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3 years ago

15

Which type of bond and how many bonds would occur as carbon dioxide (CO2) is formed from carbon in Group IVA and oxygen in Group

VIA?
A. Three metallic bonds
B. two ionic bonds
C. One covalent bond
D. Four covalent bonds

1 answer:

Rus_ich [418]3 years ago

5 0

Answer:

The answer is D. Four covalent bounds

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Answer: A Punnett square can be used to predict genotype and phenotypes of offspring from genetic crosses. ... In the P generation, one parent has a dominant yellow phenotype and the genotype YY, and the other parent has the recessive green phenotype and the genotype yy.

Explanation:

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I need help PLEASE HEEEEEEEEELPP

lana66690 [7]

Answer:

a = (v2 - v1) / t

From A to B (8 - 4) m/s / 1 s = 4 m / s^2

From A to D ( 7 - 4) m/s / 5 s = .6 m / s^2

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3 years ago

A train moving at a constant speed on a surface inclined upward at 10.0° with the horizontal travels a distance of 400 meters in

Amiraneli [1.4K]

If the velocity of the train is v=s/t, where s is the distance and t is time, then v=400/5=80m/s. To get the vertical component of the velocity we need to multiply the velocity v with a sin(α): Vv=v*sin(α), where Vv is the vertical component of the velocity and α is the angle with the horizontal. So:

Vv=80*sin(10)=80*0.1736=13.888 m/s.

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3 years ago

A steel beam of mass 1975 kg and length 3 m is attached to the wall with a pin that can rotate freely on its right side. A cable

Nuetrik [128]

Answer:

a) 29062.125 N·m

b) 0 N·m

c) $Torque, due \ to \ tension =L\cdot Tsin\theta = \frac{M\cdot L\cdot g}{2}$

d) T = 11186.02 N

Explanation:

We are given

Beam mass = 1975 kg

Beam length = 3 m

Cable angle = 60° above horizontal

a) We have the formula for torque given as follows;

Torque about the pin = Force × Perpendicular distance of force from pin

Where the force = Force due to gravity or weight, we have

Weight = Mass × Acceleration due to gravity = 1975 × 9.81 = 19374.75 N

Point of action of force = Midpoint for a uniform beam = length/2

∴ Point of action of force = 3/2 = 1.5 m

Torque due to gravity = 19374.75 N × 1.5 m = 29062.125 N·m

b) Torque about the pinned end due to the contact forces between the pin and the beam is given by the following relation;

Since the distance from pin to the contact forces between the pin and the beam is 0, the torque which is force multiplied by perpendicular distance is also 0 N·m

c) To find the expression for the tension force, T we find the sum of the moment forces about the pin as follows

Sum of moments about p is given as follows

∑M = 0 gives;

T·sin(θ) × L= M×L/2×g

Therefore torque due to tension is given by the following expression

$Torque, due \ to \ tension =L\cdot Tsin\theta = \frac{M\cdot L\cdot g}{2}$

d) Plugging in the values in the torque due to tension equation, we have;

$3\times Tsin60 = \frac{1975\times 3\times 9.81}{2} = 29062.125$

Therefore, we make the tension force, T the subject of the formula hence

$T= \frac{29062.125}{3 \times sin(60)} = 11186.02 N$

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3 years ago

Suppose I have a spherical insulating shell with inner radius r and outer radius R. The shell has a uniformly distributed charge

a_sh-v [17]

Answer:

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Explanation:

Charge density = charge / volume

volume of a spherical shell = $\frac{4}{3} \pi(R^{3} - r^{3})$

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4 years ago