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Bond [772]
2 years ago
13

As you go farther down the periodic table, the atoms get _______ and more ________.

Physics
2 answers:
Andrei [34K]2 years ago
4 0

Answer:

As we navigate down a group the atoms get bigger and bigger with more and more electrons. This means the outermost electrons get further and further away from the positively charged nucleus

Hope this helps!!!!

VMariaS [17]2 years ago
3 0

Answer:

As we navigate down a group the atoms get bigger and bigger with more and more electrons. This means the outermost electrons get further and further away from the positively charged nucleus.

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An ice cube is placed in a microwave oven.
Juli2301 [7.4K]

Answer:

It takes 266 seconds to melt the ice.

Explanation:

Given data

  • Power of the microwave oven (P): 125 Watt
  • Heat supplied to the ice (Q): 33,200 Joule
  • Time for the melting (t): to be determined

In order to determine the time required to melt the ice, we can use the following expression.

P = Q/t

t = Q / P = 33,200 J/ 125 W = 266 s

It takes 266 seconds to melt the ice.

5 0
2 years ago
The weather is warm and dry what changes would a cold front bring
MissTica
<span>A cold front separates a cold, dry air mass from a warm air mass.</span>
8 0
2 years ago
Read 2 more answers
. (a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110
7nadin3 [17]

Answer:

(a) the high of a hill that car can coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h is 47.6 m

(b) thermal energy was generated by friction is 1.88 x 10^{5} J

(C) the average force of friction if the hill has a slope 2.5º above the horizontal is 373 N

Explanation:

given information:

m = 750 kg

initial velocity, v_{0} = 110 km/h = 110 x 1000/3600 = 30.6 m/s\frac{30.6^{2} }{2x9.8}

initial height, h_{0} = 22 m

slope, θ = 2.5°

(a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h?

according to conservation-energy

EP = EK

mgh = \frac{1}{2} mv_{0} ^{2}

gh = \frac{1}{2} v_{0} ^{2}

h = \frac{v_{0} ^{2} }{2g}

  = 47.6 m

(b) If, in actuality, a 750-kg car with an initial speed of 110 km/h is observed to coast up a hill to a height 22.0 m above its starting point, how much thermal energy was generated by friction?

thermal energy = mgΔh

                         = mg (h - h_{0})

                         = 750 x 9.8 x (47.6 - 22)

                         = 188160 Joule

                         = 1.88 x 10^{5} J

(c) What is the average force of friction if the hill has a slope 2.5º above the horizontal?

f d  = mgΔh

f = mgΔh / d,

where h = d sin θ, d = h/sinθ

therefore

f = (mgΔh) / (h/sinθ)

 = 1.88 x 10^{5}/(22/sin 2.5°)

 = 373 N

8 0
3 years ago
A 5 meter long ladder leans against a wall. The bottom of the ladder slides away from the wall at the constant rate of 1 3 m/s.
Oksana_A [137]

Answer:9.75 m/s

Explanation:

Given

Length of ladder (L)=5 m

Foot the ladder is moving away with speed of \frac{\mathrm{d} x}{\mathrm{d} t}=13 m/s

From diagram

x^2+y^2=L^2------1

at x=3

y^2=25-9=16

y=4 m

Now differentiating equation 1 w.r.t time

2x\frac{\mathrm{d} x}{\mathrm{d} t}+2y\frac{\mathrm{d} y}{\mathrm{d} t}=0

x\frac{\mathrm{d} x}{\mathrm{d} t}=-y\frac{\mathrm{d} y}{\mathrm{d} t}

3\times 13=-4\times \frac{\mathrm{d} y}{\mathrm{d} t}

\frac{\mathrm{d} y}{\mathrm{d} t}=-\frac{3\times 13}{4}=-9.75 m/s

negative indicates distance is decreasing with time

5 0
2 years ago
An experiment yielded the above temperature and time information. what is the freezing point of the material in this experiment
Dmitry [639]
The answer is 0 degrees Celsius (0°C). It will be where the line flat lines the first time.  The second time would be the boiling point.  An experiment yielded the above temperature and time information. The freezing point of the material in this experiment if the material is a solid at time zero is 0 degrees Celsius (0°C) .
7 0
3 years ago
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