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3 years ago

You arrive in my class 45 seconds after leaving math which is 90 meters away. How fast did you travel?

Physics

1 answer:

pashok25 [27]3 years ago

4 0

Answer:

2m/s

Explanation:

speed = distance÷time

speed =90÷45=2m/s

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What the balance electrons for calcium

soldi70 [24.7K]

Answer:

its two valence electrons

Explanation:

Calcium is a group 2 element with two valence electrons. Therefore, it is very reactive and gives up electrons in chemical reactions.

UR WELCOME!! :)

5 0

3 years ago

Block A can slide relative to block B which, in turn, can slide on a perfectly smooth horizontal plane. If the initial velocity

Anna11 [10]

Answer:

the final velocity of the two blocks is $v = \frac{mv_o}{m+M}$

the distance that A slides relative to B is $S = \frac{v_o^2M}{2 \mu g (M+m)}$

Explanation:

From the diagram below;

acceleration of A relative to B is : $a = - ( \mu g + \frac{ \mu mg}{M})$

where

v = u + at

$0 = v_o + ( - \mu g - \frac{\mu m g }{M})t$

Making t the subject of the formula; we have:

$t = \frac{v_o M}{(\mu g )(M+m)}$

$v^2 = u^2 +2 as\\\\0^2 = v_o^2 - 2 (\mu g ) (\frac{M+m}{M})S\\\\$

$S = \frac{v_o^2M}{2 \mu g (M+m)}$ which implies the distance that A slides relative to B.

The final velocities of the two blocks can be determined as follows:

v = u + at

$v = v_o - \mu g \frac{v_oM}{\mu g (M+m)}\\\\v = \frac{\mu g mv_o}{m+M}\\\\$

$v = \frac{mv_o}{m+M}$

Thus, the final velocity of the two blocks is $v = \frac{mv_o}{m+M}$

4 0

2 years ago

Suman with a mass of 45kg climbs in 3m heigh ladder in 10 seconds. Calculate her power. Plz help me

attashe74 [19]

Answer:

work done=fdcosø

f=mg=450N

d=+3m

ø=180

450×3×cos(180°) power=work done/

time taken

1350×-1 p=135watts

7 0

2 years ago

What is the value of ΔVBA=VB−VA if the charge on the plates is 1.00 x 10-9 C, the area of the plates is 2.00 m2 and the distanc

Murljashka [212]

Answer:

The Value is $V_{AB} = 2.825V$

Explanation:

The explanation is shown on the first uploaded image

5 0

3 years ago

57. Estimate Potential Energy A boulder with a

Oksanka [162]

Answer: 4.9 x 10^6 joules

Explanation:

Given that:

mass of boulder (m) = 2,500 kg

Height of ledge above canyon floor (h) = 200 m

Gravita-tional potential energy of the boulder (GPE) = ?

Since potential energy is the energy possessed by a body at rest, and it depends on the mass of the object (m), gravitational acceleration (g), and height (h).

GPE = mgh

GPE = 2500kg x 9.8m/s2 x 200m

GPE = 4900000J

Place result in standard form

GPE = 4.9 x 10^6J

Thus, the gravita-tional potential energy of the boulder-Earth system relative to the canyon floor is 4.9 x 10^6 joules

3 0

3 years ago