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2 years ago

14

If fluorine was in ionic compounds with ions of each of the following atoms, how many fluorines would be needed?

1 answer:

nevsk [136]2 years ago

8 0

<span>You forgot to post the list of "following atoms", but

Fluorine ions have a -1 charge. So find the charges of the other ions and The number of Fluorines you'll need should match the </span>other ion's charge.

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What does law and theory have common

Ronch [10]

Answer:

It is a misconception that theories turn into laws with enough research.

Hope it helps

7 0

2 years ago

A 55-kg block, starting from rest, is pushed a distance of 5.0 m across a floor by a horizontal force Fp whose magnitude is 140

max2010maxim [7]

Answer:

The answer is "151.25 J and -547.64 J".

Explanation:

$u = 0\\\\v = 2.35\ \frac{m}{sec}\\\\d = 5.0 \ m\\\\$

Using formula:

$v^2 = u^2 + 2 \times a \times d\\\\2.35^2 = 0^2 + 2 \times a \times 5\\\\a = \frac{2.35^2}{10} \\\\$

$= 0.55 \ \frac{m}{sec^2}\\\\$

$F_{net} = m \times a\\\\F_{net} = 55 \times 0.55 = 30.25\ N\\\\$

Calculating the Work by net force

$W = F_{net}\times d\\\\W = 30.25 \times 5 = 151.25 \ J\\\\$

The above work is converted into thermal energy.

Now,

$F_{net} = F_p - F_f\\\\F_p = 140 \ N\\\\F_f = u_k\times m \times g = u_k \times 55 \times 9.81\\\\F_f = 539.55 \times u_k\\\\30.25 = 140 - u_k \times 55 \times 9.81\\\\u_k = \frac{(140 - 30.25)}{(55\times 9.81)}\\\\uk = 0.203 = \text{Coefficient of friction}\\\\W_f = -F_f \times d\\\\W_f = -0.203 \times 55 \times 9.81 \times 5\\\\Work\ done\ by\ friction = -547.64 \ J$

6 0

2 years ago

What is the time required for an object starting

PtichkaEL [24]

(3) 10.1 second Using equation of motion 500 = (0.5)(9.81)t^2. Rearranging, t = sqrt(1000/9.81) = 10.1s

8 0

3 years ago

Read the scenario and solve these two problems.

Burka [1]

Answers:

a) 5400000 J

b) 45.92 m

Explanation:

a) The kinetic energy $K$ of an object is given by:

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

Where:

$m=12000 kg$ is the mass of the train

$V=30 m/s$ is the speed of the train

Solving the equation:

$K=\frac{1}{2}(12000 kg)(30 m/s)^{2}$

$K=5400000 J$ This is the train's kinetic energy at its top speed

b) Now, according to the Conservation of Energy Law, the total initial energy is equal to the total final energy:

$E_{i}=E_{f}$

$K_{i}+P_{i}=K_{f}+P_{f}$

Where:

$K_{i}=5400000 J$ is the train's initial kinetic energy

$P_{i}=0 J$ is the train's initial potential energy

$K_{f}=0 J$ is the train's final kinetic energy

$P_{f}=mgh$ is the train's final potential energy, where $g=9.8 m/s^{2}$ is the acceleration due gravity and $h$ is the height.

Rewriting the equation with the given values:

$5400000 J=(12000 kg)(9.8 m/s^{2})h$

Finding $h$ :

$h=45.918 m \approx 45.92 m$

7 0

2 years ago

Read 2 more answers

A thin rod of length 1.4 m and mass 140 g is suspended freely from one end. It is pulled to one side and then allowed to swing l

m_a_m_a [10]

Answer:

a The kinetic energy is $KE = 0.0543 J$

b The height of the center of mass above that position is $h = 1.372 \ m$

Explanation:

From the question we are told that

The length of the rod is $L = 1.4m$

The mass of the rod $m = 140 = \frac{140}{1000} = 0.140 \ kg$

The angular speed at the lowest point is $w = 1.09 \ rad/s$

Generally moment of inertia of the rod about an axis that passes through its one end is

$I = \frac{mL^2}{3}$

Substituting values

$I = \frac{(0.140) (1.4)^2}{3}$

$I = 0.0915 \ kg \cdot m^2$

Generally the kinetic energy rod is mathematically represented as

$KE = \frac{1}{2} Iw^2$

$KE = \frac{1}{2} (0.0915) (1.09)^2$

$KE = 0.0543 J$

From the law of conservation of energy

The kinetic energy of the rod during motion = The potential energy of the rod at the highest point

Therefore

$KE = PE = mgh$

$0.0543 = mgh$

$h = \frac{0.0543}{9.8 * 0.140}$

$h = 1.372 \ m$

5 0

2 years ago

Read 2 more answers