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3 years ago

The most reactive metals are located at the of the periodic table.

Physics

2 answers:

NeX [460]3 years ago

7 0

Answer:

the elements towards the bottom left corner

Colt1911 [192]3 years ago

4 0

Explanation:

Metals are the substances which tend to lose their valence electrons in order to attain stability.

For example, metals of group 1 are lithium, sodium, potassium, rubidium and cesium.

And, each of these elements have only 1 valence electron present. So, it is much easier to lose just one electron rather than losing 2 or 3 valence electrons.

Hence, alkali metals are the most reactive substances of the periodic table and they are present at extreme left side of periodic table.

Therefore, we can conclude that the most reactive metals are located at the extreme left of the periodic table.

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The heat required to melt a piece of copper is (82 x 10 ^5 J). The heat of fusion of copper is (2.05×10 ^5 J/ kg). What is the m

Musya8 [376]

Answer:

Explanation:

Givens

Heat of Fusion = 2.05 * 10^5 J / kg watch the units.

Heat to actually melt the copper = 82 10^5 J

Formula

Mass of copper = Heat / Heat of Fusion

Solution

Mass of copper = 82*10^5 J / (2.05 * 10^5 J / kg)

Mass of copper = 40 kg

Notice that the kg is in the denominator of the second fraction. The rules of fractions would tell you the 1/1 / / 1 /kg . You take the right fraction and turn it upside down and multiply. 1 / 1 * kg/1 = 1* kg / 1*1 which is just kg.

Answer 40 kg of copper

4 0

2 years ago

A proton at location A makes an electricfield E-->1 atlocation B. A different proton, placed atlocation B, experiences a forc

luda_lava [24]

Answer:

Explanation:

|F-->1| = E-->1 x charge on proton

= 300 x 1.6 x 10⁻¹⁹

Magnitude of electric force on proton

|F-->1| = 480 x 10⁻¹⁹ N .

Inside lithium nucleus , there are three protons so charge on it

= 3 x 1.6 x 10⁻¹⁹ = 4.8 x 10⁻¹⁹ C .

Magnitude of electric force on lithium nucleus

|F-->Li| = E-->1 x charge on lithium nucleus

= 300 x 4.8 x 10⁻¹⁹

= 1440 x 10⁻¹⁹ N .

Electron has same charge as that possessed by proton so charge on electron = 1.6 x 10⁻¹⁹ C

Magnitude of force on electron

|F-->1| = E-->1 x charge on electron

= 300 x 1.6 x 10⁻¹⁹

= 480 x 10⁻¹⁹ N .

7 0

3 years ago

When the cart moves on an incline at constant speed, it is in equilibrium; i.e., the net force on it is zero. Does it require mo

Amanda [17]

Answer:

It requires more tension to pull up the track

Explanation:

Net force must be zero to maintain constant velocity.

Weight force will always be pointed down the slope. Call it W

Friction force (Call it Ff) will be down slope when movement is up slope.

Friction force will be up slope when movement is down slope.

W and Ff are always positive numbers

Call the pulling force T

If Up slope is considered the positive direction

Moving up slope

Tu - Ff - W = 0

Tu = Ff + W

Moving down slope

Td + W - Ff = 0

Td = Ff - W

Ff + W > Ff - W therefore Tu > Td

5 0

2 years ago

A 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the cable for the following case

Murljashka [212]

Answer:

The answer is below

Explanation:

A 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the

cable for the following cases:

a. The load moves downward at a constant velocity

b. The load accelerates downward at a rate 0.4 m/s??

C. The load accelerates upward at a rate 0.4 m/s??

Solution:

Acceleration due to gravity (g) = 10 m/s²

a) Given that the mass of the crane (m) is 140 kg. If the load moves downward, the tension (T) is given by:

mg - T = ma

Since the load has a constant velocity, hence acceleration (a) = 0. Therefore:

mg - T = m(0)

mg - T = 0

T = mg

T = 140(10) = 1400 N

T = 1400 N

b) If the load moves downward, the tension (T) is given by:

mg - T = ma

T = mg - ma = m(g - a)

T = 140(10 - 0.4) = 140(9.96) = 134.4

T = 134.4 N

c) If the load moves upward, the tension (T) is given by:

T - mg = ma

T = ma + mg = m(a + g)

T = 140(0.4 + 10) = 140(10.4)

T = 145.6 N

2) To find the distance (s) if the load move from rest (u= 0) and accelerates for 20 seconds (t = 20). We use:

s = ut + (1/2)gt²

s = 0(20) + (1/2)(10)(20)²

s = 2000 m

7 0

3 years ago

A piano tuner stretches a steel piano wire with a tension of 800 N. The steel wire is 0.400 m long and has a mass of 3.00 g. Wha

stepan [7]

Answer:

408.25 Hz.

Explanation:

The fundamental frequency of a stretched string is given as

f' = 1/2L√(T/m') .................... Equation 1

Note: The a steel piano wire is a string

Where f' = fundamental frequency of the wire, L = length of the wire, T = tension on the wire, m' = mass per unit length of the wire.

Given: L = 0.4 m, T = 800 N,

Also,

m' = m/L where m = mass of the steel wire = 3.00 g = 3/1000 = 0.003 kg.

L = 0.4 m

m' = 0.003/0.4 = 0.0075 kg/m.

Substituting into equation 1

f' = 1/(2×0.4)[√(800/0.0075)]

f' = 1/0.8[√(106666.67)]

f' = 326.599/0.8

f' = 408.25 Hz.

Hence the frequency of the fundamental mode of vibration = 408.25 Hz.

8 0

3 years ago