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Evgen [1.6K]
3 years ago
13

6. Draw conclusions: Newton’s first law states that an object in motion will travel at a constant velocity unless acted upon by

an unbalanced force. How do these experiments show this?
Physics
2 answers:
Nitella [24]3 years ago
7 0
I’m not sure what experiments you’re referring to. Maybe if there is a Frictional force that stops and object from sliding after a period of time instead of it continuing to slide forever.
gavmur [86]3 years ago
5 0
You can test if it’s true by holding a pencil in mid air over a table and the table is supposed to be the unbalanced forced that stopped the pencil from moving at the constant velocity it was going by.
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lana [24]
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4 0
3 years ago
What is the role of magnetism in the study of x-ray imaging?
Helga [31]
The high voltage generators which are used in X-Ray imaging system are magnetic devices. There are various magnetic X-rays imaging techniques.

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8 0
3 years ago
What is the final position of the object if its initial position is x = 0.40 m and the work done on it is equal to 0.21 J? What
r-ruslan [8.4K]

Answer:

a) Final position is x = 0.90 m

b) Final position is x = 0.133 m

Explanation:

The workdone between two points is usually approximated as the area under the force-distance curve between those two points.

From the graph,

As at the initial position, x = 0.40 m and the corresponding F = 0.8 N,

The area from that point onwards up to the end of that particular bar = 0.8 (0.5 - 0.4) = 0.08 J

The next bar has force = 0.4 N and the width of the bar = (0.75 - 0.50) = 0.25 m

Work done under this bar = 0.4 × 0.25 = 0.1 J

Total work done from the starting position up to this point now = 0.08 + 0.1 = 0.18 J, still less than 0.21 J

So, the final position has to be on the last bar. Let the position be x. The force on the last bar = 0.2 N

0.21 = 0.18 + 0.2 (x - 0.75)

0.03 = 0.2x - 0.15

0.2x = 0.18

x = 0.9 m

Therefore, the final position of the object, to do 0.21 J worth of work, starting from x = 0.4 m is 0.90 m.

b) For this part, negative work is done, this means, we will move in the negative direction to try and trace this total work done.

From the starting point where the initial position is 0.40 m, the force here is 0.80 N

The workdone under this bar to the left is

The workdone = 0.8 (0.25 - 0.4) = - 0.12 J

Since we're tracing -0.19 J, the final position has to be on the last bar (on the left), Let the position be x. The force on the last bar on the left (could also be referred to as the first bar) = 0.60 N

- 0.19 = -0.12 + 0.6 (x - 0.25)

-0.07 = 0.6x - 0.15

0.6x = 0.08

x = (0.08/0.6) = 0.133 m

Therefore, the final position of the object, after doing -0.19 J worth of work, starting from x = 0.4 m is 0.133 m.

Hope this Helps!!!

4 0
3 years ago
A 5.0-kg rock and a 3.0 × 10−4-kg pebble are held near the surface of the earth.(a)Determine the magnitude of the gravitational
a_sh-v [17]

Answer:

a). Determine the magnitude of the gravitational force exerted on each by the earth.

Rock: F = 49.06N

Pebble: F = 29.44N

(b)Calculate the magnitude of the acceleration of each object when released.

Rock: a =9.8m/s^{2}

Pebble:  a =9.8m/s^{2}

Explanation:

The universal law of gravitation is defined as:

F = G\frac{m1m2}{r^{2}}  (1)

Where G is the gravitational constant, m1 and m2 are the masses of the two objects and r is the distance between them.

<em>Case for the rock </em>m = 5.0 Kg<em>:</em>

m1 will be equal to the mass of the Earth m1 = 5.972×10^{24} Kg and since the rock and the pebble are held near the surface of the Earth, then, r will be equal to the radius of the Earth r = 6371000m.

F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(5.0 Kg)}{(6371000 m)^{2}}

F = 49.06N

Newton's second law can be used to know the acceleration.

F = ma

a =\frac{F}{m} (2)

a =\frac{(49.06 Kg.m/s^{2})}{(5.0 Kg)}

a =9.8m/s^{2}

<em>Case for the pebble </em>m = 3.0 Kg<em>:</em>

F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(3.0 Kg)}{(6371000 m)^{2}}

F = 29.44N

a =\frac{F}{m}

a =\frac{(29.44 Kg.m/s^{2})}{(3.0 Kg)}

a =9.8m/s^{2}

3 0
2 years ago
Read 2 more answers
A 4kg box is pressed into a spring. The spring constant k=220N/m. The box is released at the bottom of a frictionless ramp and r
AVprozaik [17]

Answer:

Explanation:

The stored elastic energy in the spring helps the box to rise to a height of .75 m

So stored elastic energy = potential energy attained

1/2 k d² = mgh , k is spring constant , d is compression in spring , m is mass of box , h is height attained .

Putting the values

.5 x 220 x d² = 4 x 9.8 x .75

d² = .2673

d = .517 m

51.7 cm .

8 0
3 years ago
Read 2 more answers
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