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3 years ago

What type of plate boundary interactions takes place when tectonic plates move apart?

1 answer:

8 0

Tectonic plates interactions are of three different basic types: Divergent boundaries are areas where plates move away from each other, forming either mid-oceanic ridges or rift valleys.

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A typical human consumes 2500 Kcal of energy during a day. This is the equivalent to 10,450,000 J! Say you decided to run stairs

defon

Answer:

2940.1 joules would you burn in climbing stairs all day.

Explanation:

Work = W = F $\times$ d

going up stairs would be against force of gravity

W = mgh

where h is the height

the question is not complete because we need speed or distance

h = v $\times$ t

so assuming 1 step per second

h = 86,400 steps $\times$ 7inchs/step $\times$ 0.0254 m/inch

h = 15362 m

so from this

W = 800 N $\times$ 15362

= 12289600 J

that means YOU need 12289600 J to walk 1 step per second all day

divide that by 4180 J /Kcal

Kcal = $\frac{W}{(J/Kcal)}$

= $\frac{12289600}{4180}$

= 2940.1 Kcal

if you ran faster you would use more energy 2 steps per second would mean 5880 Kcal.

8 0

3 years ago

How long must a 400w electrical engine work in order to produce 300kj of work

vladimir1956 [14]

Answer:

300000 J / 400 J/s = 750 s = 12.5 minutes

Explanation:

8 0

2 years ago

A bar of length L = 8 ft and midpoint D is falling so that, when θ = 27°, ∣∣v→D∣∣=18.5 ft/s , and the vertical acceleration of p

777dan777 [17]

Answer:

alpha=53.56rad/s

a=5784rad/s^2

Explanation:

First of all, we have to compute the time in which point D has a velocity of v=23ft/s (v0=0ft/s)

$v=v_0+at\\\\t=\frac{v}{a}=\frac{(23\frac{ft}{s})}{32.17\frac{ft}{s^2}}=0.71s$

Now, we can calculate the angular acceleration (w0=0rad/s)

$\theta=\omega_0t +\frac{1}{2}\alpha t^2\\\alpha=\frac{2\theta}{t^2}$

$\alpha=\frac{27}{(0.71s)^2}=53.56\frac{rad}{s^2}$

with this value we can compute the angular velocity

$\omega=\omega_0+\alpha t\\\omega = (53.56\frac{rad}{s^2})(0.71s)=38.02\frac{rad}{s}$

and the tangential velocity of point B, and then the acceleration of point B:

$v_t=\omega r=(38.02\frac{rad}{s})(4)=152.11\frac{ft}{s}\\a_t=\frac{v_t^2}{r}=\frac{(152.11\frac{ft}{s})^2}{4ft}=5784\frac{rad}{s^2}$

hope this helps!!

6 0

3 years ago

Read 2 more answers

A football punter wants to kick the ball so that it is in the air for 4.5 s and lands 50 m from where it was kicked. Assume that

irakobra [83]

Answer:

(a) The angle of projection is 63 degree.

(b) The velocity of projection is 24.5 m/s.

Explanation:

Height, h = 1 m

horizontal distance, d = 50 m

time, t = 4.5 s

Let the initial velocity is u and the angle is A.

(a) Horizontal distance = horizontal velocity x time

50 = u cos A x 4.5

u cos A = 11.1 .....(1)

Use second equation of motion in vertical direction

$h = u t + 0.5 gt^2\\\\- 1 = u sin A \times 4.5 - 0.5 \times 9.8\times 4.5^2\\\\u sin A = 21.8 ..... (2)$

Divide (2) by (1)

tan A = 1.97

A = 63 degree

(b) Substitute the value of A in equation (2)

u x sin 63 = 21.8

u = 24.5 m/s

7 0

2 years ago

An all-electric car (not a hybrid) is designed to run from a bank of 12.0 V batteries with total energy storage of 2.30 ✕ 107 J.

AnnyKZ [126]

Answer:

a) $I=733.33\ A$

b) $d=52272.7273\ m$

c) $d'=51948.0519\ m$

Explanation:

Given:

voltage of the battery, $V=12\ V$

energy storage capacity of the battery, $E=2.3\times 10^7\ J$

speed of the car, $v=20\ m.s^{-1}$

power drawn by the car, $P=8.8\ kW$

Now the Current delivered to the motor:

we the relation between the power and electrical current,

$P=V.I$

$8800=12\times I$

$I=733.33\ A$

Distance travelled before battery is out of juice:

we first find the time before the battery runs out,

$t=\frac{E}{P}$

$t=\frac{2.3\times 10^7}{8800}$

$t=2613.636\ s$

Now the distance:

$d=v.t$

$d=20\times 2613.636$

$d=52272.7273\ m$

When the head light of 55 W power is kept on while moving then the power consumption of the car is:

$P'=P+55$

$P'=8800+55$

$P'=8855\ W$

Now the time of operation of the car:

$t'=\frac{E}{P'}$

$t'=\frac{2.3\times 10^7}{8855}$

$t'=2597.4026\ s$

Now the distance travelled:

$d'=v.t'$

$d'=20\times 2597.4025$

$d'=51948.0519\ m$

5 0

2 years ago