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Marat540 [252]
3 years ago
11

Choose the correct statement about the periodic table's organization from the choices below.

Physics
1 answer:
Ksivusya [100]3 years ago
7 0
We need to be able to the see the choices to help you answer the question:)
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To increase the gravitational force between the two objects above, I could
egoroff_w [7]

Answer:

Option A

Explanation:

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3 years ago
Which of the following, by scientific definition is NOT work? lifting boxes of books, pedaling on your bike, holding a book over
Karolina [17]

Answer:

holding a book over your head

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3 years ago
ListenA bicycle and its rider have a combined mass of 80. kilograms and a speed of 6.0 meters per second. What is the magnitude
Setler [38]

Answer:

a) 1.2\times 10^2\ N

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

a = Acceleration

v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{0-6}{4}\\\Rightarrow a=-1.5\ m/s^2

The acceleration of the bicycle and rider is -1.5 m/s²

Force

F=ma\\\Rightarrow F=80\times -1.5\\\Rightarrow F=-120\ N=-1.2\times 10^2\ N

The magnitude of the average force needed to bring the bicycle and its rider to a stop is 1.2\times 10^2\ N

3 0
3 years ago
You are an industrial engineer with a shipping company. As part of the package- handling system, a small box with mass 1.60 kg i
Kisachek [45]

Answer:

v = 0.84m/s, v(max)= 0.997m/s

Explanation:

Initial work done by the spring, where c is the compression = 0.28m:

W_s = \frac{1}{2}kc^2

Work lost to friction:

W_f =\mu mg(c-x)

Energy:

E = W_s-W_f=\frac{1}{2}kc^2 -\mu mg(c-x)=\frac{1}{2}mv^2+\frac{1}{2}kx^2

(a) Solve for v:

v=\sqrt{\frac{k}{m}(c^2-x^2)-2\mu g(c-x)}

(b) Solve \frac{dv}{dx}=0 for x:

\frac{dv}{dx}=\frac{\mu g-\frac{k}{m}x}{\sqrt{\frac{k}{m}(c^2-x^2)-2\mu g(c-x)}}

\frac{dv}{dx}=0 if:

\mu g-\frac{k}{m}x = 0

x_{max} = \frac{\mu gm}{k}

v_{max}=\sqrt{\frac{k}{m}c^2-2\mu gc+(\mu g)^2\frac{m}{k}}

6 0
4 years ago
What caused day and night on a planet
Volgvan

Answer:

The rotation of a planet around it's sun

7 0
3 years ago
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