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Shtirlitz [24]
4 years ago
7

Match the measurements with the proper SI unit.

Physics
2 answers:
Ivan4 years ago
7 0

Answer:

Acceleration:

C. Meters per second squared

Velocity:

B. Meters per second

Distance:

A. Meters

Explanation:

We must remember that the international system of measures (SI) takes into account for the length as the main unit the meter, for the mass the kilogram, for the time the second.

The acceleration is calculated using the following expression

a = v/t = (m/s/s) = (m/s^2]

The velocity is calculated using the following expression

v = x/t = (m)/(s) = (m/s)

The distance for the SI system is given in meters

vladimir1956 [14]4 years ago
7 0

Answer:

Explanation:

⭐

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Saatlemati
vodomira [7]

Answer:

what are u asking there isnt a question

3 0
3 years ago
Question 10 of 10
Bad White [126]

Answer:

D. The Federal Reserve Bank can provide a short-term loan to banks

to prevent them from running out of money.

Explanation:

A bank run occurs when a large number of depositors withdraw their deposits simultaneously from a bank.

As the number of withdraws increases, the available cash in the bank decreases until a point that the bank can't give depositors their money.

In these situations, The Federal Reserve Bank acts as a lender of last resort that helps to reinforce the effect of deposit insurance, and to reassure bank customers that they will not lose their money.

5 0
4 years ago
A single loop of wire with an area of 0.0920 m2 is in a uniform magnetic field that has an initial value of 3.80 T, is perpendic
jekas [21]

Answer:

Induced emf in the loop is 0.02208 volt.

Induced current in the loop is 0.0368 A.

Explanation:

Given that,

Area of the single loop, A=0.092\ m^2

The initial value of uniform magnetic field, B = 3.8 T

The magnetic field is decreasing at a constant rate, \dfrac{dB}{dt}=0.24\ T/s

(a) The induced emf in the loop is given by the rate of change of magnetic flux.

\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=A\times \dfrac{dB}{dt}\\\\\epsilon=0.092\times 0.24\\\\\epsilon=0.02208\ V

(b) Resistance of the loop is 0.6 ohms. Let I is the current induced in the loop. Using Ohm's law :

\epsilon=IR\\\\I=\dfrac{\epsilon}{R}\\\\I=\dfrac{0.02208}{0.6}\\\\I=0.0368\ A

Hence, this is the required solution.

5 0
3 years ago
Calculate the heat energy released when 17.7 g of liquid mercury at 25.00 °c is converted to solid mercury at its melting point.
maria [59]

Answer;

Q = 359.2-J  

Explanation;

Given that;

Constants for mercury at 1 atm  

Heat capacity of Hg(l) is 28.0 J/(mol*K)  

melting point is 234.32 K  

Enthalpy of fusion is 2.29 kJ/mol

17.7-g Hg / 200.6g/mol = 0.0882 mol Hg;

°C + 273 = 298 K;

2.29-kJ/mol = 2290-J/mol  

Q = (m x ΔT x Cp) + (m x Hf)  

Q = 0.0882-mol x (298 - 234.32) x 28.0-J/mol*K) + (0.0882-mol x 2290-J/mol)  

Q = 157.26-J + 201.978-J  

Q = 359.2-J  

Q=359-J (3 sig fig allowed due to 17.7-g given in problem)

8 0
3 years ago
What coefficients would balance the following equation? __C2H6 + __O2 → __CO2 + __H2O A. 1C2H6 + 5O2 → 2CO2 + 3H2O B. 2C2H6 + 5O
sladkih [1.3K]
It's C, with the 2/7/4/6 in front of each reactant and product.
4 0
4 years ago
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