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aleksklad [387]
3 years ago
15

Mary needs to row her boat across a 190 m -wide river that is flowing to the east at a speed of 1.3 m/s . mary can row with a sp

eed of 3.5 m/s . part a if mary points her boat due north, how far from her intended landing spot will she be when she reaches the opposite shore?
Physics
1 answer:
Anarel [89]3 years ago
3 0
<span>If Mary is rowing at 3.5 m/s then it will take her 190 m divided by 3.5 m/s or about 54 secs to cross the river. If the river is flowing at 1.3 m/s then in 54 secs she will be carried 54 x 1.3 or around 70.5 meters downstream when she lands on the opposite shore.</span>
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Describe and name the different types of collision. In which are the linear momentum and kinetic energy conserved
ipn [44]

Answer:

1. Elastic collision

2. Inelastic collision    

Explanation:

Elastic collision: collision is said to be elastic if total kinetic energy is not conserved and if there is a rebound after collision

the collision is described by the equation bellow

m1U1+ m2U2= m1V1+m2V2

Inelastic collision: this type of collision occurs when the total kinetic energy of a body is conserved or when the bodies sticks together and move with a common velocity

the collision is described by the equation bellow

m1U1+ m2U2= V(m1+m2)

8 0
3 years ago
A train increase its speed steadily from 10m/s to 20m/s in 1minutes A what is the average speed during this time in m/s B how fa
Ipatiy [6.2K]

If it increased its speed steadily at a constant rate, then the average speed for the minute was

(1/2)(10m/s + 20m/s) = 15 m/s .

Rolling at an average speed of 15 m/s for 1 minute (60 seconds), it travels

(15 m/s) (60 sec) = 900 meters

3 0
3 years ago
2. A mixture of mercury and copper is an example of
Olin [163]

Answer:

b

Explanation:

liquid and solid

hope that helps :)

4 0
2 years ago
Read 2 more answers
Here's a basketball problem: A 87.2 kg basketball player is running in the positive direction at 7.0 m/s. She is met head-on by
Ray Of Light [21]

Answer:

2.47 m/s backwards

Explanation:

From the law of conservation of momentum,

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂...................... Equation 1

Where m₁ and m₂ = mass of the first basketball player and second basket ball player respectively, u₁ and u₂ = initial velocity of the first basket player and the second basketball player respectively, v₁ and v₂ = The final velocity of the first basket ball player and second basket ball player respectively.

Making v₁ the subject of the equation,

v₁ = (m₁u₁ + m₂u₂ - m₂v₂)/m₁.......................... Equation 2.

Given: m₁ = 87.2 kg, m₂ = 102.0 kg, u₁ = 7.0 m/s, u₂ = -5.2 m/s, v₂ = 2.9 m/s

Note: u₂ is negative because it moves towards the first basket ball player.

Substitute into equation 2

v₁ = [87.2(7.0)+102(-5.2) - (102×2.9)]/87.2

v₁ = (610.4-530-295.8)/87.2

v₁ = -215.4/87.2

v₁ = -2.47 m/s.

Thus the velocity of the 87.2 kg player = 2.47 m/s backwards.

7 0
3 years ago
In a football game, the running back takes a handoff and begins running toward midfield at 3.21 yards/s . As he moves through hi
Inessa [10]

Given Information:

Initial speed = u = 3.21 yards/s

Acceleration = α  = 1.71 yards/s²

Final speed = v = 7.54 yards/s

Required Information:

Distance = s = ?

Answer:

Distance = s = 13.61

Explanation:

We are given the speeds and acceleration of the runner and we want to find out how much distance he covered before being tackled.

We know from the equations of motion,

v² = u² + 2αs

Where u is the initial speed of the runner, v is the final speed of the runner, α is the acceleration of the runner and s is the distance traveled by the runner.

Re-arranging the above equation for distance yields,

2αs = v² - u²

s = (v² - u²)/2α

s = (7.54² - 3.21²)/2×1.71

s = 46.55/3.42

s = 13.61 yards

Therefore, the runner traveled a distance of 13.61 yards before being tackled.

8 0
3 years ago
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