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Ilia_Sergeevich [38]
3 years ago
14

An Air Force plane lands with a velocity of 125 m/s and accelerates at a maximum rate of -6.5 m/s^2.

Physics
1 answer:
Alisiya [41]3 years ago
8 0

Answer:

a) 19.2 s

b) No

Explanation:

Given:

v₀ = 125 m/s

a = -6.5 m/s²

v = 0 m/s

a) Find: t

v = at + v₀

(0 m/s) = (-6.5 m/s²) t + (125 m/s)

t ≈ 19.2 s

b) Find: Δx

v² = v₀² + 2aΔx

(0 m/s)² = (125 m/s)² + 2 (-6.5 m/s²) Δx

Δx ≈ 1200 m

An aircraft carrier that's 850 meters long won't be long enough.

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Which of the following statements about iron filings placed upon glass resting on top of a bar magnet is false?
marin [14]

Statements  A,  B,  and  C  are true.

Statement <em> D  is false</em>.

7 0
3 years ago
A thin, metallic spherical shell of radius 0.187 m has a total charge of 6.53×10−6 C placed on it. A point charge of 5.15×10−6 C
MAVERICK [17]

Answer: a) 112.88 * 10^3 N/C; b) The electric field point outward from the center of the sphere.

Explanation: In order to solve this problem we have to use the gaussian law so we use a gaussian surface at r=0.965 m and the electric flux is equal to Q inside/εo

E* 4*π*r^2= Q inside/εo

E= k*Q inside/r^2= 9*10^9*(6.53+5.15)μC/(0.965)^2=122.88 * 10 ^3 N/C

5 0
3 years ago
A tire measures 20 inches in diameter. How many revolutions will the tire have completed when it has traveled 100 yards? Express
Soloha48 [4]
<h2>Number of revolutions required to travel 100 yards is 57.</h2>

Explanation:

Diameter of tire,D = 20 inches

Perimeter of tire = πD

Perimeter of tire = π x 20 = 62.8 inches

Distance traveled = 100 yards

1 yard = 36 inches

Distance traveled = 100 x 36 = 3600 inches

In one revolution it travels 62.8 inches.

\texttt{No of revolutions = }\frac{3600}{62.8}\\\\\texttt{No of revolutions = }57.32\\\\\texttt{No of revolutions = }57

Number of revolutions required to travel 100 yards is 57.

8 0
3 years ago
A lady bug is sitting on the bottom of a can while you twirl it overhead on a string that is 65.0
MA_775_DIABLO [31]

The linear speed of the ladybug is 4.1 m/s

Explanation:

First of all, we need to find the angular speed of the lady bug. This is given by:

\omega=\frac{2\pi}{T}

where

T is the period of revolution

The period of revolution is the time taken by the ladybug to complete one revolution: in this case, since it does 1 revolution every second, the period is 1 second:

T = 1 s

Therefore, the angular speed is

\omega=\frac{2\pi}{1 s}=6.28 rad/s

Now we can find the linear speed of the ladybug, which is given by

v=\omega r

where:

\omega=6.28 rad/s is the angular speed

r = 65.0 cm = 0.65 m is the distance of the ladybug from the axis of rotation

Substituting, we find

v=(6.28)(0.65)=4.1 m/s

Learn more about angular speed:

brainly.com/question/9575487

brainly.com/question/9329700

brainly.com/question/2506028

#LearnwithBrainly

7 0
3 years ago
A wheel accelerates from rest to 34.7 rad/s at a rate of 47.0 rad/s^2. Through what angle (in radians) did the wheel turn while
dem82 [27]

12.8 rad

Explanation:

The angular displacement \theta through which the wheel turned can be determined from the equation below:

\omega^2 = \omega_0^2 + 2\alpha\theta (1)

where

\omega_0 = 0

\omega = 34.7\:\text{rad/s}

\alpha = 47.0\:\text{rad/s}^2

Using these values, we can solve for \theta from Eqn(1) as follows:

2\alpha\theta = \omega^2 - \omega_0^2

or

\theta = \dfrac{\omega^2 - \omega_0^2}{2\alpha}

\:\:\:\:= \dfrac{(34.7\:\text{rad/s})^2 - 0}{2(47.0\:\text{rad/s}^2)}

\:\:\:\:= 12.8\:\text{rad}

7 0
2 years ago
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