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ELEN [110]
3 years ago
15

Object A attracts object B with a gravitational force of 10 newtons from a given distance. If the distance between the two objec

ts is doubled,
what is the new force of attraction between them?
A. 2.5 newtons
B. 5newtons
C. 20 newtons
D. 100 newtons
Physics
2 answers:
Katyanochek1 [597]3 years ago
7 0

Answer:

20N option C

Explanation:

10× 2

taurus [48]3 years ago
6 0

C. 20 Newton’s because 10*2
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A 2.0 kg puck is at rest on a level table. It is pushed straight north with a constant force of 5N for 1.50 s and then let go. H
lukranit [14]

Answer:

d = 6.32 m

Explanation:

Given that,

The mass of a puck, m = 2 kg

It is pushed straight north with a constant force of 5N for 1.50 s and then let go.

We need to find the distance covered by the puck when move from rest in 2.25 s.

We know that,

F = ma

a=\dfrac{F}{m}\\\\a=\dfrac{5}{2}\\\\a=2.5\ m/s^2

Let d is the distance moved in 2.25 s. Using second equation of motion,

d=ut+\dfrac{1}{2}at^2\\\\d=0+\dfrac{1}{2}\times 2.5\times (2.25)^2\\\\d=6.32\ m

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4 0
3 years ago
What mRNA sequence would result from the following DNA sequence?
Nezavi [6.7K]

Answer:

UAC CUG AGG AUC

Explanation:

<em>The mRNA sequence from ATG GAC TCC TAG DNA sequence would be </em><em>UAC CUG AGG AUC.</em>

<u>According to Chargaff's base pairing rule, the purine bases always pair with pyrimidine bases. Specifically, Adenine base must pair with Thymine base while Guanine base must pair with Cytosine base. In RNA, Thymine base is replaced with Uracil base.</u>

Hence:

ATG   GAC   TCC   TAG will pair with

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3 years ago
Determine the kinetic energy of the ball immediately after it is hit. (You must provide an answer before moving to the next part
earnstyle [38]

The question is incomplete. Here is the complete question.

A baseball palyer hits a 5.1 oz baseball with an initial velocity of 140ft/sat an angle of 40° with the horizontal as shown. Determine

a) The kinetic energy of the ball immediately after it is hit

b) The kinetic energy of the ball when it reaches its maximum height

c) The maximum height above the ground reached by the ball.

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              b) KE = 0

              c) h = 126 ft

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m = 5.1(0.02835)

m = 0.1445 kg

v = 140 ft = 42.67 m/s

Then, kinetic energy is

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KE = 131.64 J

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b) At its maximum height, the ball has its highest potential energy. Because of the law of conservation of energy, when potential energy is maximum, kinetic energy is minimum and vice-versa. So, at the maximum height, kinetic energy is 0.

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h=\frac{(1820.73)(0.4132)}{19.6}

h = 38.4 m

Transforming into ft: h = 126 ft

The maximum height above the ground reached by hte ball is 126 feet.

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