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OleMash [197]
3 years ago
7

1150 g - 0.652 kg = ______________g?

Physics
2 answers:
beks73 [17]3 years ago
6 0
Answer is 498 grams
Stells [14]3 years ago
5 0

Answer:

The required answer is 498 gram

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Answer:

Biosphere

Explanation:

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What is your worldview? Explain?
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My worldview is someone who doesn't exist. Someone who isn't real but i want them to BE real. It's not possible to have someone who is alike you. Who is there for yo when you need them

5 0
2 years ago
Question C) needs to be answered, please help (physics)
Zarrin [17]

(a) Differentiate the position vector to get the velocity vector:

<em>r</em><em>(t)</em> = (3.00 m/s) <em>t</em> <em>i</em> - (4.00 m/s²) <em>t</em>² <em>j</em> + (2.00 m) <em>k</em>

<em>v</em><em>(t)</em> = d<em>r</em>/d<em>t</em> = (3.00 m/s) <em>i</em> - (8.00 m/s²) <em>t</em> <em>j</em>

<em></em>

(b) The velocity at <em>t</em> = 2.00 s is

<em>v</em> (2.00 s) = (3.00 m/s) <em>i</em> - (16.0 m/s) <em>j</em>

<em></em>

(c) Compute the electron's position at <em>t</em> = 2.00 s:

<em>r</em> (2.00 s) = (6.00 m) <em>i</em> - (16.0 m) <em>j</em> + (2.00 m) <em>k</em>

The electron's distance from the origin at <em>t</em> = 2.00 is the magnitude of this vector:

||<em>r</em> (2.00 s)|| = √((6.00 m)² + (-16.0 m)² + (2.00 m)²) = 2 √74 m ≈ 17.2 m

(d) In the <em>x</em>-<em>y</em> plane, the velocity vector at <em>t</em> = 2.00 s makes an angle <em>θ</em> with the positive <em>x</em>-axis such that

tan(<em>θ</em>) = (-16.0 m/s) / (3.00 m/s)   ==>   <em>θ</em> ≈ -79.4º

or an angle of about 360º + <em>θ</em> ≈ 281º in the counter-clockwise direction.

3 0
3 years ago
I NEED HELP PLEASE, THANKS! :)
Zina [86]

Answer:

charge C = greatest net force

charge B = the smallest net force

ratio  = 9 : 1

Explanation:

we know that in Electrostatic Forces, when 2 charges are at same sign then they repel each other and if they are different signed charges then they attract each other

so as per Coulomb's formula of Electrostatic Forces

F = \frac{k\ q_1\ q_2}{r^2}     .....................1

and here k is 9 × 10^9 N.m²/c² and we consider each charge at distance d

so two charge force at A to B is

F1 = \frac{k\ q^2}{d^2}

and force between charges at A to C, at 2d distance

F1 = \frac{k\ q^2}{(2d)^2}  =  \frac{k\ q^2}{4d^2}

force between charges at A to D,  3d distance

F1 = \frac{k\ q^2}{(3d)^2}  = \frac{k\ q^2}{9d^2}  

so

Charge a It receives force to the left from b and c and to the right from d

so at a will be

F(a)  = -F1 - F2 + F3             ....................2

put here value

F(a) = -\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}

solve it

F(a) = \frac{k\ q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})  

F(a) = -\frac{41}{36}\ F1   = 1.13 F1  

and

Charge b It  receives force to the right from a and d and to the left from c

F(b) = F1 - F1 + F2            ....................3

F(b)  =  \frac{k\ q^2}{d^2}-\frac{k\ q^2}{d^2}+\frac{k\ q^2}{4d^2}    

F(b)  = \frac{1}{4} \ F1    =  0.25 F1

and

Charge c It receives forces to the right from all charges.

F(c) = F2 + F 1 + F 1      ....................4

F(c) = \frac{k\ q^2}{4d^2}+\frac{k\ q^2}{d^2}+\frac{k\ q^2}{d^2}      

F(c) =  \frac{9}{4} \ F1   = 2.25 F1

and

Charge d It receives forces to the left from all charges

F(d) = - F3 - F2 -F 1      ....................5

F(d) = -\frac{k\ q^2}{9d^2}-\frac{k\ q^2}{4d^2}-\frac{k\ q^2}{d^2}  

so

F(d) = -\frac{49}{36} \ F1    = 1.36 F1

and

now we get here ratio of the greatest to the smallest net force that is

ratio = \frac{2.25}{0.25}

 ratio  = 9 : 1

5 0
3 years ago
a man is 9 m behind a dorr of train when it starts moving with a=2ms^-1. how far the man have to run and after what time will he
Naya [18.7K]
Is their a multiple choice to choose from I'm not sure the answer I got is even right.
That would be very helpful.
8 0
2 years ago
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