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3 years ago

Is the us warmer than antarctica?

Physics

1 answer:

Likurg_2 [28]3 years ago

7 0

Answer:

Yes

Explanation:

Antarctica is usually always cold.

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Suppose there is a uniform magnetic field, B, pointing into the page (so your index finger will point into the page). If the vel

Naily [24]

Answer:

Explanation:

We shall show all given data in vector form and calculate the direction of force with the help of following formula

force F = q ( v x B )

q is charge , v is velocity and B is magnetic field.

Given B = - Bk ( i is right , j is upwards and k is straight up the page )

v = v j

F = q ( vj x - Bk )

= -Bqvi

The direction is towards left .

a ) If velocity is down

v = - v j

F = q ( - vj x - bk )

= qvB i

Direction is right .

b ) v = v i

F = q ( vi x - Bk )

= qvB j

force is upwards

c ) v = - vi

F = q ( -vi x - Bk )

= -qvBj

force is downwards

d ) v = - v k

F = q( - vk x -Bk )

= 0

No force will be created

e ) v = v k

F = q( vk x -Bk )

= 0

No force will be created

3 0

3 years ago

What us meant by Velocity ratio of simple machine is 4?<br><br><br>

butalik [34]

The force ratio of a machine is 4 and it velocity ratio is 4 means that the load moved is four times the effort applied and the distance moved by the effort is five times the distance moved by the load at the same time interval.

Please mark as brainliest

8 0

3 years ago

A horizontal pipe of diameter 1.03m has a smooth constriction to a section of diameter 0.618 m. The density of oil flowing in th

vodka [1.7K]

Velocity of the oil in the pipe: 0.76 m/s, in the constricted section: 5.87 m/s

Explanation:

We can solve this problem by using Bernoulli's equation:

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$ (1)

where

$p_1 = 7340 N/m^2$ is the pressure in the pipe

$p_2 = 5505 N/m^2$ is the pressure in the constricted section

$\rho = 821 kg/m^3$ is the density of the oil

$v_1$ is the velocity of the oil in the pipe

$v_2$ is the velocity of the oil in the constricted section

Also, according to the continuity equation,

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the pipe, with

$r_1 = \frac{1.03}{2}=0.515 m$ is the radius

$A_2 = \pi r_2^2$ is the cross-sectional area of the constricted section, with

$r_2=\frac{0.618}{2}=0.309 m$ is the radius

So the equation becomes

$r_1^2 v_1 = r_2^2 v_2$

So we can write

$v_2=\frac{r_1^2}{r_2^2}v_1$

Substituting into eq.(1),

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho (\frac{r_1^2}{r_2^2}v_1)^2$

And solving the equation for $v_1$ :

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho \frac{r_1^4}{r_2^4}v_1^2\\v_1=\sqrt{\frac{p_2-p_1}{\frac{1}{2}\rho-\frac{1}{2}\rho \frac{r_1^4}{r_2^4}}}=0.76 m/s$