How many valence electrons are in an atom of fluorine

The local high school is installing new bleachers at the stadium and must also add handrails to meet code. The students know the

Advocard [28]

Answer:

The handrails must be approximately 10.63 meters long

Explanation:

The given parameters are;

The height of the bleachers, h = 8 m

The depth of the bleachers, d = 7 m

The length of the hand rails to go along the bleachers from bottom to top is given by Pythagoras' Theorem as follows;

The length of the hand rail = √(d² + h²)

∴ The length of the hand rail = √(7² + 8²) = √113 ≈ 10.63

In order for the handrails to go along the bleachers from top to bottom, they must be approximately 10.63 meters long.

5 0

3 years ago

A paper clip that has a mass of 1.5 grams is thrown into the air and initially has a kinetic energy

Vsevolod [243]

Answer:

$v = 4.2 \ m/s$

Explanation:

Given data:

Mass of the paper clip, $m = 1.5 \ g = 0.0015 \ kg$

Kinetic energy, $K = 0.013 \ \rm J$

Let the velocity of the paper clip when it is thrown be v.

Thus,

$K = \frac{1}{2}mv^{2}$

$0.013 = 0.5 \times 0.0015 \times v^{2}$

$\Rightarrow \ v = 4.16 \ m/s$

$v = 4.2 \ m/s.$ (rounding to nearest tenth)

3 0

3 years ago

A hawk in level flight 135m above the ground drops the fish it caught. If the hawk horizontal speed is 20m/s, how far ahead of t

Kaylis [27]

Find how long it will take the fish to hit the ground

8 0

3 years ago

Ship A is located 4.1 km north and 2.3 km east of ship B. Ship A has a velocity of 22 km/h toward the south and Ship B has a vel

castortr0y [4]

Answer:

a) x component = -31.25 km/hr

b) y component = 46.64 km/hr

Explanation:

Given data:

A position is 4km north and 2.5 km east to B

Ship A velocity = 22 km/hr

ship B velocity = 40 km/hr

A velocity wrt to velocity of B

$\vec{V_{AB}} =\vec{V_A} - \vec{V_B}$

$\vec{V_A} = 22 km/hr$

$\vec{V_B} = 40 cos38\hat{i} + 40sin 38 \hat{j}$

$= 31.52\hat{i} + 24.62 \hat{j}$

putting respective value to get velocity of A with respect to B

$\vec{V_{AB}} = -22 \hat {j} - (31.52\hat{i} + 24.62 \hat{j})$

$\vec{V_{AB}} = -31.52\hat{i} - 46.62\hat{j}$

a) x component = -31.25 km/hr

b) y component = 46.64 km/hr

6 0

3 years ago

Object A has mass 83.0 g and hangs from an insulated thread. When object B, which has a charge of +140 nC, is held nearby, A is

erastova [34]

Answer:

a) -238 nC

b) 0.889 N

Explanation:

Concepts and Principles

Particle in Equilibrium: If a particle maintains a constant velocity (so that a = 0), which could include a velocity of zero, the forces on the particle balance and Newton's second law reduces to:

∑F = 0 (1)

Coulomb's Law: the magnitude of the electrostatic force exerted by a point charge q1 on a second point charge q2 separated by a distance r is directly proportional to the product of the two charges and is inversely proportional to the square of the distance between them:

F_12 = k*| q1 |*| q2 |/r^2 (2)

where k = 8.99 x 10^9 N m^2/C^2 is Coulomb constant.

Given Data

mA (mass object A) = (83 g)*(1/1000g)=0.09 kg

qB (charge of object B) = (140 nC)*(1/10^9 nC) = 130 x 10^-9 C

Object A is attracted to object B.

Ф(angle made by object A with the vertical) = 7.2°

( r (distance between the two objects) = (5 cm) * (1 m/ 100 cm) =0.05 m

Object A is in equilibrium.

Required Data

In part (a), we are asked to determine the charge qA of object A.

In part (b), we are asked to determine the tension T in the thread.

(a) The FBD in Figure 1 shows the forms acting on object A; Fe is the electric force exerted on object A by object B, T is the tension force exerted on the thread, and m_a*g is the gravitational force exerted on object A.

Model object A as a particle in equilibrium in the horizontal and vertical direction and apply Equation (1) to it:

∑F_x = F_e-Tsin = 0 F_e=TsinФ (3)

∑F_y = TcosФ - m_a*g= 0 m_a*g=TsinФ (4)

Divide Equation (3) by Equation (4) to eliminate T:

F_e/m_a*g=tanФ

F_e=m_a*g*tanФ

Substitute for F_e by using Coulomb's law from Equation (2):

k*| q_A |*| q_B |/r = m_a*g*tanФ

Solve for q_A :

| q_A | = m_a*g*tanФ_r/k*| q_B |

Substitute numerical values from given data:

| q_A | = 238 nC

Because object A is attracted to object B. it has an opposite negative charge. Therefore, the charge on object A is | q_A | = -238 nC

(b)

Solve Equation (4) for T:

T = m_a*g/cosФ

Substitute numerical values from given data:

T = (0.09 kg)(9.8 m/s^2) /cos 7.2°

= 0.889 N

4 0

3 years ago