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miv72 [106K]
3 years ago
9

The most likely models of the planet Mercury indicate that more than half the planet may be composed of________.

Physics
1 answer:
Oxana [17]3 years ago
4 0

Answer:

iron

Explanation:

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A jogger runs 5.0 km on a straight trail at an angle of 60° south of west. What is the southern component of the run rounded to
geniusboy [140]

Answer:

4.3 km

Explanation:

5 0
2 years ago
If a high jumper needs to make his center of gravity rise 1.50 m, how fast must he be able to sprint? Assume all of his kinetic
sergiy2304 [10]

Answer:

v = 5.42 m/s

Explanation:

given,

height of the jumper = 1.5 m

velocity of sprinter = ?

kinetic energy can be transformed into potential energy

m g h = \dfrac{1}{2}mv^2

g h = \dfrac{1}{2}v^2

v =\sqrt{2gh}

v =\sqrt{2\times 9.8 \times 1.5}

v = 5.42 m/s

Speed of the sprinter is equal to v = 5.42 m/s

7 0
3 years ago
A particle moving on a circle has a velocity of 5 m/s and a normal acceleration of 10 m/s^2. What is the radius of the circle?
dybincka [34]

Answer:

Radius of the circle will be 2.5 m

Explanation:

We have given velocity of particle moving in the circle v = 5 m/sec

Acceleration of particle in the circle a=10m/sec^2

We have to find the radius of the circle

We know that acceleration is given by a=\frac{v^2}{r}

So 10=\frac{5^2}{r}

r=\frac{25}{10}=2.5m

So radius of the circle will be 2.5 m

3 0
3 years ago
Velocity vector and acceleration vector in a uniform circular motion are related as.
mr_godi [17]

They are related as \bold{\underline {v}\,.\,\underline a }= \bold{0}

  • In a uniform circular motion, the magnitude of the speed does not change during the travel and only the instantaneous direction changes.
  • This speed is always directed along the tangent to the circle at a given point. (refer to the figure attached)
  • For any circular motion, the must-have acceleration is the centripetal acceleration that is directed towards the centre of the circular locus (if the motion has a tangential acceleration, it has a tangential acceleration additionally).
  • Therefore, both the directions of the tangential speed and the centripetal acceleration are orthogonal to each other (perpendicular: one is 90 degrees apart from the other).
  • In mathematics, 2 vectors (\underline p , \underline q) that are perpendicular to each other have a quality that their dot product (\underline p\,.\, \underline q) equal to zero vector (\bold 0) which is written as \undeline p\,.\, \underline q = \bold 0.
  • This quality can be considered when dealing with the velocity vector and the acceleration vector in a manner \underline v\,.\, \underline a =\bold 0.

#SPJ4

8 0
1 year ago
If a 1.00 kg body has an acceleration of 2.44 m/s2 at 53° to the positive direction of the x axis, then what are (a) the x comp
Ilia_Sergeevich [38]

(a) Fx = 1.464 N

(b) Fy = 1.952 N

(c) F(x, y) = 1.464 i + 1.952 j

Given

Mass = 1kg

Acceleration = 2.44 m/s2

Angle with positive X axis = 53°

As we know

F = ma

By substituting value

F= 1×2.44 N

F= 2.44 N

(a)   Component of force in X direction

Fx = F Cosθ

Fx = 2.44 Cos(53°)

Fx = 2.44 × 0.60 = 1.464 N

(b) Component of force in Y direction

Fy = F Sinθ

Fy = 2.44 Sin(53°) = 2.44 × 0.80 = 1.952 N

(c) Net force in vector notation

F(x, y) = 1.464 i + 1.952 j

Thus we got net force.

#SPJ4

For details visit www.brainly.com

6 0
2 years ago
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