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Anarel [89]
3 years ago
9

A 1520 kg car is rolling down a frictionless 14.4° slope. What is its acceleration?

Physics
1 answer:
Umnica [9.8K]3 years ago
7 0

Answer:

a = 2.44 m/s²

Explanation:

The Net force acting on the car downwards the plane will be equal to the component of the weight that is parallel to the plane.

Thus;

mg(sin θ) = ma

Thus;. g(sin θ) = a

g is acceleration due to gravity = 9.81 m/s²

θ = 14.4°

Thus;

9.81 × sin 14.4 = a

a = 2.44 m/s²

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4. If a 75 kg astronaut moves from Earth to Mars, how much weight did he lose?
zalisa [80]
He doesn't lose weight he stays the same weight it's just gravity that changes
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1 year ago
I NEED HELP WITH THIS PLEASE
Ann [662]

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D

Explanation:

One of the factors of increasing the rate of a reaction is increasing concentration. Therefore adding more people increases the number of people on the dance floor , therefore increasing the concentration increases the rate of reaction.

6 0
3 years ago
As shown in the diagram, two forces act on an object. The forces have magnitudes F1 = 5.7 N and F2 = 1.9 N. What third force wil
galina1969 [7]

Answer:

Second option 6.3 N at 162° counterclockwise from  

F1->

Explanation:

Observe the attached image. We must calculate the sum of all the forces in the direction x and in the direction y and equal the sum of the forces to 0.

For the address x we have:

-F_3sin(b) + F_1 = 0

For the address and we have:

-F_3cos(b) + F_2 = 0

The forces F_1 and F_2 are known

F_1 = 5.7\ N\\\\F_2 = 1.9\ N

We have 2 unknowns (F_3 and b) and we have 2 equations.

Now we clear F_3 from the second equation and introduce it into the first equation.

F_3 = \frac{F_2}{cos (b)}

Then

-\frac{F_2}{cos (b)}sin(b)+F_1 = 0\\\\F_1 = \frac{F_2}{cos (b)}sin(b)\\\\F_1 = F_2tan(b)\\\\tan(b) = \frac{F_1}{F_2}\\\\tan(b) = \frac{5.7}{1.9}\\\\tan^{-1}(\frac{5.7}{1.9}) = b\\\\b= 72\°\\\\m = b +90\\\\\m= 162\°

Then we find the value of F_3

F_3 = \frac{F_1}{sin(b)}\\\\F_3 =\frac{5.7}{sin(72\°)}\\\\F_3 = 6.01 N

Finally the answer is 6.3 N at 162° counterclockwise from  

F1->

7 0
3 years ago
A man attempts to pick up his suitcase of weight ws by pulling straight up on the handle. However, he is unable to lift the suit
kvasek [131]

Answer:

b)

Explanation:

Normal force, is always directed upward the surface over which is placed the object, and can adopt any value, as required to meet Newton's 2nd Law.

In this case, as the external force on the suitcase pulls upward, in order  to counteract the influence of gravity, normal force is less than the weight of the suitcase, as follows:

F + Fn = m*g

⇒ Fn = m*g - F

So, the normal force is equal to the magnitude of the weight of the suitcase (m*g) minus the magnitude of the force of the pull (F) which is the same expressed by the statement b.

4 0
3 years ago
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