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BARSIC [14]
3 years ago
13

6. A 4 kg object hangs below a 6 kg object by a string of negligible mass. If the 6 kg object is pulled upward by a force of 440

N, what is the tension in the string?

Physics
1 answer:
MrRissso [65]3 years ago
7 0

Answer:

T =176 N

Explanation:

from diagram

F -(m_1+m_2_g) = (m_1+m_2_g)a

440 - (6+4)g = (6+4)a

a =\frac{440-10*9.8}{10}

a =34.2 m/s^2

frrom free body diagram of mass m2 = 4kg

T -m_2g =m_2a

T = m_2(g +a)

T = 4(9.81+34.2)

T =176 N

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A convex lens can produce a real image but not a viral image<br> a. true<br> b. false
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The answer is true a convex lens can produce a real image but not a viral image
5 0
3 years ago
Young's experiment is performed with light of wavelength 502 nm from excited helium atoms. Fringes are measured carefully on a s
PSYCHO15rus [73]

Answer:

Explanation:

wave length of light λ = 502 nm

screen distance D = 1.2 m

width of one fringe = 10.2 mm / 20

= .51 mm

fringe width = λ D / a  , a is separation of slits

Puting the values given

.51 x 10⁻³ =  502 x 10⁻⁹ x 1.2 / a

a = 502 x 10⁻⁹ x 1.2 / .51 x 10⁻³

= 1181.17 x 10⁻⁶ m

1.18 x 10⁻³ m

= 1.18 mm .

8 0
3 years ago
An object on a planet has a mass of 243 kg. What is the acceleration of the
Vesna [10]

Answer:

The gravitational acceleration of a planet of mass M and radius R

a = G*M/R^2.

In this case we have:

G = 6.67 x 10^-11 N (m/kg)^2

R = 2.32 x 10^7 m

M = 6.35 x 10^30 kg

Now we can compute:

a = (6.67*6.35/2.32^2)x10^(-11 + 30 - 2*7) m/s^2 = 786,907.32 m/s^2

The acceleration does not depend on the mass of the object.

3 0
3 years ago
A 5.0-kg block of wood is placed on a 2.0-kg aluminum frying pan. How much heat is required to raise the temperature of both the
Shalnov [3]

Heat required to raise the temperature of a given system is

Q = ms\Delta T

here we know that

m = mass

s = specific heat capacity

\Delta T = change in temperature

now as we know that

mass of wood = 5 kg

mass of aluminium pan = 2 kg

change in temperature = 45 - 20 = 25 degree C

specific heat capacity of wood = 1700 J/kg C

specific heat capacity of aluminium = 900 J/kg C

now here we will find the total heat to raise the temperature of both

Q = m_1s_1\Delta T_1 + m_2s_2\Delta T_2

Q = 5 * 1700 * 25 + 2 * 900 * 25

Q = 212500 + 45000

Q = 257500 J

So heat required to raise the temperature of the system is 257500 J

4 0
3 years ago
One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction
Sophie [7]

Answer:

F = 156.3 N

Explanation:

Let's start with the top block, apply Newton's second law

         F - fr = 0

         F = fr

         fr = 52.1 N

Now we can work  with the bottom block

In this case we have two friction forces, one between the two blocks and the other between the block and the surface. In the exercise, indicate that the two friction coefficients are equal

we apply Newton's second law

Y axis

        N - W₁ -W₂ = 0

        N = W₁ + W₂

as the two blocks are identical

        N = 2W

X axis

        F - fr₁ - fr₂ = 0

        F = fr₁ + fr₂

indicates that the lower block is moving below block 1, therefore the upper friction force is

          fr₁ = 52.1 N

          fr₁ = μ N

a

s the normal in the lower block of twice the friction force is

          fr₂ = μ 2N

          fr₂ = 2 μ N

          fr₂ = 2 fr₁

we substitute

          F = fr₁ + 2 fr₁

          F = 3 fr₁

          F = 3  52.1

          F = 156.3 N

7 0
3 years ago
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