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kiruha [24]
3 years ago
8

What is its maximum altitude above the ground? The answer is the maximum height above the ground

Physics
1 answer:
Kisachek [45]3 years ago
3 0

Answer:

Maximum altitude above the ground = 1,540,224 m = 1540.2 km

Explanation:

Using the equations of motion

u = initial velocity of the projectile = 5.5 km/s = 5500 m/s

v = final velocity of the projectile at maximum height reached = 0 m/s

g = acceleration due to gravity = (GM/R²) (from the gravitational law)

g = (6.674 × 10⁻¹¹ × 5.97 × 10²⁴)/(6370000²)

g = -9.82 m/s² (minus because of the direction in which it is directed)

y = vertical distance covered by the projectile = ?

v² = u² + 2gy

0² = 5500² + 2(-9.82)(y)

19.64y = 5500²

y = 1,540,224 m = 1540.2 km

Hope this Helps!!!

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4 0
2 years ago
Suppose a 48-N sled is resting on packed snow. The coefficient of kinetic friction is 0.10. If a person weighing 660 N sits on t
Annette [7]

Assume the snow is uniform, and horizontal.

Given:

coefficient of kinetic friction = 0.10 = muK

weight of sled = 48 N

weight of rider = 660 N

normal force on of sled with rider = 48+660 N = 708 N = N

Force required to maintain a uniform speed

= coefficient of kinetic friction * normal force

= muK * N

= 0.10 * 708 N

=70.8 N


Note: it takes more than 70.8 N to start the sled in motion, because static friction is in general greater than kinetic friction.


8 0
3 years ago
A particle that carries a net charge of -41.8 μc is held in a region of constant, uniform electric field. the electric field vec
miss Akunina [59]
The total work done by the electric field on the charge is given by the scalar product between the electric force acting on the charge and the displacement of the charge:
W=F d cos \theta
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3 0
3 years ago
The diagram shows a car moving along a road.
crimeas [40]

Answer:

A constant speed

correct me if I'm wrong

3 0
3 years ago
Read 2 more answers
5. Principal Rodriguez has asked all students exit the building by 2:45 pm! If you want to exit
Kryger [21]

With the Pythagoras Theorem we can find that the distance traveled is:

      39 m

The distance is the length of the path between two points and it is a scalar magnitude so if the path changes direction the Pythagorean theorem should be used

               d = \sqrt{(x-x_o)^2 + (y-y_o)^2 + (z-z_o)^2 }

Where d is the distance, (x,y,z) is the interes point, (x₀,y₀,z₀) is de reference point.

In this case, let's set a reference system in the lower part of the school, take the z-axis as vertical and set the point of arrival at as the reference (0, 0, 0).

The distance that the students descend is d₁ = 30 k ^ m, when they arrive from the bottom of the school they travel d₂ = 15 j ^ m and d₃ = 20 i ^ m

let's calculate

              d =\sqrt{(20-0)^2 + (15-0)^2 + ( 30-0)^2 }  

              d = 39.05 m

Notice that the distance by being a scalar does not have unit vectors

In conclusion using the Pythagoras Theorem we can find that the distance traveled is 39 m

Learn more about distance here:

brainly.com/question/7942332

3 0
3 years ago
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