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givi [52]
3 years ago
10

In an experiment, you find the mass of a cart to be 250 grams. What is the mass of the cart in kilograms?

Physics
2 answers:
postnew [5]3 years ago
7 0
The answer is:  0.25 kg  .
______________________________

250 g * ( 1 kg / 1000 g) = (250/ 1000) kg = 0.250 kg = 0.25 kg .
gizmo_the_mogwai [7]3 years ago
5 0
It should be 0.25kg because you converter from g to kg and since 1g<1kg so you move the decimal to the left
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Which of these is NOT a possible type of energy transformation?
Orlov [11]

to be franc i really think the answer is B

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3 years ago
An eight-turn coil encloses an elliptical area having a major axis of 40.0 cm and a minor axis of 30.0 cm. The coil lies in the
Darina [25.2K]

Answer:

9.25 x 10^-4 Nm

Explanation:

number of turns, N = 8

major axis = 40 cm

semi major axis, a = 20 cm = 0.2 m

minor axis = 30 cm

semi minor axis, b = 15 cm = 0.15 m

current, i = 6.2 A

Magnetic field, B = 1.98 x 10^-4 T

Angle between the normal and the magnetic field is 90°.

Torque is given by

τ = N i A B SinФ

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5 0
3 years ago
Distance versus Displacement Worksheet
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6 0
3 years ago
Read 2 more answers
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Liono4ka [1.6K]

Answer:

The electric potential is approximately 5.8 V

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Explanation:

The two protons can be considered as point charges. Therefore, the electric potential is given by the point charge potential:

\displaystyle{U=\frac{q}{4\pi \epsilon_0r}} (1)

where q is the charge of the particle, \epsilon_0 the electric permittivity of the vacuum (I assuming the two protons are in a vacuum) and r is the distance from the point charge to the point where the potential is being measured. Because the electric potential is an scalar, we can simply add the contribution of the two potentials in the midpoint between the protons. Thus:

\displaystyle{U_{midpoint}=\frac{q}{4\pi \epsilon_0r}}+\frac{q}{4\pi \epsilon_0r}}=\frac{q}{2\pi \epsilon_0r}}}

Substituting the values q=1.602 \cdot10^{-19}\ C, \displaystyle{\frac{1}{4\pi\epsilon_0}=8.99\cdot 10^9 N\cdot m^2\cdot C^{-2}} and r=0.5 \cdot 10^{-9} m we obtain:

\displaystyle{U_{midpoint}=\frac{q}{2\pi \epsilon_0r}}=5.759 \approx 5.8 V}

The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero.

6 0
3 years ago
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DochEvi [55]

Answer:

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Explanation:

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4 0
3 years ago
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