Why are yet planes faster tharn. cargo planes?

ExtremeBDS [4]

Answer:

<h2>work = mgh</h2><h2> = 125 × 2</h2><h2> = 250 joules</h2>

7 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

17. A 25 kg block is initially at rest on a rough, horizontal surface. A horizontal force of 75 N is required to set

lapo4ka [179]

Answer:

0.30581

0.24464

Explanation:

$\mu_s$ = Coefficient of static friction

$\mu_k$ = Coefficient of kinetic friction

$F_f$ = 75 N

$F_k$ = 60 N

Normal force

$F_n=mg\\\Rightarrow F_n=25\times 9.81\\\Rightarrow F_n=245.25\ N$

Frictional force

$F_f=\mu_sF_n\\\Rightarrow \mu_s=\frac{F_f}{F_n}\\\Rightarrow \mu_s=\frac{75}{245.25}\\\Rightarrow \mu_s=0.30581$

The coefficient of static friction is 0.30581

Kinetic force

$F_k=\mu_kF_n\\\Rightarrow \mu_k=\frac{F_k}{F_n}\\\Rightarrow \mu_s=\frac{60}{245.25}\\\Rightarrow \mu_s=0.24464$

The coefficient of kinetic friction is 0.24464

4 0

3 years ago

A ball with a weight of 0.5 N is submerged under water and then released. There is a net force of 5 N upwards. what is the buoya

Natasha2012 [34]

Answer:

4.5 N upward

Explanation:

You take the net force and subtract it from the weight

~DjMia~

6 0

3 years ago

Everyone open this please!

notsponge [240]

Answer:

yeah i knowwwwwwwwwwwwww

Explanation:

7 0

3 years ago

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