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2 years ago

5

Un vagón de 250 kg situado en la cima de una montaña rusa inicia su descenso por una rampa inclinado 60º sobre la horizontal. Si

no tenemos en cuenta el rozamiento: a. Dibuja todas las fuerzas que actúan sobre el vagón. b. Calcula la fuerza y la aceleración con que desciende.

1 answer:

earnstyle [38]2 years ago

6 0

Um I don’t understand what your saying

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An ore sample weighs 17.50 N in air. When the sample is suspended by a light cord and totally immersed in water, the tension in

valkas [14]

Answer:

Volume of the sample: approximately $\rm 0.6422 \; L = 6.422 \times 10^{-4} \; m^{3}$ .

Average density of the sample: approximately $\rm 2.77\; g \cdot cm^{3} = 2.778 \times 10^{3}\; kg \cdot m^{3}$ .

Assumption:

$\rm g = 9.81\; N \cdot kg^{-1}$ .
$\rho(\text{water}) = \rm 1.000\times 10^{3}\; kg \cdot m^{-3}$ .
Volume of the cord is negligible.

Explanation:

<h3>Total volume of the sample</h3>

The size of the buoyant force is equal to $\rm 17.50 - 11.20 = 6.30\; N$ .

That's also equal to the weight (weight, $m \cdot g$ ) of water that the object displaces. To find the mass of water displaced from its weight, divide weight with $g$ .

$\displaystyle m = \frac{m\cdot g}{g} = \rm \frac{6.30\; N}{9.81\; N \cdot kg^{-1}} \approx 0.642\; kg$ .

Assume that the density of water is $\rho(\text{water}) = \rm 1.000\times 10^{3}\; kg \cdot m^{-3}$ . To the volume of water displaced from its mass, divide mass with density $\rho(\text{water})$ .

$\displaystyle V(\text{water displaced}) = \frac{m}{\rho} = \rm \frac{0.642\; kg}{1.000\times 10^{3}\; kg \cdot m^{-3}} \approx 6.42201 \times 10^{-4}\; m^{3}$ .

Assume that the volume of the cord is negligible. Since the sample is fully-immersed in water, its volume should be the same as the volume of water it displaces.

$V(\text{sample}) = V(\text{water displaced}) \approx \rm 6.422\times 10^{-4}\; m^{3}$ .

<h3>Average Density of the sample</h3>

Average density is equal to mass over volume.

To find the mass of the sample from its weight, divide with $g$ .

$\displaystyle m = \frac{m \cdot g}{g} = \rm \frac{17.50\; N}{9.81\; N \cdot kg^{-1}} \approx 1.78389 \; kg$ .

The volume of the sample is found in the previous part.

Divide mass with volume to find the average density.

$\displaystyle \rho(\text{sample, average}) = \frac{m}{V} = \rm \frac{1.78389\; kg}{6.42201 \times 10^{-4}\; m^{3}} \approx 2.778\; kg \cdot m^{-3}$ .

3 0

3 years ago

PLEASE ASSIST 20 POINTS

adelina 88 [10]

Answer:

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7 0

2 years ago

In which electric circuit would the voltmeter read 10 volts ?

ki77a [65]

Given that,

Voltage = 10 volt

Suppose, The three resistance is connected in parallel and each resistance is 12 Ω. find the current in the electric circuit.

We need to calculate the equivalent resistance

Using formula of parallel

$\dfrac{1}{R}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}$

Put the value into the formula

$\dfrac{1}{R}=\dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}$

$\dfrac{1}{R}=\dfrac{1}{4}$

$R=4\ \Omega$

We need to calculate the current in the circuit

Using ohm's law

$V=IR$

$I=\dfrac{V}{R}$

Where, V = voltage

R = resistance

Put the value into the formula

$I=\dfrac{10}{4}$

$I=2.5\ A$

Hence, The current in the circuit is 2.5 A

4 0

2 years ago

Velocities of two bodies A and B are given in vectors notation as va =i+2j-3k and Vb=3i+2j-k what will be the relative velocity

ArbitrLikvidat [17]

Answer:

$V_{B/A}=2i+2k$

Explanation:

The relative velocity can be calculated by means of the difference between vector B minus vector A.

$V_{A}=i+2j-3k\\V_{B}=3i+2j-k\\V_{B}-V_{A}=(3-1)i + (2-2)j+(-1-(-3))k\\V_{B/A}=2i+2k$

5 0

3 years ago

How many joules of work are done on an object when a force of 10 N pushes it 5 m?

zhenek [66]

Answer:

option C

Explanation:

given,

Force on the object = 10 N

distance of push = 5 m

Work done = ?

we know,

work done is equal to Force into displacement.

W = F . s

W = 10 x 5

W = 50 J

Work done by the object when 10 N force is applied is equal to 50 J

Hence, the correct answer is option C

5 0

3 years ago