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Bumek [7]
3 years ago
15

Can someone pls help ! I’m sorry it’s small

Physics
1 answer:
solmaris [256]3 years ago
7 0

Answer:

The velocity time graph which shows a negative constant velocity (horizontal line below the horizontal axis) Labeled as Graph #2

Explanation:

Notice that the position-time graph shows the position decreasing in the form of a line with negative slope. Recall that the velocity graph is associated with the change in position with time, and this change in position is in the form of a line of constant negative slope. The slope of the position-time graph is in fact the velocity of the object, which therefore is negative and constant.

Such behavior is represented by the graph labeled as Graph 2 of a constant (horizontal) line below the horizontal axis.  

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Elanso [62]
There are 6 atoms total:
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2 Jupiter orbits the sun in a nearly circular path with radius 7.8x10^11m. The orbital period of Jupiter is 12 years.
bazaltina [42]

Mass of Jupiter=1.9×10

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Mean distance of Jupiter from Sun=7.8×10

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Gravitational Force, F=

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4 0
3 years ago
If it takes 150 n of force to accelerate an object at 30 m/s what is the mass of the object
iris [78.8K]
F  =  ma.         

Note Force should  be = 150 N.  Acceleration = 30 m/s2   ( Am presuming for your question you meant to write 30m/s2  and not 30m/s as you wrote)

150 = m ( 30)

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Mass of the object = 5 Kg.
3 0
3 years ago
A vector is parallel to the y axis, what is it x component?<br>​
slavikrds [6]

Answer:

Explanation:

A vector is parallel to the y axis .

Let its magnitude be A . So the vector can be represented as A j .

where i and j are unit vectors in x and y axis direction .

The x component of A j will be dot product of A j with i

The x component of A j = A j . i

= A x 0      [  Since j . i = 0 ]

= 0

4 0
3 years ago
Two charges are located in the x–y plane. If q1 = -2.90 nC and is located at x = 0.00 m, y = 0.840 m and the second charge has m
Lunna [17]

Answer:

Epx= - 21.4N/C

Epy= 19.84N/C

Explanation:

Electric field theory

The electric field at a point P due to a point charge is calculated as follows:

E= k*q/r²

E= Electric field in N/C

q = charge in Newtons (N)

k= electric constant in N*m²/C²

r= distance from load q to point P in meters (m)

Equivalences

1nC= 10⁻⁹C

known data

q₁=-2.9nC=-2.9 *10⁻⁹C

q₂=5nC=5  *10⁻⁹C

r₁=0.840m

r_{2} =\sqrt{1^{2} +0.8^{2} } =\sqrt{1.64}

sin\beta =\frac{0.8}{\sqrt{1.64} } =0.6246

cos\beta =\frac{1}{\sqrt{1.64} } =0.7808

Calculation of the electric field at point P due to q1

Ep₁x=0

Ep_{1y} =\frac{k*q_{1} }{r_{1}^{2}  } =\frac{8.99*10^{9}*2.9*10^{-9}  }{0.84^{2} } =36.95\frac{N}{C}

Calculation of the electric field at point P due to q2

Ep_{2x} =-\frac{k*q_{2} *cos\beta }{r_{2}^{2}  } =-\frac{8.99*10^{9}*5*10^{-9} *0.7808 }{(\sqrt{1.64})^{2}  } =-21.4\frac{N}{C}

Ep_{2y} =-\frac{k*q_{2} *sin\beta }{r_{2}^{2}  } =-\frac{8.99*10^{9}*5*10^{-9} *0.6242 }{(\sqrt{1.64})^{2}  } =-17.11\frac{N}{C}

Calculation of the electric field at point P(0,0) due to q1 and q2

Epx= Ep₁x+ Ep₂x==0 - 21.4N/C =- 21.4N/C

Epy= Ep₁y+ Ep₂y=36.95 N/C-17.11N =19.84N/C

7 0
4 years ago
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