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3 years ago

5

A 13-N weight and a 12-N weight are connected by a massless string over a massless, frictionless pulley. The 13-N weight has a d

ownward acceleration with magnitude equal to that of a freely falling body times:
A. 1
B. 1/12
C. 1/13
D. 1/25
E. 13/25

1 answer:

drek231 [11]3 years ago

4 0

Answer:

D

Explanation:

We are given that

Weight, $w_1=13 N$

$w_2=12 N$

We have to find the acceleration of 13 N weight .

$m_1=\frac{w_1}{g}=\frac{13}{g}$ kg

$m_2=\frac{w_2}{g}=\frac{12}{g} kg$

Net force, $w_1-w_2=13-12=1 N$

Total mass, $m=m_1+m_2=\frac{13}{g}+\frac{12}{g}=\frac{25}{g}$ kg

Acceleration, $a=\frac{Net\;force}{total\;mass}$

Using the formula

$a=\frac{1}{\frac{25}{g}}=\frac{1}{25}g$

Where g=Acceleration due to gravity

Hence, option D is true.

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If an unknown element has a mass of 17 and contains 6 neutrons, how many protons does it have ?

lbvjy [14]

The mass number is the total number of protons and neutrons within an atom and since we know that the unknown element has 6 neutrons, we can simply subtract the number of neutrons from the mass number to get the number of protons.

17 - 6 = 11

There are 11 protons in this unknown element.

Extra:

The number of protons (+) and electrons (-) are equal in a neutral atom so since you know that there are 11 protons you also know that there are 11 electrons. On the periodic table, the element with 11 electrons is Na or Sodium.

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3 years ago

A space vehicle of mass m has a speed v. At some instant, it separates into two pieces, each of mass 0.5m. One of the pieces is

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Answer:

W = ½ m v²

Explanation:

In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation

We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved

initial instant. before separation

p₀ = m v

final attempt. after separation

$p_{f}$ = m /2 0 + m /2 v_{f}

p₀ = p_{f}

m v = m /2 $v_{f}$

v_{f}= 2 v

this is the speed of the second part of the ship

now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body

initial energy

K₀ = ½ m v²

final energy

$K_{f}$ = ½ m/2 0 + ½ m/2 v_{f}²

K_{f} = ¼ m (2v)²

K_{f} = m v²

the expression for work is

W = ΔK = K_{f} - K₀

W = m v² - ½ m v²

W = ½ m v²

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3 years ago

Ship A is located 4.1 km north and 2.3 km east of ship B. Ship A has a velocity of 22 km/h toward the south and Ship B has a vel

castortr0y [4]

Answer:

a) x component = -31.25 km/hr

b) y component = 46.64 km/hr

Explanation:

Given data:

A position is 4km north and 2.5 km east to B

Ship A velocity = 22 km/hr

ship B velocity = 40 km/hr

A velocity wrt to velocity of B

$\vec{V_{AB}} =\vec{V_A} - \vec{V_B}$

$\vec{V_A} = 22 km/hr$

$\vec{V_B} = 40 cos38\hat{i} + 40sin 38 \hat{j}$

$= 31.52\hat{i} + 24.62 \hat{j}$

putting respective value to get velocity of A with respect to B

$\vec{V_{AB}} = -22 \hat {j} - (31.52\hat{i} + 24.62 \hat{j})$

$\vec{V_{AB}} = -31.52\hat{i} - 46.62\hat{j}$

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4 years ago

What are three areas of science that rely on physical science?

yanalaym [24]

"physical science" is a branch of science that is based on practical tests and explanations of the different phenomena. It is based on scientific evidence and tests/experiments.

Some of the branches that are based on physical science are:
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3 years ago

Read 2 more answers

What is the average emf induced in a circular wedding ring of diameter 2.2 cm if the ring is at the center of the MRI magnet coi

Minchanka [31]

Missing information

Consider a superconducting MRI magnet for which the magnet field decreases from 8.0 T to nearly 0 in 20 s

Answer:

0.152 mV

Explanation:

Change in magnetic field, dB=8-0=8 T

Time interval, dt=20 s

Diameter of ring= 2.2cm hence radius is 1.1 cm equivalent to 0.011

$\epsilon=nA\frac {dB}{dt}$ where n is the number of terms and for this question n=1 and A is area which is given by $\pi r^{2}$ hence

$\epsilon=n\pi r^{2} \frac {dB}{dt}$ and substituting the values

$\epsilon=1\pi 0.011^{2}\times \frac {8}{20}=1.52\times 10^{-4} V=0.152 mV$

Therefore, induced emf is 0.152 mV

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4 years ago