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harkovskaia [24]
3 years ago
8

Please select the word from the list that best fits the definition the name for continents in one landmass surrounded by giganti

c ocean
Physics
1 answer:
frez [133]3 years ago
8 0
Hello!

The name for continents in one landmass surrounded by gigantic ocean is Pangaea.

This name is originated from the Greek words Pan ("all") and Gaia ("Earth"). It describes the supercontinent that would have existed in the Cenozoic and Mesozoic periods by the joining of all continents that we know them today.

The gigantic ocean that surrounded it is called Panthalassa, from Greek Pan ("all") and thálassa ("sea"). 

Have a nice day!
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A box falls off of a tailgate and slides along the street for a distance of 62.5 m. Friction slows the box at –5.0 m/s2. At what
Mila [183]

Answer:

25 m/s

Explanation:

This question can be solved using equation of motion

v^2 = u^2 + 2as

where

v is the final velocity

u is the initial velocity

s is the distance covered while moving from initial to final velocity

a is the acceleration

_____________________________________________

Given

box moved for distance of 62.5 m

Friction slows the box at –5.0 m/s2----> this statement means that there is deceleration , speed of truck decreases by 5 m/s in every second until the box comes to rest. Friction causes this deceleration.

thus in this problem

a = -5.0 m/s2

V = 0   as body came to rest due to friction deceleration

u the initial velocity we have to find

the initial velocity of box will be the same as speed of truck, as the box was in the truck and hence box will pick the speed of truck.

so if we find speed of box, we will be able get sped of truck as well.

using equation of motion

v^2 = u^2 + 2as\\0^2 = u^2 + 2*-5* 62.5\\0 = u^2 - 625\\u^2 = 625\\\sqrt{u^2} = \sqrt{625} \\u = 25

Thus, initial speed with the truck was travelling was 25 m/s.

3 0
3 years ago
What Kinetic energy is exactly equal to Gravitational Potential Energy why is height halfway between the maximum height?
prohojiy [21]

Explanation:

Let us calculate the work done in lifting an object of mass m through a height h, such as in Figure 1. If the object is lifted straight up at constant speed, then the force needed to lift it is equal to its weight mg. The work done on the mass is then W = Fd = mgh. We define this to be the gravitational potential energy (PEg) put into (or gained by) the object-Earth system. This energy is associated with the state of separation between two objects that attract each other by the gravitational force

Potential energy is a property of a system rather than of a single object—due to its physical position. An object’s gravitational potential is due to its position relative to the surroundings within the Earth-object system. The force applied to the object is an external force, from outside the system. When it does positive work it increases the gravitational potential energy of the system. Because gravitational potential energy depends on relative position, we need a reference level at which to set the potential energy equal to 0. We usually choose this point to be Earth’s surface, but this point is arbitrary; what is important is the difference in gravitational potential energy, because this difference is what relates to the work done. The difference in gravitational potential energy of an object (in the Earth-object system) between two rungs of a ladder will be the same for the first two rungs as for the last two rungs.

5 0
2 years ago
IMPORTANT!!! DUE IN 5 MINS I NEED HELP PLEASE
Ket [755]

Answer:

684.5 is the weight on mars and 1813 on earth

Explanation:

185*3.7=684.5 185*9.8=1813 you multiply for earth by 9.8 because that's the gravity on earth and you multiply by 3.7 because that's the gravity on mars

6 0
2 years ago
Some of the highest tides in the world occur in the Bay of Fundy on the Atlantic Coast of Canada. At Hopewell Cape the water dep
Nady [450]

Answer:

(a) 1.939 m/h

(b) 0.926 m/h

(c) -0.315 m/h

(d) -1.21 m/h

Explanation:

Here, we have the water depth given by the function of time;

D(t) = 7 + 5·cos[0.503(t-6.75)]

Therefore, to find the velocity of the depth displacement with time, we differentiate the given expression with respect to time as follows;

D'(t) = \frac{d(7 + 5\cdot cos[0.503(t-6.75)])}{dt}

= 5×(-sin(0.503(t-6.75))×0.503

= -2.515×(-sin(0.503(t-6.75))

= -2.515×(-sin(0.503×t-3.395))

Therefore we have;

(a) at 5:00 AM = 5 -  0:00 = 5

D'(5) =  -2.515×(-sin(0.503×5-3.395)) = 1.939 m/h

(b) at 6:00 AM = 6 -  0:00 = 6

D'(5) =  -2.515×(-sin(0.503×6-3.395)) = 0.926 m/h

(c) at 7:00 AM = 7 -  0:00 = 7

D'(5) =  -2.515×(-sin(0.503×7-3.395)) = -0.315 m/h

(d) at Noon 12:00 PM = 12 -  0:00 = 12

D'(5) =  -2.515×(-sin(0.503×12-3.395)) = -1.21 m/h.

4 0
3 years ago
Find Vxl and Vyl of a pumpkin launched at a velocity of 55 m/s at an angle of 20 degrees
Vinvika [58]

Answer:

             

Explanation:

is  A projectile is any object on which the only force acting is gravity and air resistance (drag).

Examples of projectiles are:

baseballs and softballs in the air after being hit by the bat

golf balls hit by a club

objects dropped from aircraft, such as people (skydivers), bombs, crates of food being dropped to refugees

objects launched by cannons, such as cannonballs, shells, and circus performers

Once the baseball, softball, golf ball, skydiver, bomb, crate, cannonball, shell, or clown are no longer touching the bat, club, aircraft, or cannon, and are in the air with only gravity and slight air resistance acting on it, then it is a projectile.

Here is an online projectile motion applets to play with, just for fun.

Unless otherwise stated in a particular problem or discussion, we will be ignoring the effects of air resistance.

The key to understanding the motion of projectiles is that the horizontal motion and the vertical motion of the projectile are independent of each other. So we can write separate equations for the displacement of the projectile in the horizontal (x) and vertical (y) directions.

                         

The only common variable between these two equations is t, the time. Because in projectile problems there is usually no acceleration (i.e. we ignore air resistance) in the horizontal direction, we can write

           

The velocity components follow the same equations we used for one-dimensional motion.

                             

Because there is usually no acceleration in the x direction, the x-velocity is constant.

3 0
3 years ago
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