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Sunny_sXe [5.5K]
3 years ago
11

The farther apart elements are on the periodic table the more likely they are to form_____bonds.

Physics
1 answer:
irga5000 [103]3 years ago
4 0
The answer is A, ionic bonds. Hope this helped you 

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Please help ASAP it’s really nedded
iren2701 [21]

Answer:120

Explanation:

3 0
3 years ago
Read 2 more answers
The chart shows the power ratings of 3 clothes dryers and the time needed to dry a load of 10 towels.
atroni [7]

Answer:

12010

Explanation:

Given:

power    time

4,370     25

4,420     22

4,150      26

To find: difference in energy used between the dryer that uses the greatest amount of energy and the dryer that uses the least amount

Solution:

Energy = power × time

For clothes dryer x:

energy = 4370 × 25 = 109250

For clothes dryer y:

energy = 4420 × 22 = 97240

For clothes dryer z:

energy = 4150 × 26 = 107900

Dryer x uses the greatest amount of energy and dryer y uses the least amount of energy.

Difference in energy used between the dryer that uses the greatest amount of energy and the dryer that uses the least amount = 109250 - 97240 = 12010

6 0
3 years ago
A boy coasts down a hill on a sled, reaching a level surface at the bottom with a speed of 5.9 m/s. if the coefficient of fricti
bagirrra123 [75]
Frictional force = vertical force x coeff of friction = 590N x 0.045 = 26.55N

Mass of boy and sled = 590N / g = 590N / 9.8m/s^2 = 60.20 kg

Deceleration due to friction = 26.55N / 60.20kg = 0.44 m/s^2.

For constant acceleration we have:

v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and ti is time. In this case v = 0, u = 5.9 m/s, a = -0.44 m/s^2. So we have
0 = 5.9 - 0.44t
which gives t = 5.9 / 0.44 = 13.409 s.

Distance traveled in this time d = ut + 0.5at^2 = 5.9 x 13.409 - 0.5 x 0.44 x 13.409^2 = 39.56 m

<span>Answer: </span>40 meters
8 0
3 years ago
A 15-µF capacitor and a 25-µF capacitor are connected in parallel, and charged to a potential difference of 60 V. How much energ
Alborosie

Answer:

Energy stored, E = 0.072 J

Explanation:

Given that,

Capacitance, C_1=15\ \mu F

Capacitance, C_2=25\ \mu F

These two capacitor are connected in parallel, and charged to a potential difference of, V = 60 volts

We know that in parallel combination of capacitor, the equivalent capacitance is given by :

C=C_1+C_2\\\\C=(15+25)\ \mu F\\\\C=40\times 10^{-6}\ F

The energy stored in the capacitor is given by :

E=\dfrac{1}{2}CV^2\\\\E=\dfrac{1}{2}\times 40\times 10^{-6}\times (60)^2\\\\E=0.072\ J

So, the energy stored in the capacitor in this capacitor combination is 0.072 J.

4 0
3 years ago
A 2.5 kg , 20-cm-diameter turntable rotates at 150 rpm on frictionless bearings. Two 500 g blocks fall from above, hit the turnt
emmainna [20.7K]

The concepts required to solve this problem are those related to the conservation of the angular momentum and the moment of inertia of the disk. We will begin by calculating the moment of inertia of the disc, then the moment of inertia of the disc after the two two blocks hits and sticks to the edges of the turn table. In the end we will apply the conservation theorem.

The radius is given as,

R = \frac{20cm}{2} = 10cm = 0.1m

When a block falls from above and sticks to the turn table, the moment inertia of the turntable increases.

Since two blocks are stick to the turn table, the total final moment of inertia of the turntable is the sum moment of inertias of individual turntable, and two blocks.

I_1 = \frac{1}{2} MR^2

I_1 = \frac{1}{2} (2.5)(0.1)^2

I_1 = 0.0125kg \cdot m^2

The moment of inertia of each block is

I_0 = mR^2

Total moment of inertia of two block is

I_0' = 2mR^2

The final moment of inertia of the turn table is

I_2 = I_1 +I'0

I_2 = I_1 +2mR^2

I_2 = 0.01kg\cdot m^2 + 2(500*10^{-3}kg)(0.1m)^2

I_2 = 0.0225kg\cdot m^2

From the conservation of the angular momentum, the initial angular momentum of the system is equal to final angular momentum of the system,

Rearrange the equation we have that

I_1\omega_2 = I_2\omega_2

\omega_2 = \frac{I_1\omega_2}{I_2}

\omega_2 = \frac{0.01*150rpm}{0.0225}

\omega_2 = 66.67rpm

The magnitude of the turntable's angular velocity is 66.67rpm

3 0
3 years ago
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