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3 years ago

The number of outer shell electrons determines the chemical properties of an element?

1 answer:

Ivan3 years ago

3 0

Yes, it is true to a certain that the number of outer shell electrons determines the chemical properties of an element, only because this determines how the element interacts with other elements.

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A car of mass 500 kg is moving at a speed of 1.2 m/s. A man pushes the car, increasing the speed to 2 m/s. How much work did the

Vikki [24]

Answer:

Option B - 360 J

Explanation:

One of the formulas used to find out how much work was done is :

E = 0.5×m×v² (where m is the mass and v is the velocity)

Now we substitute values from the question and simplify :

E = 0.5×500×1.2²

E = 250×1.2²

E = 250×1.44

E = 360 J

So our answer is Option B

Hope this helped and have a good day

5 0

2 years ago

The current that charges a capacitor transfers energy that is stored in the capacitor’s electric field. Consider a 2.0 μF capaci

lapo4ka [179]

Answer:

the capacitor voltage is V = 20 V

Explanation:

Given,

Capacitance of the capacitor = 2.0 μF

energy stored = 200 W

time (t) =2.0 μs

Capacitor voltage = ?

$\dfrac{dE}{dt}=200 W$

$E =200\times 2 \times 10^{-6}$

$E =400 \times 10^{-6}\ J$

we know,

$E = \dfrac{1}{2}CV^2$

$V =\sqrt{\dfrac{2E}{C}}$

$V =\sqrt{\dfrac{2\times 400 \times 10^{-6}}{2\times 10^{-6}}}$

$V =\sqrt{400}$

V = 20 V

so , the capacitor voltage is V = 20 V

7 0

3 years ago

Which type of environmental science career involves collecting information about how human events impact the environment?

SVEN [57.7K]

C. environmental engineer

4 0

2 years ago

Read 2 more answers

Which of the following is an example of a conductor?

nordsb [41]

Answer: B. a gold chain

7 0

2 years ago

you have been called to testify as a as an expert witness in a trial involving a head-on collision Car A weighs 1515 pounds and

GrogVix [38]

Answer:

70.6 mph

Explanation:

Car A mass= 1515 lb

Car B mass=1125 lb

Speed of car B is 46 miles/h

Distance before locking, d=19.5 ft

Coefficient of kinetic friction is 0.75

Initial momentum of car B=mv where m is mass and v is velocity in ft/s

46 mph*1.46667=67.4666668 ft/s

$Momentum_B=1125*67.4666668 ft/s$

Initial momentum of car A is given by

$Momentum_A=1515v_a$ where $v_a$ is velocity of A

Taking East as positive and west as negative then the sum of initial momentum is

$1515v_a-(1125*67.4666668 ft/s)$

The common velocity is represented as $v_c$ hence after collision, the final momentum is

$Momentum_final=(m_a+m_b)v_c=(1515+1125)v_c=2640v_c$

From the law of conservation of linear momentum, sum of initial and final momentum equals each other hence

$1515v_a-(1125*67.4666668 ft/s)= 2640v_c$

The acceleration of two cars $a=-\mug=-0.75*32.17=-24.1275 ft/s^{2}$

From kinematic equation

$v^{2}=u^{2}+2as$ hence

$v^{2}-u^{2}=2as$

$0^{2}-(v_c)^{2}=2*-24.1275*19.5$

$v_c=\sqrt{2*24.1275*19.5}=30.67 ft/s$

Substituting the value of $v_c$ in equation $1515v_a-(1125*67.4666668 ft/s)= 2640v_c$

$1515v_a-(1125*67.4666668 ft/s)= 2640*30.67$

$1515v_a=(1125*67.4666668 ft/s)+2640*30.67$

$v_a=\frac {156868.8}{1515}=103.5438 ft/s$

$\frac {103.5438}{1.46667}=70.59787 mph\approx 70.60 mph$

3 0

3 years ago