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Ket [755]
3 years ago
5

A paper airplane is thrown horizontally with a velocity of 20 mph. The plane is in the air for 7.63 s before coming to a standst

ill on the ground. What is the acceleration of the plane?
Physics
1 answer:
Inessa05 [86]3 years ago
7 0

Answer:

-1.17 m/s²

Explanation:

Given:

v₀ = 20 mph = 8.94 m/s

v = 0 m/s

t = 7.63 s

Find: a

v = at + v₀

0 m/s = a (7.63 s) + 8.94 m/s

a = -1.17 m/s²

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Submarine a travels horizontally at 11.0 m/s through ocean water. it emits a sonar signal of frequency f 5 5.27 3 103 hz in the
xeze [42]
Velocity of submarine A is vs = 11.0m/s
frequency emitted by submarine A. F = 55.273 × 10∧3HZ
Velocity of submarine B = vO = 3.00m/s
The given equation is
f' = ((V + vO) ((v - vS)) × f
The observer on submarine detects the frequency f'.
The sign of vO should be positive as the observer of submarine B is moving away from the source of submarine A.
The speed of the sound used in seawater is 1533m/s
The frequency which is detected by submarine B is 
fo = fs (V -vO/ v +vs)
= 53.273 × 10∧3hz) ((1533 m/s - 4.5 m/s)/ (1533 m/s +11 m/s)
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6 0
4 years ago
You are at the top of the Empire State Building on the 102nd floor, which is located 373 m above the ground, when your favorite
Liula [17]

Answer:

1.5 m

Explanation:

H = actual height of the superhero = ?

H₀ = height of the superhero as observed = 1.73 m

v = speed of the superhero = 0.50 c

Using the equation

H = H_{o} \sqrt{1 - \left ( \frac{v}{c} \right )^{2}}

Inserting the values

H = 1.73 \sqrt{1 - \left ( \frac{0.50 c}{c} \right )^{2}}

H = 1.5 m

3 0
4 years ago
A softball is thrown perfectly vertically (straight-up) at 18 m/s.
inessss [21]

Answer:

a) y = 16.51 [m]

b) t = 1.83 [s]

Explanation:

To solve this problem we must use two kinematics equations, the first to determine the height to which the ball reaches, and the second equation to determine how long it lasts in the air.

v_{f} ^{2}= v_{i} ^{2} - (2*g*t)\\

where:

Vf = final velocity = 0

Vi = initial velocity = 18 [m/s]

g = gravity acceleration = 9.81[m/s^2]

t = time [s]

Note: the negative sign of the Equation indicates that the acceleration of gravity acts in the opposite direction to the movement of the ball.  The final velocity is zero, since the ball reaches its maximum altitude when the velocity is zero.

Now replacing:

0 = (18)^2 - (2*9.81*y)

y = 16.51 [m]

b)

v_{f} = v_{i} - (g*t)

0 = 18 - (9.81*t)

t = 1.83 [s]

8 0
3 years ago
59. (II) The crate shown in Fig. 4-60 lies on a plane tilted at an angle A = 25.0° to the horizontal, with Mk 0.19. (a) Determin
Daniel [21]

Explanation:

a) We need to write down first Newton's 2nd law as applied to the given system. The equations of motion for the x- and y-axes can be written as follows:

x:\;\;\;\;\;mg\sin 25° - \mu_kN = ma\;\;\;\;\;\;(1)

y:\;\;\;\;\;N - mg\cos 25° = 0\;\;\;\;\;\;\;\;\;(2)

From Eqn(2), we see that

N = mg\cos 25°\;\;\;\;\;\;\;(3)

so using Eqn(3) on Eqn(1), we get

mg\sin 25° - \mu_kmg\cos 25° = ma

Solving for the acceleration, we see that

a = g(\sin 25° - \mu_k\cos 25°)

\;\;\;\;= 2.45\:\text{m/s}^2

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v^2 = v_0^2 + 2ax

Since the crate started from rest, v_0 = 0. Thus our equation reduces to

v^2 = 2ax \Rightarrow v = \sqrt{2ax}

v = \sqrt{2(2.45\:\text{m/s}^2)(8.15\:\text{m})}

\;\;\;\;= 6.32\:\text{m/s}

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