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3 years ago

Plz answer this! I am stumped

Physics

1 answer:

Gwar [14]3 years ago

6 0

MgCl2
Mg = magnesium
Cl = chlorine

Magnesium + chlorine = magnesium chloride.

This is because compounds are always written with the METAL FIRST and the NON METAL SECOND. the non metal ends in - ide when it reacts with a metal.

So ur answer would be magnesium chloride. :)

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Caleb is filling up water balloons for the Physics Olympics balloon tosscompetition. Caleb sets a 0.50-kg spherical water balloo

Mashcka [7]

• P = F/A

P = pressure = 630 N/m^2

F = force

A = area

F = mg = 0.50 kg x 9.8 m/s^2 = 4.9 N

m= mass

g= gravity

P = F/A

A = F/P

A = 4.9 N / 630 N/m^2 = 7.778 x 10^-3 m^2

• Area of a circle = pi* radius ^2

7.778 x 10^-3 m^2 = pi* radius ^2

√(7.778 x 10^-3 m^2 / pi ) = radius

radius = 0.04976 m

Answers:

a ) 7.778 x 10^-3 m^2

b) 0.04976 m

8 0

1 year ago

A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A cons

Virty [35]

Complete Question

A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.What will be the equilibrium height of the mass?

Answer:

$H_m=1.65m$

$H_E=1.16307m$

Explanation:

From the question we are told that

Mass of ball $M=2kg$

Length of string $L= 2m$

Wind force $F=13.2N$

Generally the equation for $\angle \theta$ is mathematically given as

$tan\theta=\frac{F}{mg}$

$\theta=tan^-^1\frac{F}{mg}$

$\theta=tan^-^1\frac{13.2}{2*2}$

$\theta=73.14\textdegree$

Max angle = $2*\theta= 2*73.14=>146.28\textdegree$

Generally the equation for max Height $H_m$ is mathematically given as

$H_m=L(1-cos146.28)$

$H_m=0.9(1+0.8318)$

$H_m=1.65m$

Generally the equation for Equilibrium Height $H_E$ is mathematically given as

$H_E=L(1-cos73.14)$

$H_E=0.9(1+0.2923)$

$H_E=1.16307m$

8 0

2 years ago

A horizontal 810-N merry-go-round of radius 1.60 m is started from rest by a constant horizontal force of 55 N applied tangentia

Sloan [31]

Answer:

576 joules

Explanation:

From the question we are given the following:

weight = 810 N

radius (r) = 1.6 m

horizontal force (F) = 55 N

time (t) = 4 s

acceleration due to gravity (g) = 9.8 m/s^{2}

K.E = 0.5 x MI x ω^{2}

where MI is the moment of inertia and ω is the angular velocity

MI = 0.5 x m x r^2

mass = weight ÷ g = 810 ÷ 9.8 = 82.65 kg

MI = 0.5 x 82.65 x 1.6^{2}

MI = 105.8 kg.m^{2}

angular velocity (ω) = a x t

angular acceleration (a) = torque ÷ MI

where torque = F x r = 55 x 1.6 = 88 N.m

a= 88 ÷ 105.8 = 0.83 rad /s^{2}

therefore

angular velocity (ω) = a x t = 0.83 x 4 = 3.33 rad/s

K.E = 0.5 x MI x ω^{2}

K.E = 0.5 x 105.8 x 3.33^{2} = 576 joules

6 0

3 years ago

14.A 90 kg quarterback gets tackled by being hit by a 120 kg lineman backwards

quester [9]

The acceleration of the quarterback and the lineman is 5.55m/s² and 4.16m/s² respectively in the same direction.

As, we know, the 120 Kg lineman is moving with a force of 500N.

His net acceleration will be in the same direction as his motion.

It is already known that, If M is the mass of the body and a is the acceleration of the body, then the force F on the body can be calculated by using the formula,

F = Ma.

The weight of the quarterback is 90 Kg. He is being hit by a force of 500N.

So, the acceleration can be calculated using the formula,

500N = 90kg x a

a = 5.55 m/s².

Now, the weight if the lineman is 120kg, the force applied by him is 500N.

So, from the formula, his acceleration A will be,

500N = 120Kg x A

A = 4.16 m/s².

both of them will have acceleration in the same direction,

To know more about Force, visit,

brainly.com/question/25239010

#SPJ9

8 0

11 months ago

Monochromatic light of wavelength 687 nm is incident on a narrow slit. On a screen 1.65 m away, the distance between the second

Sophie [7]

Answer:

a ) 1.267 radian

b ) 1.084 10⁻³ mm

Explanation:

Distance of screen D = 1.65 m

Width of slit d = ?

Wave length of light λ = 687 nm.

Distance of second minimum fro centre y = 2.09 cm

Angle of diffraction = y / D

= 2.09 /1.65

= 1.267. radian

Angle of diffraction of second minimum

= 2 λ / d

so 2 λ / d = 1.267

d = 2 λ / 1.267 = (2 x 687 ) /1.267 nm

=1084.45 nm = 1.084 x 10⁻³ mm.

3 0

3 years ago