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SCORPION-xisa [38]
2 years ago
9

Carol tries to push a bookshelf across the room. As much as she tries, she can not get the bookshelf to move. Which statement be

st describes why the
bookshelf is not moving?
O
A Carol did not apply enough force to the bookshelf and so the forces are unbalanced.
B. Carol did not apply enough force to accelarate the bookshelf and so the forces are still balanced.
0
C. Carol applied too much force to the bookshelf and now the forces on the bookshelf are balanced
D. Carol applied too much force to the bookshelf and now the forces on the bookshelf are unbalanced
Physics
1 answer:
Anettt [7]2 years ago
7 0

Answer:

B.

Explanation:

Carol did not apply enough force to accelarate the bookshelf and so the forces are still balanced.

Hope this helps!! :)

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A ________ ion is created when an electron is removed.
Oksanka [162]
Anion is created when electrons are lost.
5 0
3 years ago
If the units of your answer are kg ∙ m2/s3, which of the following types of quantities could your answer be?
kirza4 [7]

Answer:

  D.  power

Explanation:

  kg represents mass

  (m/v)² represents velocity squared

Then kg·m²/s² represents mass·velocity² = <em>kinetic energy</em> or <em>potential energy</em> or <em>work</em>.

  kg·m²/s³ will be the <em>rate of doing work</em>, which is power

4 0
2 years ago
Read 2 more answers
(1) find the density of a substance if the mass of the substance is 150kg and dimension 20mby10mby5m
Vikentia [17]

Answer:

0.15kg/m³

Explanation:

Density = mass/ volume

Given that

Mass = 150kg

Note that volume = length x breadth x height

Volume = 20 x 10 x 5

Volume = 1000m³

Density = mass ➗ volume

Density = 150kg ➗ 1000m³

Density = 0.15kg/m³

I hope this was helpful, Please mark as brainliest  

3 0
3 years ago
Telephone signals are often transmitted over long distances by microwaves. What is the frequency of microwave radiation with a w
amm1812

Answer:

1) f= 8.6 GHz

2) t= 0.2 ms

Explanation:

1)

  • Since microwaves are electromagnetic waves, they move at the same speed as the light in vacuum, i.e. 3*10⁸ m/s.
  • There exists a fixed relationship between the frequency (f) , the wavelength (λ) and the propagation speed in any wave, as follows:

        v = \lambda * f (1)

  • Replacing by the givens, and solving for f, we get:

       f =\frac{c}{\lambda} =\frac{3e8m/s}{0.035m} = 8.57e9 Hz (2)

⇒     f = 8.6 Ghz (with two significative figures)

2)

  • Assuming that the microwaves travel at a constant speed in a straight line (behaving like rays) , we can apply the definition of average velocity, as follows:

       v =\frac{d}{t} (3)

       where v= c= speed of light in vacuum = 3*10⁸ m/s

       d= distance between mountaintops = 52 km = 52*10³ m

  • Solving for t, we get:

       t = \frac{d}{c} = \frac{52e3m}{3e8m/s} = 17.3e-5 sec = 0.173e-3 sec = 0.173 ms (4)

       ⇒  t = 0.2 ms (with two significative figures)

6 0
2 years ago
A prismatic bar AB of length L and solid circular cross section (diameter d) is loaded by a distributed torque of constant inten
Lyrx [107]

Answer:

a) the maximum shear stress τ_{max} the bar is 16T_{max} /πd³

b) the angle of twist between the ends of the bar is 16tL² / πGd⁴  

Explanation:

Given the data in the question, as illustrated in the image below;

d is the diameter of the prismatic bar of length AB

t is the intensity of distributed torque

(a) Determine the maximum shear stress tmax in the bar

Maximum Applied torque  T_max = tL

we know that;

shear stress τ = 16T/πd³

where d is the diameter

so

τ_{max} = 16T_{max} /πd³

Therefore, the maximum shear stress τ_{max} the bar is 16T_{max} /πd³

(b) Determine the angle of twist between the ends of the bar.

let theta (\theta) be the angle of twist

polar moment of inertia I_p} = πd⁴/32

now from the second image;

lets length dx which is at distance of "x" from "B"

Torque distance x

T(x) = tx

Elemental angle twist = d\theta = T(x)dx / GI_{p}

so

d\theta = tx.dx / G(πd⁴/32)

d\theta = 32tx.dx / πGd⁴

so total angle of twist \theta will be;

\theta =  \int\limits^L_0  \, d\theta

\theta =  \int\limits^L_0  \, 32tx.dx / πGd⁴

\theta = 32t / πGd⁴  \int\limits^L_0  \, xdx

\theta = 32t / πGd⁴ [ L²/2]

\theta = 16tL² / πGd⁴  

Therefore,  the angle of twist between the ends of the bar is 16tL² / πGd⁴  

7 0
3 years ago
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