What will sunshine help produce in the body?

How can a dictatorship best be classified

telo118 [61]

think of north korea the people are controlled by the leader

3 0

3 years ago

The charges Q1=Q and Q2=4Q that are a distance d apart, repel each other with a force of 1.60 N. What would be the force between

gladu [14]

Answer:

50.4 N

Explanation:

Q1 = Q

Q2 = 4 Q

Distance = d

The force is given by

$F = \frac{KQ_{1}Q_{2}}{d^{2}}$

$1.60 = \frac{4KQ^{2}}{d^{2}}$ .... (1)

Now,

Q3 = 2 Q

Q4 = 7 Q

distance = d/3

$F' = \frac{9KQ_{3}Q_{4}}{d^{2}}$

$F' = \frac{126KQ^{2}}{d^{2}}$ .... (2)

Divide equation (2) by equation (1), we get

F' / 1.60 = 126 / 4

F' = 50.4 N

Thus, the force is 50.4 N.

7 0

3 years ago

Which planet moves the fastest

aksik [14]

Answer:

Explanation:

Mercury moves the fastest.

8 0

2 years ago

Read 2 more answers

If the speed of a wave doubles while the frequency remains the same, it means

sattari [20]

Answer:

Explanation:

Velocity of a wave is describe as

velocity =Frequency × Wavelength

Mathematically

v = fλ

Hence, Frequency, F = v / λ

Wavelength λ = v/f

So, if the frequency is kept constant, wavelength of the wave becomes directly proportional to velocity of the wave.

And this implies that, as the speed double, the wavelength is double.

5 0

3 years ago

Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2

Bess [88]

Answer:

The acceleration of $M_2$ is $a = 0.7156 m/s^2$

Explanation:

From the question we are told that

The mass of first block is $M_1 = 2.25 \ kg$

The angle of inclination of first block is $\theta _1 = 43.5^o$

The coefficient of kinetic friction of the first block is $\mu_1 = 0.205$

The mass of the second block is $M_2 = 5.45 \ kg$

The angle of inclination of the second block is $\theta _2 = 32.5^o$

The coefficient of kinetic friction of the second block is $\mu _2 = 0.105$

The acceleration of $M_1 \ and\ M_2$ are same

The force acting on the mass $M_1$ is mathematically represented as

$F_1 = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

=> $M_1 a = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

Where T is the tension on the rope

The force acting on the mass $M_2$ is mathematically represented as

$F_2 = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

$M_2 a = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

At equilibrium

$F_1 = F_2$

So

$T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

making a the subject of the formula

$a = \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}$

substituting values $a = \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}$

=> $a = 0.7156 m/s^2$

3 0

3 years ago