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Elodia [21]
2 years ago
14

Two blocks are attached to opposite ends of a massless rope that goes over a massless, frictionless, stationary pulley. One of t

he blocks, with a mass of 1.0 kg accelerates downward at 34g. What is the mass of the other box?

Physics
1 answer:
Marysya12 [62]2 years ago
3 0

Answer:

<h2>1/7 kg</h2>

Explanation:

Find the diagram attached for better understanding of the question.

Given the mass of one of the blocks to be 1.0kg and accelerates downward at 3/4g.

g = acceleration due to gravity.

Let the block accelerating downward be M, mass of the other body be 'm' and the acceleration of the body M be 'a'.

M = 1.0 kg and a = 3.4g

According to newton's second law; \sum fy = ma_y

For body of mass m;

T - mg = ma ... (1)

For body of mass M;

Mg - T = Ma ... (2)

Adding equation 1 ad 2;

+Mg -mg = ma + Ma

Ma-Mg = -mg-ma

M(a-g) = -m(a+g)

Substituting M = 1.0 kg and a = 3/4g into the resulting equation;

3/4 g-g = -m(3/4 g+g)

3/4 g-g = -m(7/4 g)

-g/4 = -m(7/4 g)

1/4 = 7m/4

28m = 4

m = 1/7 kg

Therefore the mass of the other box is 1/7 kg

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The magnitude J(r) of the current density in a certain cylindrical wire is given as a function of radial distance from the cente
kipiarov [429]

Answer:

I=68.31\times 10^{-6}\ A

Explanation:

Given that

J(r) = Br

We know that area of small element

dA = 2 π dr

I = J A

dI = J dA

Now by putting the values

dI = B r . 2 π dr

dI= 2π Br² dr

Now by integrating above equation

\int_{0}^{I}dI= \int_{r_1}^{r_2}2\pi Br^2 dr

I={2\pi B}\times \dfrac{r_2^3-r_1^3}{3}

Given that

B= 2.35 x 10⁵ A/m³

r₁ = 2 mm

r₂ = 2+ 0.0115 mm

r₂ = 2.0115 mm

I={2\pi B}\times \dfrac{r_2^3-r_1^3}{3}

By putting the values

I={2\pi \times 2.35 \times 10^5 }\times \dfrac{(2.0115\times 10^{-3})^3-(2\times 10^{-3})^3}{3}\ A

I=68.31\times 10^{-6}\ A

7 0
2 years ago
Read 2 more answers
The sun transfers energy to the earth by radiation at a rate of approximately 1.00 kW per square meter of surface.
Mashutka [201]

Answer:

1320336992.2512 m²

1320.33 kilometers or 509.79 miles

Explanation:

Energy transferred by the sun

W=0.24\times 1\times 10^3=240\ W/m^2

Energy required by the United States is 1\times 10^{19}\ J/yr (assumed)

Power

P=\frac{E}{t}\\\Rightarrow P=\frac{1\times 10^{19}}{365.25\times 24\times 3600}\\\Rightarrow P=316880878140.2895\ W

Area

A=\frac{P}{W}\\\Rightarrow A=\frac{316880878140.2895}{240}\\\Rightarrow A=132033699.2512\ m^2

Area of the solar collector would be 1320336992.2512 m²

Converting to km²

1\ m^2=\frac{1}{1000\times 1000}\ km^2

1320336992.2512\ m^2=1320336992.2512\times \frac{1}{1000\times 1000}\ km^2=1320.33\ km^2

Converting to mi²

1\ m^2=\frac{1}{1609.34\times 1609.34}\ mi^2

1320336992.2512\ m^2=1320336992.2512\times \frac{1}{1609.34\times 1609.34}\ mi^2=509.79\ mi^2

Each side of the square would be 1320.33 kilometers or 509.79 miles

4 0
2 years ago
A fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)=2.30mmcos[(6
Shalnov [3]

Answer:

a. y(x,t)= 2.05 mm cos[( 6.98 rad/m)x + (744 rad/s).

b. third harmonic

c. to calculate frequency , we compare with general wave equation

y(x,t)=Acos(kx+ωt)

from ωt=742t

ω=742

ω=2*pi*f

742/2*pi

f=118.09Hz

Explanation:

A fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)=2.30mmcos[(6.98rad/m)x+(742rad/s)t]. Being more practical-minded, you measure the rope to have a length of 1.35 m and a mass of 3.38 grams. Assume that the ends of the rope are held fixed and that there is both this traveling wave and the reflected wave traveling in the opposite direction.

A) What is the wavefunction y(x,t) for the standing wave that is produced?

B) In which harmonic is the standing wave oscillating?

C) What is the frequency of the fundamental oscillation?

a. y(x,t)= 2.05 mm cos[( 6.98 rad/m)x + (744 rad/s).

b. lambda=2L/n

when comparing the wave equation with the general wave equation , we get the wavelength to be

2*pi*x/lambda=6.98x

lambda=0.9m

we use the equation

lambda=2L/n

n=number of harmonics

L=length of string

0.9=2(1.35)/n

n=2.7/0.9

n=3

third harmonic

c. to calculate frequency , we compare with general wave equation

y(x,t)=Acos(kx+ωt)

from ωt=742t

ω=742

ω=2*pi*f

742/2*pi

f=118.09Hz

8 0
2 years ago
Milk containing 3.7% fat and 12.8% total solids is to be evaporated to produce a product containing 7.9% fat. What is the yield
ratelena [41]

Answer:

the yield of product is YP=46.835 % and the concentration of solids is

Cs = 27.33%

Explanation:

Assuming that all the solids and fats remains in the milk after the evaporation, then the mass of product mP will be

Mass of fat in 100 kg of milk = 100 kg* 0.037 = mP* 0.079

mP = 100 kg* 0.037/0.079  =  46.835 kg

then the yield YP of the product is

YP= mP / 100 kg =  46.835 kg / 100 kg = 46.835 %

YP= 46.835 %

the concentration of solids Cs is

Mass of solids in 100 kg of milk = 100 kg* 0.128 = 46.835 kg * Cs

Cs = 100 kg* 0.128 / 46.835 kg  = 0.2733 = 27.33%

Cs = 27.33%

3 0
2 years ago
Water is contained in a closed, rigid 0.2 m 3 tank at an initial pressure of 5 bar and a quality of 50%. Heat transfer occurs un
elena55 [62]

Answer:

Final mass=0.89kg

Final pressure=5.6bar

Explanation:

To find mass,m=v/v1

But v1=vf + x(vg-vf)

Vf= 0.001093m^3/kg

Vg= 0.3748m^3/kg

V1= 0.001093+0.5(0.3748-0.001093)

V1= 0.225m^3/kg

M= 0.20/0.225 =0.89kg

Final pressure will be:

V/V1= P/P1

Cross multiply

VP1=V1P

P1= 0.225×5/0.2

P1=:5.6 bar

7 0
2 years ago
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