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klemol [59]
3 years ago
12

A man jogs at a velocity of 2.5 m/s west for 1,200 s. He then walks at a velocity 1.0 m/s east for 500 s and then stops to rest

. What is the displacement of man when he stops to rest?
Physics
1 answer:
olganol [36]3 years ago
5 0

Speed of man towards West =2.5 m/s

Time to travel in the given direction = 1200 s

Total displacement towards west

d_1 = v_1 t

d_1 = 2.5 \times 1200 = 3000 m

Now speed of man towards East = 1 m/s

time to travel in east direction = 500 s

total displacement towards East

d_2 = v_2 t

d_2 = 1 \times 500 = 500 m

Now total displacement will be

d = d_1 - d_2

d = 3000 - 500 = 2500 m

so it is 2500 m towards west

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rusak2 [61]

Answer : The volume of a sample of 4.00 mol of copper is 28.5cm^3

Explanation :

First we have to calculate the mass of copper.

\text{ Mass of copper}=\text{ Moles of copper}\times \text{ Molar mass of copper}

\text{ Mass of copper}=(4.00moles)\times (63.5g/mole)=254g

Now we have to calculate the volume of copper.

Formula used :

Density=\frac{Mass}{Volume}

Now put all the given values in this formula, we get:

8.92\times 10^3kg/m^3=\frac{254g}{Volume}

Volume=\frac{254g}{8.92\times 10^3kg/m^3}=2.85\times 10^{-2}L=2.85\times 10^{-2}\times 10^3cm^3=28.5cm^3

Conversion used :

1kg/m^3=1g/L\\\\1L=10^3cm^3

Therefore, the volume of a sample of 4.00 mol of copper is 28.5cm^3

7 0
3 years ago
Read 2 more answers
A hollow cylinder with an inner radius of and an outer radius of conducts a 3.0-A current flowing parallel to the axis of the cy
Artemon [7]

Complete Question:

A hollow cylinder with an inner radius of 4.0mm and an outer radius of 30mm conducts a 3.0-A current flowing parallel to the axis of the cylinder. If the current density is uniform throughout the wire, what is the magnitude of the magnetic field at a point 12mm from its center?

Answer:

The magnitude of the magnetic field = 7.24 μT

Explanation:

Inner radius, a = 4.0 mm = 0.004 m

Outer radius, b = 30 mm = 0.03 m

Radius, r = 12 mm = 0.012 m

let h² = b² - a²

h² = 0.03² - 0.004²

h² = 0.000884

Let d² = r² - a²

d² = 0.012² - 0.004²

d² = 0.000128

Current I = 3A

μ = 4π * 10⁻⁷

The magnitude of the magnetic field is given by:

B = \frac{\mu I d^{2} }{2\pi r h^{2} } \\B =  \frac{4\pi * 10^{-7}   * 3* 0.000128^{2} }{2\pi *0.012* 0.000884^{2} }

B = 7.24 * 10⁻⁶T

B = 7.24 μT

7 0
3 years ago
Two scales on a nondigital voltmeter measure voltages up to 20.0 and 30.0 V, respectively. The resistance connected in series wi
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Answer:

Resistance of the circuit is 820 Ω

Explanation:

Given:

Two galvanometer resistance are given along with its voltages.

Let the resistance is "R" and the values of voltages be 'V' and 'V1' along with 'G' and 'G1'.

⇒ V=20\ \Omega,\ V_1=30\ \Omega

⇒ G=1680\ \Omega,\  G_1=2930\ \Omega

Concept to be used:

Conversion of galvanometer into voltmeter.

Let G be the resistance of the galvanometer and I_g the maximum deflection in the galvanometer.

To measure maximum voltage resistance R is connected in series .

So,

⇒ V=I_g(R+G)

We have to find the value of R we know that in series circuit current are same.

For G=1680                                    For G_1=2930

⇒ I_g=\frac{V}{R+G}   equation (i)                ⇒ I_g=\frac{V_1}{R+G_1} equation (ii)

Equating both the above equations:

⇒ \frac{V}{R+G} = \frac{V_1}{R+G_1}

⇒ V(R+ G_1) = V_1 (R+G)

⇒ VR+VG_1 = V_1R+V_1G

⇒ VR-V_1R = V_1G-VG_1

⇒ R(V-V_1) = V_1G-VG_1

⇒ R =\frac{V_1G-VG_1}{(V-V_1)}

⇒ Plugging the values.

⇒ R =\frac{(30\times 1680) - (20\times 2930)}{(20-30)}

⇒ R =\frac{(50400 - 58600)}{(-10)}

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3 years ago
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attashe74 [19]

Answer:

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Explanation:

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The time period of pendulum is defined as time taken by the pendulum to complete one full oscillation . it is denoted by T.

By <em>Huygens law of period of pendulum</em>,

T = 2π\sqrt{\frac{L}{g} }   eqn 1

where L is the length of pendulum,

          g is acceleration due to gravity

<em>Period of pendulum is independent of the mass of pendulum,</em>

<em />

Substituting values in eqn 1

T = 2π \sqrt{\frac{1.02}{9.8\\} }

T =   2.02 s

<em>Time period of pendulum is 2.02 s.</em>

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Answer:

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Explanation:

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