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Nostrana [21]
2 years ago
12

There are 11 steps one should follow when shooting true or false(Archery)​

Physics
1 answer:
kkurt [141]2 years ago
4 0

Answer:

true

Explanation:

1. Stance

2. Nock

3. Set Draw Hand

4. Set Bow Hand

5. Pre-Draw

6. Draw

7. Anchor

8. Aim

9. Shot Set Up

10. Release

11. Follow-Through & Reflect

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Does the fastest car always travel the farthest?
Fiesta28 [93]

Answer:

No because it could run out of gas quicker or get into a reck.

Explanation:

4 0
2 years ago
A 2,000-kilogram car uses a braking force of 12,000 Newton to stop in 5 seconds. What is its initial speed of the car?
AnnZ [28]
First, solve for the acceleration of the car. You know the mass of the car and the braking force, so you can use the equation Force = Mass x Acceleration. This gives you 12,000 = 2,000 x A. Divide 12,000 by 2,000 to find the acceleration equal to 6 m/s^2. This is the rate that the car is slowing down at. Velocity is equal to accleration x time (rate x time), so you multiply 6 by the time of 5 seconds. This leaves you with a velocity of 30 m/s or about 67.1 mph.
5 0
3 years ago
Read 2 more answers
In an elastic collision, a 400-kg bumper car collides directly from behind with a second, identical bumper car that is traveling
kirza4 [7]

Answer:

v₁ = 3.5 m/s

v₂ = 6.4 m/s

Explanation:

We have the following data:

m₁ = mass of trailing car = 400 kg

m₂ = mass of leading car = 400 kg

u₁ = initial speed of trailing car = 6.4 m/s

u₂ = initial speed of leading car = 3.5 m/s

v₁ = final speed of trailing car = ?

v₂ = final speed of leading car = ?

The final speed of the leading car is given by the following formula:

v_2=\frac{2m_1}{m_1+m_2}u_1-\frac{m_1-m_2}{m_1+m_2}u_2\\\\v_2=\frac{(2)(400\ kg)}{400\ kg+400\ kg}(6.4\ m/s)-\frac{400\ kg-400\ kg}{400\ kg + 400\ kg}(3.5\ m/s)

<u>v₂ = 6.4 m/s</u>

The final speed of the leading car is given by the following formula:

v_1=\frac{m_1-m_2}{m_1+m_2}u_1+\frac{2m_2}{m_1+m_2}u_2\\\\v_1=\frac{400\ kg-400\ kg}{400\ kg + 400\ kg}(6.4\ m/s)+\frac{(2)(400\ kg)}{400\ kg+400\ kg}(3.5\ m/s)

<u>v₁ = 3.5 m/s</u>

4 0
3 years ago
A motorcycle accelerates uniformly from rest at 7.9\,\dfrac{\text{m}}{\text{s}^2}7.9 s 2 m ​ 7, point, 9, space, start fraction,
8090 [49]

Answer:

t = 3.516 s

Explanation:

The most useful kinematic formula would be the velocity of the motorcylce as a function of time, which is:

v(t) = v_0 +at

Where v_0 is the initial velocity and a is the acceleration. However the problem states that the motorcyle start at rest therefore v_0 = 0

If we want to know the time it takes to achieve that speed, we first need to convert units from km/h to m/s.

This can be done knowing that

1 km = 1000 m

1 h = 3600 s

Therefore

1 km/h = (1000/3600) m/s = 0.2777... m/s

100 km/h = 27.777... m/s

Now we are looking for the time t, for which v(t) = 27.77 m/s. That is:

27.777 m/s = 7.9 m/s^2 t

Solving for t

t = (27.7777 / 7.9) s = 3.516 s

6 0
3 years ago
A scientist notices that an oil slick floating on water when viewed from above has many different colors reflecting off the surf
Sati [7]

Answer:

a) The minimum thickness of the oil slick at the spot is 313 nm

b) the minimum thickness be now will be 125 nm

Explanation:  

Given the data in the question;

a) The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot?

t_{min = λ/2n

given that; wavelength λ = 750 nm and  index of refraction of the oil n = 1.20

we substitute

t_{min = 750 / 2(1.20)

t_{min = 750 / 2.4

t_{min = 312.5 ≈ 313 nm

Therefore, The minimum thickness of the oil slick at the spot is 313 nm

b)

Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now?

minimum thickness of the oil slick at the spot will be;

t_{min = λ/4n

given that; wavelength λ = 750 nm and  index of refraction of the oil n = 1.50

we substitute

t_{min = 750 / 4(1.50)

t_{min = 750 / 6

t_{min = 125 nm

Therefore, the minimum thickness be now will be 125 nm

4 0
2 years ago
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