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Luda [366]
3 years ago
13

The net horizontal force on a car is 981 N. The car has a mass of 1550 kg and the force is applied when the car has a speed of 2

5 mi/h. If the force continues to be applied to the car, how far will the car go in 35 s?
Physics
1 answer:
viktelen [127]3 years ago
5 0

Answer:

Distance, d = 778.05 m                          

Explanation:

Given that,

Force acting on the car, F = 981 N

Mass of the car, m = 1550 kg

Initial speed of the car, v = 25 mi/h = 11.17 m/s

We need to find the distance covered by car if the force continues to be applied to the car. Firstly, lets find the acceleration of the car:

F=ma\\\\a=\dfrac{F}{m}\\\\a=\dfrac{981}{1550}\\\\a=0.632\ m/s^2

Let d is the distance covered by car. Using second equation of motion as :

d=ut+\dfrac{1}{2}at^2\\\\d=11.17\times 35+\dfrac{1}{2}\times 0.632\times (35)^2\\\\d=778.05\ m

So, the car will cover a distance of 778.05 meters.

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Draw the following vector quantity Using the coordinate system.
DiKsa [7]

The given vectors quantities can be described by their properties of both

magnitude and direction.

  • a. The drawing of the vector extending from point (0, 0) to (190, 0) on the coordinate plane is attached.
  • b. The velocity vector extending from  (0, 0) to (108.76, 50.714) on the coordinate plane is attached.
  • c. The displacement vector extending from (0, 0) to (30·√2, 30·√2) is attached.

Reasons:

a. The magnitude of the vector = 190 N

The direction in which the vector acts = East

Therefore, in vector form, we have;

\vec{F} = 190 × cos(0)·i + 190 × sin(0)·j = 190·i

The vector can be represented by an horizontal line, 190 units long

Coordinate points on the vector = (0, 0) and (190, 0)

The drawing of the vector with the above points using MS Excel is attached.

b. Magnitude of the velocity vector = 120 km/hr. 25° North of east

Solution;

The vector form of the velocity is; \vec{v} = 120 × cos(25)·i + 120×sin(25)·j, which gives;

\vec{v} = 120 × cos(25)·i + 120×sin(25)·j ≈ 108.76·i + 50.714·j

\vec{v} ≈ 108.76·i + 50.714·j

Therefore, points that define the vector are; (0, 0) and (108.76, 50.714)

The drawing of the vector is attached

c. The magnitude of the vector = 60 m

The direction of the vector is southwest = West 45° south

The vector form of the displacement is \vec{d} = 60 × cos(45°)·i + 60 × sin(45°)·j

Which gives;

\vec{d} = 30·√2·i + 30·√2·j

Points on the vector are therefore; (0, 0), and (30·√2, 30·√2)

The drawing of the vector is attached

Learn more about vectors here:

brainly.com/question/10409036

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2 years ago
Object A and object B are moving with the same momentum. Which of the following statements must be true?
nordsb [41]

The FIRST statement on the list is the definition of momentum, so that's the one that must be true.

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3 years ago
What is the best example of a compound machine A ax. B ramp. C car D screw
liberstina [14]

Answer: I believe the answer is C

Explanation:

Cars are composed of hundreds of simple machines

3 0
2 years ago
Read 2 more answers
a) A unit of time sometimes used in microscopic physics is the shake. One shake equals 10–8 s. Are there more shakes in a second
lawyer [7]

Answer:

a) Yes, there are 10^8 shakes in a second (1 s \frac{1 shake}{10^{-8}s}=10^8 shake) in a year there are 31,536,000 seconds... that is 3.1536x10^7 (1year\frac{365 day}{1 year} \frac{24 h}{1 day} \frac{60 min}{1 h}\frac{60s}{1min}=3.1536 *10^7)

b) Note that the defined universe day will have 60*60*24=86400 universe seconds if the seconds are defined as normal seconds.

Also one universe day (or 86400 universe seconds) is equivalent to 1010 years. Now using a rule of three we can know the seconds that humans have existed.

106 years * \frac{86,400 universe-seconds}{1,010 years}=9,067.728 s

That is about 2 and half hours.

3 0
3 years ago
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
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