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ELEN [110]
3 years ago
13

Which expression is equal to the number of grams (g) in 2.43 kilograms (kg)?

Physics
2 answers:
Mazyrski [523]3 years ago
5 0

Answer:

A

Explanation:

edge 2020

Kryger [21]3 years ago
4 0

Answer:

A. 2.43 kg x 1000 g/1 kg

Explanation:

We know that the equivalence between grams and kilograms is

1 kg = 1000 g

So, if we want to convert a mass m from kg to g, we just need to multiply the value by the factor \frac{1000 g}{1 kg}.

In this problem, the mass is

m = 2.43 kg

Therefore, multiplying by the factor above, we get

m = 2.43 kg \cdot \frac{ 1000 g}{1 kg}

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Which material would you take, steel or bricks/stone to design a bridge. Why? ( Question related to poisson's ratio)​
Ann [662]

i  would choose steel because the steel used for normal construction have several hundred Mega Pascal strength. one of the special feature of steel is its ductility. it is the deformation capability before the final breakage.

8 0
4 years ago
In a double-slit interference experiment, the slit separation is 2.41 μm, the light wavelength is 512 nm, and the separation bet
Elina [12.6K]

Answer:

Using equation 2dsinФ=n*λ

given d=2.41*10^-6m

λ=512*10^-12m

θ=52.64 degrees

7 0
4 years ago
Read 2 more answers
-2 m
Nana76 [90]
Report this clown who put the first answer he’s trying to get your ip
4 0
3 years ago
Find A and effective resistance.<br> (Ill give Brainliest if you provide explaination)
Alex777 [14]

Answer:

A = 2.4 A

R_{eq}  = 5 \ \Omega

Explanation:

The voltage in the circuit, V = 12 V

The given circuit shows four resistors with R₁ and R₂ arranged in series with both in parallel to R₃ and R₄ which are is series to each other

R₁ = 4 Ω

R₂ = 6 Ω

R₄ = 5 Ω

The voltage across R₃ = 6 V

Voltage across parallel resistors are equal, therefore;

The total voltage across R₃ and R₄ = 12 V

The total voltage across R₁ and R₂ = 12 V

The voltage across R₃ + The voltage across R₄ = 12 V

∴ The voltage across R₄ = 12 V - 6 V = 6 V

The current flowing through  R₄ = 6V/(5 Ω) = 1.2 A

The current flowing through R₃ = The current flowing through R₄ = 1.2 A

The resistor, R₃ = 6 V/1.2 A = 5 Ω

Therefore, we have;

The sum of resistors in series are R₁ + R₂ and R₃ + R₄, which gives;

R_{series \, 1} = R₁ + R₂ = 4 Ω + 6 Ω = 10 Ω

R_{series \, 2} = R₃ + R₄ = 5 Ω + 5 Ω = 10 Ω

The sum of the resistors in parallel is given as follows;

\dfrac{1}{R_{eq}}  = \dfrac{1}{R_{series \, 1}} + \dfrac{1}{R_{series \, 2}} = \dfrac{R_{series \, 2} + R_{series \, 1}}{R_{series \, 1} \times R_{series \, 2}}

Therefore;

R_{eq} =  \dfrac{R_{series \, 1} \times  R_{series \, 2}}{R_{series \, 1} + R_{series \, 2}}

Therefore;

R_{eq} =  \dfrac{10\times  10}{10 + 10} \ \Omega = 5 \ \Omega

R_{eq}  = 5 \ \Omega

The value of the current, <em>A</em>, in the circuit, I = V/R_{eq}

A = I = 12 V/(5 Ω) = 2.4 A

A = 2.4 A

3 0
3 years ago
PLEASE HELP !!
LenKa [72]

Explanation:

It is given that

The kinetic energy of  Mr. Thomson’s prius, K = 641300 J

He is moving with a velocity, v = 22 m/s

It is required to find the mass of Mr. Thomson’s prius.

Due to his motion, a kinetic energy will possess by him. The mathematical form of kinetic energy is given by :

K=\dfrac{1}{2}mv^2

Solving for m

m=\dfrac{2K}{v^2}

Plugging all the values, we get :

m=\dfrac{2K}{v^2}\\\\m=\dfrac{2\times 641300}{(22)^2}\\\\m=2650\ kg

So, the mass of Mr. Thomson’s prius is 2650 kg.

8 0
4 years ago
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