All categories

3 years ago

Fill in the blank

Physics

2 answers:

elixir [45]3 years ago

8 0

Is there any types of answer to get an idea

Naya [18.7K]3 years ago

5 0

The reactivity of an alkali metal depends on the metals ability to lose electrons.

You might be interested in

When all group iia elements lose their valence electrons, the remaining electron configurations are the same as for what family

Orlov [11]

Noble Gases

I hope I helped

ΩΩΩΩΩΩΩΩΩΩ

8 0

3 years ago

Read 2 more answers

Hufflepuff and Ravenclaw are having their annual tug of war contest. Each team can have 20 students. What is the most important

Troyanec [42]

It should be S since S stands for students. Giving the team with more members the upper hand. Hope this helps.

3 0

3 years ago

Read 2 more answers

A 5.0-kg rock and a 3.0 × 10−4-kg pebble are held near the surface of the earth.(a)Determine the magnitude of the gravitational

a_sh-v [17]

Answer:

a). Determine the magnitude of the gravitational force exerted on each by the earth.

Rock: $F = 49.06N$

Pebble: $F = 29.44N$

(b)Calculate the magnitude of the acceleration of each object when released.

Rock: $a =9.8m/s^{2}$

Pebble: $a =9.8m/s^{2}$

Explanation:

The universal law of gravitation is defined as:

$F = G\frac{m1m2}{r^{2}}$ (1)

Where G is the gravitational constant, m1 and m2 are the masses of the two objects and r is the distance between them.

Case for the rock $m = 5.0 Kg$ :

m1 will be equal to the mass of the Earth $m1 = 5.972×10^{24} Kg$ and since the rock and the pebble are held near the surface of the Earth, then, r will be equal to the radius of the Earth $r = 6371000m$ .

$F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(5.0 Kg)}{(6371000 m)^{2}}$

$F = 49.06N$

Newton's second law can be used to know the acceleration.

$F = ma$

$a =\frac{F}{m}$ (2)

$a =\frac{(49.06 Kg.m/s^{2})}{(5.0 Kg)}$

$a =9.8m/s^{2}$

Case for the pebble $m = 3.0 Kg$ :

$F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(3.0 Kg)}{(6371000 m)^{2}}$

$F = 29.44N$

$a =\frac{F}{m}$

$a =\frac{(29.44 Kg.m/s^{2})}{(3.0 Kg)}$

$a =9.8m/s^{2}$

3 0

3 years ago

Read 2 more answers

A pickup truck starts from rest and maintains a constant acceleration a0. After a time t0, the truck is moving with speed 25 m/s

MA_775_DIABLO [31]

Answer:

B v1= 12.5m/s

Explanation:

As the acceleration is constant and time taken is also same and distance covered is halfed so the speed will also be halfed.

7 0

3 years ago

The basic barometer can be used to measure the height of a building. If the barometric readings at the top and at the bottom of

LekaFEV [45]

Answer:

(h₁-h₂) = 2.30 10² m

Explanation:

The pressure depends on the height with the formula

P = P_atm + rho g h

Let's apply this expression for the building

P₁ = P_atm + rho_air g h₁

P₂ = P_atm + rho_air g h₂

Subtract

P₁ - P₂ = roh_air g (h₁ –h₂)

The measured pressure is in mm Hg to take this unit to units of pressure must be multiplied by the density of mercury and the acceleration of gravity

P₁- P₂ = rho_Hg g (h₁-h₂) _Hg

rho_Hg g (h₁-h₂) _Hg = roh_air g (h₁ –h₂)

(h₁ –h₂) = rho_Hg / rho_air (h₁-h₂) _ Hg

Let's calculate

(h₁-h₂) = 13600 / 1.18 (695-675)

(h₁-h₂) = 2.30 10⁵ mm

Let's reduce to meter

(h₁-h₂) = 2.30 10⁵ mm (1 m / 10³ mm)

(h₁-h₂) = 2.30 10² m

4 0

3 years ago