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Zina [86]
2 years ago
9

A sled that has a mass of 8 kg is pulled at a 50 degree angle with a force of 20 N. The force of friction acting on the sled is

2.4 N. The free-body diagram shows the forces acting on the sled.
What is the acceleration of the sled and the normal force acting on it, to the nearest tenth?

Physics
1 answer:
ratelena [41]2 years ago
8 0

Answer:

a = 1.3 m/s2; fn = 63.1 n :)

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Two black holes (the remains of exploded stars), separated by a distance of
jolli1 [7]

The largest mass is 4.7 x 10³⁰ kg and the smallest mass is 5 x 10²⁹ kg.

The given parameters;

  • <em>distance between the two black holes, r = 10 AU = 1.5 x 10¹² m</em>
  • <em>gravitational force between the two black holes, F = 6.9 x 10²⁵ N.</em>
  • <em>combined mass of the two black holes = 5.20 x 10³⁰ kg</em>

The product of the two masses is calculated from Newton's law of universal gravitational as follows;

F = \frac{Gm_1m_2}{r^2} \\\\m_1m_2 = \frac{Fr^2}{G} \\\\m_1m_2 = \frac{(6.9\times 10^{25}) \times (1.5\times 10^{12})^2}{6.67\times 10^{-11}} \\\\m_1m_2 = 2.328 \times 10^{60} \ kg^2

The sum of the two masses is given as;

m₁ + m₂ = 5.2 x 10³⁰ kg

m₂ = 5.2 x 10³⁰ kg - m₁

The first mass is calculated as follows;

m₁(5.2 x 10³⁰ - m₁) = 2.328 x 10⁶⁰

5.2 x 10³⁰m₁ - m₁² = 2.328 x 10⁶⁰

m₁² - 5.2 x 10³⁰m₁  + 2.328 x 10⁶⁰ = 0

<em>solve the quadratic equation using formula method</em>;

a = 1, b =-  5.2 x 10³⁰, c = 2.328 x 10⁶⁰

m_1 = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\m_1 = \frac{-(-5.2\times 10^{20})  \ \ +/- \ \ \sqrt{(-5.2\times 10^{20})^2 - 4(1\times 2.328\times 10^{60})} }{2(1)} \\\\m_1 = 4.7 \times 10^{30} \ kg \ \ or \ \ 4.9 \times 10^{29} \ kg

The second mass is calculated as follows;

m₂ = 5.2 x 10³⁰ kg - m₁

m₂ = 5.2 x 10³⁰ kg  -  4.7 x 10³⁰ kg

m₂ = 5 x 10²⁹ kg

or

m₂ = 5.2 x 10³⁰ kg  -  4.9 x 10²⁹ kg

m₂ = 4.7 x 10³⁰ kg

Thus, the largest mass is 4.7 x 10³⁰ kg and the smallest mass is 5 x 10²⁹ kg.

Learn more here:brainly.com/question/9373839

3 0
2 years ago
1. Calcular la masa de mercurio que pasó de 30 °C hasta 120 °C y absorbió 4400 cal. Calor específico del
timofeeve [1]

Answer:

Masa, m = 0.088 kg

Explanation:

Given the following data;

Temperatura inicial = 30°C

Temperatura final = 120°C

Capacidad calorífica específica = 138J/kg.K

Calor absorbido, Q = 4400 cal.

Para encontrar la masa;

La capacidad calorífica viene dada por la fórmula;

Q = mct

Dónde;

Q representa la capacidad calorífica o la cantidad de calor.

m representa la masa de un objeto.

c representa la capacidad calorífica específica del agua.

dt representa el cambio de temperatura.

dt = T2 - T1

dt = 120 - 30

dt = 90°C to kelvin = 273 + 90 = 363K

Sustituyendo en la fórmula, tenemos;

4400 = m*138*363

4400 = 50094m

m = \frac {4400}{50094}

Masa, m = 0.088 kg

7 0
2 years ago
If a resistor draws 1.2 x 10^-3 a of current from a 12 v battery, then what is the value of the resistor?
bixtya [17]
Value of resistor = (12V) /(1.2 x 10^-3A)=10000ohms=10k ohms
8 0
3 years ago
A buoy is anchored tho the oceanfloor. A large wave approaches the buoy. How will the buoy move as the wave goes by?
nadezda [96]

Answer:

The buoy will move upwards and downwards and well as left and right

Explanation:

As the wind is storing so had the capability to move the buoy in all directions as it is a light object despite being anchored down

PLEASE GIVE BRAINLIEST AND THANKS :-)

5 0
3 years ago
How do I find deceleration
Mars2501 [29]
Deceleration isn't a term in physics; you can have positive acceleration which is speeding up and negative acceleration which is slowing down
Use newton's law F = ma
a = F/m
4 0
3 years ago
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