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MariettaO [177]
3 years ago
13

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

Physics
1 answer:
grigory [225]3 years ago
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. <span />
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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}

a)

power drawn by the car, P=8.8\ kW

<u>Now the Current delivered to the motor:</u>

we the relation between the power and electrical current,

P=V.I

8800=12\times I

I=733.33\ A

b)

<u>Distance travelled before battery is out of juice:</u>

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

c)

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

<u>Now the time of operation of the car:</u>

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

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

t'=2597.4026\ s

<u>Now the distance travelled:</u>

d'=v.t'

d'=20\times 2597.4025

d'=51948.0519\ m

5 0
2 years ago
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