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Minchanka [31]
2 years ago
12

A bicyclist travels $4.5\text{ km}$ west, then travels $6.7\text{ km}$ at an angle $27.0^\circ$ South of West.

Physics
1 answer:
MatroZZZ [7]2 years ago
7 0

Answer:

10.88 km

Explanation:

We shall represent displacement in terms of i , j  unit vectors in the direction of east and north .

4.5 km due west

D₁ = - 4.5 i

6.7 km at an angle of 27° south of west

D₂ = - 6.7 cos27 i - 6.7 sin27j

= - 6.7 x .89 i - 6.7 x .45 j

= - 5.96i - 3 j

Total displacement

= D₁ + D₂

=  - 4.5 i - 5.96i - 3 j

= -10.46 i - 3j

Magnitude = √ ( 10.46² + 3²)

= √ ( 109.41 + 9)

= √ 118.41

= 10.88 km .

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You can describe the motion of an object by its position, speed, direction, and acceleration
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2 years ago
On February 15, 2013, Asteroid 2012 DA14 passed within 17,200 miles [mi] of the surface of the Earth at a relative speed of 7.8
xz_007 [3.2K]

Answer:

The total amount of energy that would have been released if the asteroid hit earth = The kinetic energy of the asteroid = 1.29 × 10¹⁵ J = 1.29 PetaJoules = 1.29 PJ

1 PJ = 10¹⁵ J

Explanation:

Kinetic energy = mv²/2

velocity of the asteroid is given as 7.8 km/s = 7800 m/s

To obtain the mass, we get it from the specific gravity and diameter information given.

Density = specific gravity × 1000 = 3 × 1000 = 3000 kg/m³

But density = mass/volume

So, mass = density × volume.

Taking the informed assumption that the asteroid is a sphere,

Volume = 4πr³/3

Diameter = 30 m, r = D/2 = 15 m

Volume = 4π(15)³/3 = 14137.2 m³

Mass of the asteroid = density × volume = 3000 × 14137.2 = 42411501 kg = 4.24 × 10⁷ kg

Kinetic energy of the asteroid = mv²/2 = (4.24 × 10⁷)(7800²)/2 = 1.29 × 10¹⁵ J

6 0
3 years ago
An Earth satellite moves in a circular orbit 680 km above Earth's surface with a period of 98.15 min. What are (a) the speed and
kotykmax [81]

The solution is in the attachment

6 0
2 years ago
Experimenting with free fall, Mariana observes that her baseball takes 1.5 s to travel the last 30m before hitting the ground. F
Art [367]

Answer:

37.8 m

Explanation:

At point 0, the ball is at height y₀.

At point 1, the ball is at height 30 m.

At point 2, the ball is at height 0 m.

Given:

y₁ = 30 m

y₂ = 0 m

v₀ = 0 m/s

a = -10 m/s²

t₂ − t₁ = 1.5 s

Find: y₀

Use constant acceleration equation.

y = y₀ + v₀ t + ½ at²

Evaluate at point 1.

y₁ = y₀ + v₀ t₁ + ½ at₁²

30 m = y₀ + (0 m/s) t₁ + ½ (-10 m/s²) t₁²

30 = y₀ − 5t₁²

Evaluate at point 2.

y₂ = y₀ + v₀ t₂ + ½ at₂²

0 m = y₀ + (0 m/s) t₂ + ½ (-10 m/s²) t₂²

0 = y₀ − 5t₂²

y₀ = 5t₂²

Substitute:

y₀ = 5 (1.5 + t₁)²

y₀ = 5 (2.25 + 3t₁ + t₁²)

y₀ = 11.25 + 15t₁ + 5t₁²

30 = 11.25 + 15t₁ + 5t₁² − 5t₁²

30 = 11.25 + 15t₁

t₁ = 1.25

30 = y₀ − 5t₁²

30 = y₀ − 5(1.25)²

y₀ ≈ 37.8

4 0
3 years ago
A distance of 2.00 mm separates two objects of equal mass. If the gravitational force between them is 0.0104 N, find the mass of
Reptile [31]
F = \frac{gm,m2}{r2}
r = 0.002

0.0104 = \frac{(6.673 x  10^{-11} }{ 0.002)^{2} }

m = 25.0kg
4 0
2 years ago
Read 2 more answers
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