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tangare [24]
3 years ago
8

The center of mass of a 0.30-kg (non-uniform) meter stick is located at its 45-cm mark. What is the magnitude of the torque (in

N⋅m) due to gravity if it is supported at the 28-cm mark? (Use g = 9.79 m/s2). (NEVER include units with the answer to ANY numerical question.)
Physics
1 answer:
lbvjy [14]3 years ago
6 0

Answer:

The magnitude of the torque due to gravity if it is supported at the 28-cm mark is 0.5 N-m.

Explanation:

Given that,

Mass of the meter stick, m = 0.3 kg

Center of mass is located at its 45 cm mark.

We need to find the magnitude of the torque due to gravity if it is supported at the 28-cm mark. Torque acting on the object is given by :

\tau=r\times F\\\\\tau=(45-28)\times 10^{-2}\times 0.3\times 9.79\\\\\tau=0.499\ N-m

or

\tau=0.5\ N-m

So, the magnitude of the torque due to gravity if it is supported at the 28-cm mark is 0.5 N-m.

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The energy your body gets from food is originally provided by which nuclear
Naily [24]

Answer:

Nuclear Fusion

Explanation:

The process releases energy because the total mass of the resulting single nucleus is less than the mass of the two original nuclei. The leftover mass becomes energy.

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3 years ago
Three positive charges A, B, and C, and a negative charge D are placed in a line as shown in the diagram. All four charges are o
polet [3.4K]

Answer:

a. charge C experiences the greatest net force, and charge B receives the smallest net force

b. ratio=9

Explanation:

<u>Electrostatic Force</u>

Two point-charges q_1 and q_2 separated a distance d will exert a force on each other of a magnitude given by the Coulomb's formula

\displaystyle F=\frac{k\ q_1\ q_2}{r^2}

Where k is the proportional constant of value

k=9*10^9\ N.m^2/c^2

The diagram provided in the question shows four identical charges (let's assume their value is Q) separated by identical distance (of value d). The force between the charges next to others is

\displaystyle F_1=\frac{k\ Q\ Q}{d^2}

\displaystyle F_1=\frac{k\ Q^2}{d^2}

The force between charges separated 2d is

\displaystyle F_2=\frac{k\ Q^2}{(2d)^2}

\displaystyle F_2=\frac{k\ Q^2}{4d^2}

And the force between the charges A and D is

\displaystyle F_3=\frac{k\ Q^2}{(3d)^2}

\displaystyle F_3=\frac{k\ Q^2}{9d^2}

Now, let's analyze each charge and the force applied to them by the others

Let's recall equally signed charges repel each other and differently signed charges attrach each other

Charge A. It receives force to the left from B and C and to the right from D

\displaystyle F_A=-F_1-F_2+F_3=-\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}

\displaystyle F_A=\frac{k\ Q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})

\displaystyle F_A=-\frac{41}{36}F_1

Charge B. It receives force to the right from A and D and to the left from C

\displaystyle F_B=F_1-F_1+F_2=\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{4d^2}

\displaystyle F_B=\frac{1}{4}F_1

Charge C. It receives forces to the right from all charges.

\displaystyle F_C=F_2+F_1+F_1=\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{d^2}

\displaystyle F_C=\frac{9}{4}F_1

Charge D. It receives forces to the left from all charges

\displaystyle F_D=-F_3-F_2-F_1=-\frac{k\ Q^2}{9d^2}-\frac{k\ Q^2}{4d^2}-\frac{k\ Q^2}{d^2}

\displaystyle F_D=-\frac{49}{36}F_1

Comparing the magnitudes of each force is just a matter of computing the fractions

\displaystyle \frac{41}{36}=1.13,\ \frac{1}{4}=0.25,\ \frac{9}{4}=2.25,\ \frac{49}{36}=1.36

a.

We can see the charge C experiences the greatest net force, and charge B receives the smallest net force

b.

The ratio of the greatest to the smallest net force is

\displaystyle \frac{\frac{9}{4}}{\frac{1}{4}}=9

The greatest force is 9 times the smallest net force

7 0
3 years ago
MULTIPLE CHOICE
Juli2301 [7.4K]

Answer:

answers d

Explanation:

hopes its healp

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2 years ago
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C, they didn't know any better
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What does the diameter in meters need to be of a round parachute that the designer has determined needs to have an area of 500 s
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Area of a circle is
A= pi r^2
so
500m^2 = 3.14 r2
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to 3 sig fig
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