

3 years ago