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VikaD [51]
3 years ago
15

1. An ideal ammeter should have a resistance of______________ while an ideal voltmeter should have a resistance of _____________

____.
2. Explain why these are desirable attributes for the respective measuring instruments.
Physics
1 answer:
Masja [62]3 years ago
4 0

Answer:

Explanation:

1. An ideal ammeter has very small or almost zero resistance. As the resistance is small the maximum current can pass through the ammeter which it can read it.

2. An ideal voltmeter has very large or infinite resistance. As the resistance is very large so the maximum voltage drops across the resistor and gives the accurate reading.

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Which change would result in a stronger electromagnet?
Sonbull [250]
I'd say B.) Increasing the voltage of the battery.
6 0
3 years ago
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How do you find the kinetic energy of an object? ​
Sergeeva-Olga [200]

Answer:

KE = 1/2 * m * v^{2}

Explanation:

use the formula:

KE = 1/2 * m * v^{2}

KE = kinetic energy in joules (J)

m = mass in kg

v = velocity in m/s

5 0
2 years ago
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If a 1000-pound capsule weighs only 165 pounds on the moon, how much work is done in propelling this capsule out of the moon's g
Bas_tet [7]

Answer:

  W = 1,307 10⁶ J

Explanation:

Work is the product of force by distance, in this case it is the force of gravitational attraction between the moon (M) and the capsule (m₁)

              F = G m₁ M / r²

              W = ∫ F. dr

              W = G m₁ M ∫ dr / r²

we integrate

             W = G m₁ M (-1 / r)

                 

We evaluate between the limits, lower r = R_ Moon and r = ∞

           W = -G m₁ M (1 /∞ - 1 / R_moon)

            W = G m1 M / r_moon

Body weight is

             W = mg

             m = W / g

The mass is constant, so we can find it with the initial data

For the capsule

            m = 1000/32 = 165 / g_moon

            g_moom = 165 32/1000

            .g_moon = 5.28 ft / s²

I think it is easier to follow the exercise in SI system  

           W_capsule = 1000 pound (1 kg / 2.20 pounds)

           W_capsule = 454 N

           W = m_capsule g

           m_capsule = W / g

           m = 454 /9.8

           m_capsule = 46,327 kg

Let's calculate

          W = 6.67 10⁻¹¹ 46,327   7.36 10²² / 1.74 10⁶

          W = 1,307 10⁶ J

7 0
3 years ago
Express the equilibrium rule in words ???
spin [16.1K]
Whan object is at equilibrium, then the forces are balanced. Balanced is the key word that is used to describe equilibrium situations.
6 0
3 years ago
A cylindrical tank of methanol has a mass of 70 kg and a volume of 75 L. Determine the methanol’s weight, density, and specific
dem82 [27]

Answer:

Weight=686.7N, \rho=933kg/m^{3}, S.G.=0.933, F=17.5N

Explanation:

So, the first value the problem is asking us for is the weight of methanol. (This is supposing there is a mass of methanol of 70kg inside the tank). We can find this by using the formula:

W=mg

so we can substitute the data the problem provided us with to get:

W=70kg(9.81m/s^{2})

which yields:

W=686.7N

Next, we need to find the density of methanol, which can be found by using the following formula:

\rho=\frac{m}{V}

we know the volume of methanol is 75L, so we can convert that to m^{3} like this:

75L*\frac{0.001m^{3}}{1L}=0.075m^{3}

so we can now use the density formula to find our the methanol's density, so we get:

\rho=\frac{m}{V}

\rho=\frac{70kg}{0.075m^{3}}

\rho=933.33kg/m^{3}

Next, we can us these values to find the specific gravity of methanol by using the formula:

S.G.=\frac{\rho_{sample}}{\rho_{H_{2}O}}

when substituting the known values we get:

S.G.=\frac{933.33kg/m^{3}}{1000kg/m^{3}}

so:

S.G.=0.933

We can now find the force it takes to accelerate this tank linearly at 0.25m/s^{2}

F=ma

F=(70kg)(0.25m/s^{2})

F=17.5N

6 0
3 years ago
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