Can i get some help please

Name an element in the same family as sodium

AlladinOne [14]

Answer:

The Alkali Metals

sodium is a member of the alkali metal family with potassium (K) and lithium (Li). Sodium's big claim to fame is that it's one of two elements in your table salt. When bonded to chlorine (Cl), the two elements make sodium chloride (NaCl).

7 0

3 years ago

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In a natural gas power plant, natural gas is burned to heat steam, which turns a turbine. What energy conversion is happening in

Irina18 [472]

<h2>Hello!</h2>

The answer is A. Chemical energy to heat energy to kinetic energy to electrical energy.

Let's check the energy conversion given in the statement:

Chemical energy: Chemical energy is usually obtained from a natural resource combustion. For this particular case is the result of natural gas combustion.

Heat energy: Heat energy comes from chemical reactions such as natural gas combustions.

Kinetic energy: Kinetic energy is obtained by turbines transforming the pressurized steam on mechanical movement. Turbines are formed by a turbine that absorbs the heat energy to do mechanical work and transmit it to an output rotating shaft.

Electrical energy: Electrical energy is generated by gas turbines when the rotating output shaft is connected to an electrical generator. Electrical generators use mechanical energy to generate electrical energy by using electromagnetic induction. Electrical generators are basically formed by a rotor and a stator.

Have a nice day!

5 0

3 years ago

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At the equator earth rotates with a velocity of about 465 m/s.

Dafna1 [17]

The given velocity is 465 m/s.

Part a.
$465 \, \frac{m}{s} =(465 \times 10^{-3} \, \frac{km}{s})*( 3600 \, \frac{s}{h} ) = 1674 \, \frac{km}{h}$
Answer: 1674 km/h

Part b.
$1674 \frac{km}{h} = (1674 \, \frac{km}{h})*(24 \, \frac{h}{day} ) = 40176 \, \frac{km}{day}$
Answer: 40,176 km/day.

3 0

3 years ago

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If earth's mass were half its actual value but its radius stayed the same, the escape velocity of earth would be:________

siniylev [52]

If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be $V_e = \sqrt{\dfrac{GM}{r}}$ .

<h3>What is an escape velocity?</h3>

The ratio of the object's travel distance over a specific period of time is known as its velocity. As a vector quantity, the velocity requires both the magnitude and the direction. the slowest possible speed at which a body can break out of the gravitational pull of a certain planet or another object.

The formula to calculate the escape velocity of earth is given below:-

$V_e=\sqrt{\dfrac{2GM}{r}}$

Given that earth's mass was half its actual value but its radius stayed the same. The escape velocity will be calculated as below:-

$V_e=\sqrt{\dfrac{2GM}{r\times 2}}$

$V_e = \sqrt{\dfrac{GM}{r}}$ .

Therefore, If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be $V_e = \sqrt{\dfrac{GM}{r}}$ .

To know more about escape velocity follow

brainly.com/question/14042253

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8 0

2 years ago

A cart loaded with bricks has a total mass of 9.13 kg and is pulled at constant speed by a rope. The rope is inclined at 24.7 ◦

blagie [28]

Answer:

W = 0.63 KJ

Explanation:

Work (W) is defined as the point product of force (F) by the distance (d)the body travels due to this force.

W= F*d *cosα Formula (1)

F : force (N)

d : displacement (m)

α : angle between force and displacement

Newton's second law:

∑F = m*a Formula (2)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration (m/s²)

We define the x-axis in the direction parallel to the movement of the cart on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the cart

W: Weight of the cart : In vertical direction

FN : Normal force : perpendicular to the floor

f : Friction force: parallel to the floor

T : tension Force, inclined at θ=24.7° above the horizontal

Calculated of the W

W= m*g

W= 9.13 kg* 9.8 m/s² = 89.47 N

x-y components o the tension force (T)

Tx = Tcosθ = T*cos 24.7° (N)

Ty = Tsin θ = T*sin 24.7° (N)

Calculated of the FN

We apply the formula (2)

∑Fy = m*ay ay = 0

FN +Ty- W = 0

FN = W-Ty

FN = 89.47-T*sin 24.7°

Calculated of the friction force (f)

f = μk*FN

f =(0.597)*( 89.47-T*sin 24.7° )

f= 53.41-0.249T

Calculated of the tension force of the rope (f)

We apply the formula (2) :

∑Fx = m*ax , ax= 0 ,because the speed of the cart is constant

Tx - f = 0

T*cos 24.7°-( 53.41 - 0.249T )= 0

T*cos 24.7° + 0.249T = 53.41

(1.1575)T = 53.41

T= (53.41) / (1.1575)

T= 46.14 N

Work done on the cart by the rope

We apply the formula (1)

W=T*d *cosα

W= (46.14 N)*(15.1 m) *(cos24.7)

W = 632.97 (N*m) = 632.97 (J)

W = 0.63 KJ

6 0

4 years ago