Ask question

Ask question

All categories

4 years ago

11

What types of coast line match up with continental drifts

1 answer:

zalisa [80]4 years ago

3 0

All continental coastlines match up with continental drifts. As a matter of fact, you could literally cut out the continents from a map and put them together like a puzzle and they would fit. Easiest to notice is the west of Africa and the east of South America.

You might be interested in

The plates on a vacuum capacitor have a radius of 3.0 mm and are separated by a distance of 1.5 mm. What is the capacitance of t

Mkey [24]

Link provided in other answer is a SCAM, don’t click on the link !!!!

6 0

3 years ago

A 1-lb block and a 100-lb block are placed side by side at the top of a frictionless hill. Each is given a very light tap to beg

qwelly [4]

Answer:

(c). The two blocks end in a tie

Explanation:

the reason being the absence of any resistance offered to both of the blocks.

if the slope of the hill is for instance 60 deg.

then the acceleration in absence of any resistance is a= 9.81sin(60)

since the acceleration is same then both of the blocks will reach at the same instant

4 0

4 years ago

An astronaut landed on a far away planet that has a sea of water. To determine the gravitational acceleration on the planet's su

ra1l [238]

Answer:

Acceleration due to gravity will be $g=17.3m/sec^2$

Explanation:

We have given gauge pressure P = 3.8 atm = 3.8×101325 = 385035 Pa

Depth h = 24.3 m

Density $\rho =1000kg/m^3$

We have find the acceleration due to gravity at the surface of planet

We know that pressure is given by

$P=\rho gh$

So $385035=1000\times g\times 24.3$

$g=17.3m/sec^2$

Acceleration due to gravity will be $g=17.3m/sec^2$

4 0

4 years ago

What is the net torque on the square plate, with sides 0.2 m, from each of the three forces? F1=18 N, F2=26 N, and F3=14 N. Use

marshall27 [118]

Answer:

The net torque on the square plate is 2.72 N-m.

Explanation:

Given that,

Side = 0.2 m

Force $F_{1}=18\ N$

Force $F_{2}=26\ N$

Force $F_{3}=14\ N$

We need to calculate the torque due to force F₁

Using formula of torque

$\tau_{1}=-F_{1}d_{1}$

$\tau_{1}=-F_{1}\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{1}=-18\times\dfrac{0.2}{2}$

$\tau_{1}=-1.8\ N-m$

We need to calculate the torque due to force F₂

Using formula of torque

$\tau_{2}=F_{2}d_{2}$

$\tau_{2}=F_{2}\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{2}=26\times\dfrac{0.2}{2}$

$\tau_{2}=2.6\ N-m$

We need to calculate the torque due to force F₃

Using formula of torque

$\tau_{3}=F_{3}d_{3}$

$\tau_{3}=(F_{3}\sin45+F_{3}\cos45)\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{3}=0.1(14\sin45+14\cos45)$

$\tau_{3}=1.92\ N-m$

We need to calculate the net torque on the square plate

$\tau=\tau_{1}+\tau_{2}+\tau_{3}$

$\tau=-1.8+2.6+1.92$

$\tau=2.72\ N-m$

Hence, The net torque on the square plate is 2.72 N-m.

3 0

4 years ago

Which of the options below has the correct number of each element for the compound?

slamgirl [31]

Answer:

there are no options buddy

7 0

3 years ago

Read 2 more answers