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Fed [463]
4 years ago
8

How could glacial marks on one continent be similar to glacial marks on another continent far away?

Physics
1 answer:
BigorU [14]4 years ago
6 0

Answer:

Plate Tectonics

Explanation:

Plate tectonics is the scientific theory that the continents move around due to due plate tectonics and the movement of magma under the crust. This explains why some fossils of the same kind are found on other continents, and same goes for the Glacial Marks.

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Define density and give its SI unit​
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Answer:

Density is defined as mass per unit volume. its SI unit is: kg/m³

Explanation:

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2 years ago
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If a bouncing ball has a total energy of 20 J and a kinetic energy of 5 J, the ball’s potential energy is
navik [9.2K]

Answer:

15 and Increasing

Explanation:

Hope this helps

Have a wonderful day and many more to come

Alexis~

~A.K.A Moon~

7 0
2 years ago
What is the magnitude of the kinetic frictional force
Effectus [21]

The magnitude of the kinetic friction force, ƒk, on an object is. Where μk is called the kinetic friction coefficient and |FN| is the magnitude of the normal force of the surface on the sliding object. The kinetic friction coefficient is entirely determined by the materials of the sliding surfaces. hope it helps

4 0
3 years ago
A box is initially sliding across a frictionless floor toward a spring which is attached to a wall. the box hits the end of the
Serjik [45]
The elastic potential energy of a spring is given by
U= \frac{1}{2}kx^2
where k is the spring's constant and x is the displacement with respect to the relaxed position of the spring.

The work done by the spring is the negative of the potential energy difference between the final and initial condition of the spring:
W=-\Delta U =  \frac{1}{2}kx_i^2 -  \frac{1}{2}kx_f^2

In our problem, initially the spring is uncompressed, so x_i=0. Therefore, the work done by the spring when it is compressed until x_f is
W=- \frac{1}{2}kx_f^2
And this value is actually negative, because the box is responsible for the spring's compression, so the work is done by the box.
8 0
4 years ago
A basketball is held over head at a height of 2.4 m. The ball is lobbed to a teammate at 8 m/s at an angle of 40'. If the ball i
cupoosta [38]

Explanation:

since both the teammates are of the same height, their height won't matter. Because now the basketball won't cover any vertical distance.

We have to calculate its range the horizontal distance covered by it when tossed from one teammate to the other.

range can be calculated by the formula :-

\boxed{\mathfrak{range =  \frac{  u  {}^{2}   \sin 2\theta }{g} }}

u is the velocity during its take off and \theta is the angle at which its thrown

Given that

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  • \theta = 40°

calculating range using the above formula

= \frac{ {8}^{2} \sin2(40)  }{10}

=  \frac{64 \times  \sin(80) }{10}

value of sin 80 = 0. 985

=  \frac{64 \times 0.985}{10}

=  \frac{63.027}{10}

= 6.3027

Hence,

\mathfrak { \blue{the \: teammate \: is \:  \red{\underline{6.3027 \: meters} }\: away } }

7 0
3 years ago
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