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Harlamova29_29 [7]
3 years ago
11

A student solving for the acceleration of an object has applied appropriate physics principles and obtained the expression a=a1

+ F/m where a1 = 3.00 meter/ second2 F=12.0 kilogram.meter/second2 and m=7.00 kilogram. First which of the following is the correct step for obtaining a common denominator for the two fractions in the expression in solving for a?a. (m/m times a1/1) + (1/1 times F/m)b. (1/m times a1/1) + (1/m times F/m)c. (m/m times a1/1) + (F/F times F/m)d. (m/m times a1/1) +(m/m times F/m )
Physics
1 answer:
AlekseyPX3 years ago
8 0

Explanation:

A student solving for the acceleration of an object has applied appropriate physics principles and obtained the expression :

a=a_1+\dfrac{F}{m}

Where

a_1=3\ m/s^2

F=12\ kg-m/s^2

m = 7 kg

So, the correct step for obtaining a common denominator for the two fractions in the expression in solving for a is (a) and the value of a is :

a=3+\dfrac{12}{7}

a=4.71\ m/s^2

Hence, the correct option is (a).

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At what position in its elliptical orbit is the speed of a planet a maximum? when it is closest to the sun when it is farthest f
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Answer:

1.when it is closest to the sun

2.when it is midway between its farthest

Explanation:

According to the law of  Kepler's

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We also know that time period T given as

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v\alpha \dfrac{r}{T}

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1. What are valence electrons used for by an element?
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Answer:

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3 years ago
A block rests on a frictionless table on Earth. After a 40-N horizontal force is applied to the block, it accelerates at 9.2 m/s
Paul [167]

Answer:

4.6 \frac{m}{s^2}

Explanation:

Since the table is frictionless, there is no force of dynamic  friction between table an block when the horizontal force is applied to it on Earth. Exactly the same is true when the table is taken to the Moon. Therefore, the Net Force acting on the object in both cases when the object accelerates, is the external horizontal force.

Notice that on Earth and on the Moon, the weight of the object (vertical and pointing up) is compensated by the normal force of the table on the object (pointing up and of the same magnitude as the weight) that precludes movement in the vertical direction. So in both cases, its acceleration will only be due to the horizontal force.

We use the equation for Net Force to find the mass of the object:

F=m*a\\40 N =m * 9.2 \frac{m}{s^2}\\\frac{40}{9.2} kg=m\\m=\frac{40}{9.2} kg

We use this mass (since the mass of the object is a constant independent of where the object is) to find the acceleration the object will experience when the 20 N horizontal force is applied on it on the Moon:

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3 years ago
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