Spiders cannot actually propel their bodies through the water as a swimmer does, but they can use objects to get across the water and some can run across the water.

Explanation:

3 0

3 years ago

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Balance the equation

mote1985 [20]

Is this a chemistry question ?

7 0

3 years ago

A proton enters a region of constant magnetic field, perpendicular to the field and after being accelerated from rest by an elec

kicyunya [14]

Answer:

B = 1.1413 10⁻² T

Explanation:

We use energy concepts to calculate the proton velocity

starting point. When entering the electric field

Em₀ = U = q V

final point. Right out of the electric field

em_f = K = ½ m v²

energy is conserved

Em₀ = Em_f

q V = ½ m v²

v = $\sqrt{2qV/m}$

we calculate

v = $\sqrt{\frac{ 2 \ 1.6 \ 10^{-19} \ 300}{1.67 \ 0^{-27}} }$

v = $\sqrt{632.3353 \ 10^8}$

v = 25.15 10⁴ m / s

now enters the region with magnetic field, so it is subjected to a magnetic force

F = m a

the force is

F = q v x B

as the velocity is perpendicular to the magnetic field

F = q v B

acceleration is centripetal

a = v² / r

we substitute

qvB =1/2 m v² / r

B = v $\frac{m v}{2 q r}$

we calculate

B = $\frac{1.67 \ 10^{-27} 25.15 \ 10^4 }{1.6 \ 10^{-19} 0.23}$

B = 1.1413 10⁻² T

8 0

3 years ago

A ball is thrown downward at 12 m/s from a windowsill 35 m above the ground. At the same time, another ball is thrown upward at

wariber [46]

Answer:

The second ball lands 1.5 s after the first ball.

Explanation:

Given;

initial velocity of the ball, u = 12 m/s

height of fall, h = 35 m

initial velocity of the second, v = 12 m/s

Time taken for the first ball to land;

$t = \sqrt{\frac{2h}{g} }\\\\t =\sqrt{ \frac{2*35}{9.8}}\\\\t = 2.67 \ s$

determine the maximum height reached by the second ball;

v² = u² -2gh

at maximum height, the final velocity, v = 0

0 = 12² - (2 x 9.8)h

19.6h = 144

h = 144 / 19.6

h = 7.35 m

time to reach this height;

$t_1 = \sqrt{\frac{2h}{g} }\\\\t_1 = \sqrt{\frac{2*7.35}{9.8}}\\\\t_1 = 1.23 \ s$

Total height above the ground to be traveled by the second ball is given as;

= 7.35 m + 35m

= 42.35 m

Time taken for the second ball to fall from this height;

$t_2 = \sqrt{\frac{2h}{g} }\\\\t_2 = \sqrt{\frac{2*42.35}{9.8} }\\\\t_2 = 2.94 \ s$

total time spent in air by the second ball;

T = t₁ + t₂

T = 1.23 s + 2.94 s

T = 4.17 s

Time taken for the second ball to land after the first ball is given by;

t = 4.17 s - 2.67 s

T = 1.5 s

Therefore, the second ball lands 1.5 s after the first ball.

4 0

3 years ago