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egoroff_w [7]
4 years ago
10

A cannonball is fired on flat ground at 420 m/s at a 53.0° angle. how far away does it land?

Physics
1 answer:
Anastaziya [24]4 years ago
3 0

Answer:

17,300 m

Explanation:

Using kinematic equations, first find the time it takes to land.

Δy = v₀ t + ½ at²

0 m = (420 sin 53.0° m/s) t + ½ (-9.8 m/s²) t²

t = 0 s or 68.5 s

The horizontal distance it moves in that time is:

Δx = v₀ t + ½ at²

Δx = (420 cos 53.0° m/s) (68.5 s) + ½ (0 m/s²) (68.5 s)²

Δx = 17,300 m

Alternatively, you can use the range equation:

R = v₀² sin(2θ) / g

R = (420 m/s)² sin(2 × 53.0°) / (9.8 m/s²)

R = 17,300 m

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Alchemists searched for ways to change lead into gold. Which type of change
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A fish swimming in a horizontal plane has velocity i = (4.00 + 1.00 ) m/s at a point in the ocean where the position relative to
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Answer:

a. The horizontal component of acceleration a₁ = 0.68 m/s²

The vertical component of acceleration a₂ = -0.11 m/s²

b. -9.19° = 350.81° from the the positive x-axis

Explanation:

The initial velocity v₁ of the fish is v₁ = 4.00i + 1.00j m/s. Its final velocity after accelerating for t = 19.0 s is v₂ =  17.0i - 1.00j m/s

a. The acceleration a = (v₂ - v₁)/t = [17.0i - 1.00j - (4.00i + 1.00j)]/19 = [(17.0 -4.0)i - (-1.0 -1.0)j]/19 = (13.0i - 2.0j)/19 = 0.68i - 0.11j m/s²

The horizontal component of acceleration a₁ = 0.68 m/s²

The vertical component of acceleration a₂ = -0.11 m/s²

b. The direction of the acceleration relative to the unit vector i,

tanθ = a₂/a₁ = -0.11/0.68 = -0.1618

θ = tan⁻¹(-0.1618) = -9.19° ⇒ 360 + (-9.19) = 350.81° from the the positive x-axis

6 0
3 years ago
A car and a train move together along straight, parallel paths with the same constant cruising speed v0. At t=0 the car driver n
satela [25.4K]

Answer:

a) t1 = v0/a0

b) t2 = v0/a0

c) v0^2/a0

Explanation:

A)

How much time does it take for the car to come to a full stop? Express your answer in terms of v0 and a0

Vf = 0

Vf = v0 - a0*t

0 = v0 - a0*t

a0*t = v0

t1 = v0/a0

B)

How much time does it take for the car to accelerate from the full stop to its original cruising speed? Express your answer in terms of v0 and a0.

at this point

U = 0

v0 = u + a0*t

v0 = 0 + a0*t

v0 = a0*t

t2 = v0/a0

C)

The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed v0 again? Express the separation distance in terms of v0 and a0 . Your answer should be positive.

t1 = t2 = t

Distance covered by the train = v0 (2t) = 2v0t

and we know t = v0/a0

so distanced covered = 2v0 (v0/a0) = (2v0^2)/a0

now distance covered by car before coming to full stop

Vf2 = v0^2- 2a0s1

2a0s1 = v0^2

s1 = v0^2 / 2a0

After the full stop;

V0^2 = 2a0s2

s2 = v0^2/2a0

Snet = 2v0^2 /2a0 = v0^2/a0

Now the separation between train and car

= (2v0^2)/a0 - v0^2/a0

= v0^2/a0

8 0
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