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4 years ago

1. Go onto the Newsround website or article

Physics

2 answers:

poizon [28]4 years ago

5 0

Answer:

I read the article - the story was really sad.

Explanation:

- sincerelynini

Alecsey [184]4 years ago

4 0

I cant find a article with that name? Are you sure its right?

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• Jason thinks he's got a faster are

GarryVolchara [31]

T = (36 - 27) / 2.5
t = 3.6 s

6 0

3 years ago

A 12.5843-gram sample of metal bromide, MBr4, was dissolved and, after reaction with silver nitrate, AgNO3, all of the bromide w

barxatty [35]

Answer:

zirconium

Explanation:

Given, Mass of AgBr(s) = 23.0052 g

Molar mass of AgBr(s) = 187.77 g/mol

The formula for the calculation of moles is shown below:

$Moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles\ of\ AgBr= \frac{23.0052\ g}{187.77\ g/mol}$

$Moles\ of\ AgBr= 0.1225\ mol$

The reaction taking place is:

$MBr_4+4AgNO_3\rightarrow 4AgBr+M(NO_3)__4$

From the reaction,

4 moles of AgBr is produced when 1 mole of $MBr_4$ undergoes reaction

1 mole of AgBr is produced when 1 / 4 mole of $MBr_4$ undergoes reaction

0.1225 mole of AgBr is produced when $\frac {1}{4}\times 0.1225$ mole of $MBr_4$ undergoes reaction

Moles of $MBr_4$ got reacted = 0.030625 moles

Mass of the sample taken = 12.5843 g

Let the molar mass of the metal = x g/mol

So, Molar mass of $MBr_4$ = x + 4 × 79.904 g/mol = 319.616 + x g/mol

Thus,

$0.030625 = \frac{12.5843}{319.616 + x}$

Solve for x,

we get, x = 91.2999 g/mol

The metal shows +4 oxidation state and has mass of 91.2999 g/mol . The metals is zirconium.

8 0

3 years ago

A car with the mass of 18,000kg accelerates at the rate of 9m/s. what is the force being applied to the car?

Vika [28.1K]

Hope it cleared your doubt.

5 0

3 years ago

A 4.00 kg object is moving at 5.00 m/s NORTH. It strikes a 6.00 kg object that is moving WEST at 2.00 m/s. The objects undergo a

Veronika [31]

We are given that an object is moving north at a speed of 5 m/s, lets call this object 1. We have another object moving west at a speed of 2 m/s, this is object 2. A diagram of the problem is the following:

To determine the loss in kinetic energy we need to determine the difference in kinetic energy before the collision and after the collision:

$\Delta K=K_f-K_0$

The final kinetic energy is:

$K_f=\frac{1}{2}(m_1+m_2)v^2_f$

We use the sum of the masses because the objects are stuck together after the collision. The initial kinetic energy is:

$K_0=\frac{1}{2}m_1v^2_{01}+\frac{1}{2}m_2v^2_{02}$

Substituting we get:

$\Delta K=\frac{1}{2}(m_1+m_2)v^2_f-(\frac{1}{2}m_1v^2_{01}+\frac{1}{2}m_2v^2_{02})$

The only missing variable is the final velocity. To determine the final velocity we will use the conservation of momentum.

We will use the conservation of momentum in the horizontal direction (west) and the conservation of momentum in the vertical direction (north).

In the horizontal direction we have:

$m_1v_{h1}+m_2v_{h2}=(m_1+m_2)v_{hf}$

Since the object 1 has no velocity in the horizontal direction we have that:

$\begin{gathered} m_1(0)+m_2v_{h2}=(m_1+m_2)v_{hf} \\ m_2v_{h2}=(m_1+m_2)v_{hf} \end{gathered}$

Now we solve for the final horizontal velocity:

$\frac{m_2v_{h2}}{\mleft(m_1+m_2\mright)}=v_{hf}$

Now we substitute the values:

$\frac{(6kg)(2\frac{m}{s})}{(4kg+6kg)}=v_{hf}$

Solving the operations we get:

$1.2\frac{m}{s}=v_{hf}$

Now we use the conservation of momentum in the vertical direction, we get:

$m_1v_{v1}+m_2v_{v2}=(m_1+m_2)v_{vf}$

Since the second object has no vertical velocity we get:

$m_1v_{v1}=(m_1+m_2)v_{vf}$

Now w solve for the final vertical velocity, we get:

$\frac{m_1v_{v1}}{\mleft(m_1+m_2\mright)}=v_{vf}$

Now we substitute the values:

$\frac{(4kg)(5\frac{m}{s})}{(4kg+6kg)}=v_{vf}$

Now we solve the operations:

$2\frac{m}{s}=v_{vf}$

Now we determine the magnitude of the final velocity using the following formula:

$v_f=\sqrt[]{v^2_{hf}+v^2_{vf}}$

Substituting the values:

$v_f=\sqrt[]{(1.2\frac{m}{s})^2+(2\frac{m}{s})^2}$

Solving the operations:

$v_f=2.53\frac{m}{s}$

Now we substitute this in the formula for the kinetic energy and we get:

$\Delta K=\frac{1}{2}(4kg+6kg)(2.53\frac{m}{s})^2-(\frac{1}{2}(4kg)(5\frac{m}{s})^2+\frac{1}{2}(6kg)(2\frac{m}{s})^2)$

Solving the operations:

$\Delta K=32J-62J=-30J$

Therefore, there was a loss of 30J of kinetic energy.

6 0

1 year ago

In nuclear physics, what units are used to measure the radius of an atom? A. liters B. mililiters C. fermi D. grams?

RSB [31]

It's quite easy to get the answer. Think about it: liters and mililiters both are measure of volume - withdraw, grams are used to measure mass.
The one remained is C. fermi( femtometer) which is obviously correct answer.

7 0

3 years ago