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3 years ago

In an object's net charge positive or negative if it loses electrons?

1 answer:

7 0

Answer: If a neutral atom gains electrons, then it will become negatively charged. If a neutral atom loses electrons, then it become positively charged. 6. Determine the quantity and type of charge on an object that has 3.62 x 1012 more protons than electrons.

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A particle is projected from a point on a horizontal plane and has an initial velocity of 28root 3 m/s at an angle of 60 degree

Lelechka [254]

Answer:

uujjjjjctc7tox7txr9ll8rz8lr5xl8r6l8dl85x8rl5x8rl5x8rl5xrx8l58rk5xr8l5xr6l8xr68lc

4 0

3 years ago

The rigid beam is supported by the three suspender bars. bars ab and ef are made of aluminum and bar cd is made of steel. if eac

faltersainse [42]

Answer:

Pmax = 67.5 KN

Explanation:

We need to calculate the maximum allowable value of P for both aluminum and steel bars.

FOR STEEL BARS:

Since,

(σallow)st = (Pmax)st/A

where,

(σallow)st = maximum allowable stress of steel bar = 200 MPa = 2 x 10⁸ Pa

A = Cross-sectional area of steel bar = 450 mm² = 0.45 x 10⁻³ m²

(Pmax)st = Maximum allowable force for steel bar = ?

Therefore,

2 x 10⁸ Pa = (Pmax)st/0.45 x 10⁻³ m²

(Pmax)st = (2 x 10⁸ Pa)(0.45 x 10⁻³ m²)

(Pmax)st = 9 x 10⁴ N = 90 KN

FOR Aluminum BARS:

Since,

(σallow)al = (Pmax)al/A

where,

(σallow)al = maximum allowable stress of Al bar = 150 MPa = 1.5 x 10⁸ Pa

A = Cross-sectional area of Aluminum bar = 450 mm² = 0.45 x 10⁻³ m²

(Pmax)al = Maximum allowable force for Aluminum bar = ?

Therefore,

1.5 x 10⁸ Pa = (Pmax)al/0.45 x 10⁻³ m²

(Pmax)al = (1.5 x 10⁸ Pa)(0.45 x 10⁻³ m²)

(Pmax)al = 6.75 x 10⁴ N = 67.5 KN

Since,

(Pmax)al < (Pmax)st

Therefore,

The maximum allowable force will be:

Pmax = (Pmax)al

Pmax = 67.5 KN

3 0

4 years ago

Why can't the teacher test the amount of water and the amount of sunlight in the same experiment? The plants would not grow At t

iren [92.7K]

Answer:the answer is a

Explanation:

6 0

3 years ago

How much kinetic energy does a system containing 3 moles of an ideal gas at 300 K possess? What is the heat capacity at constant

ElenaW [278]

Answer:

1. K.E = 11.2239 kJ ≈ 11.224 kJ

2. $C_{V} = 37.413 JK^{- 1}$

3. $Q = 10.7749 kJ$

Solution:

Now, the kinetic energy of an ideal gas per mole is given by:

K.E = $\frac{3}{2}mRT$

where

m = no. of moles = 3

R = Rydberg's constant = 8.314 J/mol.K

Temperature, T = 300 K

Therefore,

K.E = $\frac{3}{2}\times 3\times 8.314\times 300$

K.E = 11223.9 J = 11.2239 kJ ≈ 11.224 kJ

Now,

The heat capacity at constant volume is:

$C_{V} = \frac{3}{2}mR$

$C_{V} = \frac{3}{2}\times 3\times 8.314 = 37.413 JK^{- 1}$

Now,

Required heat transfer to raise the temperature by $15^{\circ}$ is:

$Q = C_{V}\Delta T$

$\Delta T = 15^{\circ} = 273 + 15 = 288 K$

$Q = 37.413\times 288 = 10774.9 J = 10.7749 kJ$

8 0

3 years ago

WILL GIVE BRAINLIEST!! Part 1: Other than distance from the sun, which factor affects the temperature of a planet?

Svet_ta [14]

The thickness of the ozone layer (for earth) and the atmosphere.
Mercury is the closest planet to the sun but it isn’t the hottest because it has no atmosphere. Although Venus is the second planet from the sun, it is the hottest because of it’s thick atmosphere and the clouds that trap heat in.

8 0

4 years ago