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3 years ago

What is a situation when you might travel at a high velocity, but with low acceleration?

2 answers:

7 0

Rocket in space, once it gets into space and we're to just orbit the earth you're traveling at a high velocity because velocity is change in direction or speed, and you're constantly changing in direction relative to earth

Alexandra [31]3 years ago

5 0

Cruising at 35,000 feet in the air, straight toward the east at 500 miles per hour. Since your speed is not changing and your moving in a straight line, your acceleration is 0.

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During a circus performance, a 72-kg humancannonball is shot out of an 18-m-long cannon. If thehuman cannonball spends 0.95 s in

andreyandreev [35.5K]

Answer:

2872.8 N

Explanation:

We have the following information

m =n72kg

Δy = 18m

t = 0.95s.

From here we use the equation

Δy=1/2at2 in order to solve for the acceleration.

So a

=( 2x 18m)/(0.95s²)

= 36/0.9025

= 39.9m/s2.

From there we use the equation

F = ma

F=(72kg) x (39.9)

= 2872.8N.

2872.8N is the average net force exerted on him in the barrel of the cannon.

Thank you!

7 0

3 years ago

4. Using the bone density of 2.0 kg/m3, calculate the mass of an adult femur bone that has a volume of 0.00027 m3.

kolbaska11 [484]

Answer:

$\boxed{\sf Mass \ of \ an \ adult \ femur \ bone = 0.00054 \ kg}$

Given:

Bone density = 2.0 kg/m³

Volume of bone (V) = 0.00027 m³

To Find:

Mass of an adult femur bone (m).

Explanation:

$\sf \implies Density = \frac{Mass (m)}{Volume (V)} \\ \\ \sf \implies \frac{Mass}{Volume} = Density \\ \\ \sf \implies Mass = Density \times Volume \\ \\ \sf \implies Mass = 2.0 \ kg/ \cancel{m^{3}} \times 0.00027 \ \cancel{m^{3}} \\ \\ \sf \implies Mass = 0.00054 \ kg$

6 0

3 years ago

A baseball is hit at ground level. The ball reaches its maximum height above ground level 3.0 s after being hit. Then 2.5 s afte

vaieri [72.5K]

Answer:

Part a)

$H = 44.1 m$

Part b)

$y = 13.48 m$

Part c)

$d = 8.86 m$

Explanation:

Part a)

As we know that ball will reach at maximum height at

t = 3 s

now we will have

$t = \frac{v sin\theta}{g}$

now we have

$3 = \frac{vsin\theta}{9.8}$

$v sin\theta = 29.4 m/s$

Now maximum height above ground is given as

$H = \frac{v^2sin^2\theta}{2g}$

$H = \frac{29.4^2}{2(9.8)}$

$H = 44.1 m$

Part b)

Height of the fence is given as

$y = (vsin\theta) t - \frac{1}{2}gt^2$

$y = (29.4)(5.5) - \frac{1}{2}(9.8)(5.5^2)$

$y = 13.48 m$

Part c)

As we know that its horizontal distance moved by the ball in 5.5 s is given as

$x = v_x t$

$97.5 = v_x (5.5)$

$v_x = 17.72 m/s$

now total time of flight is given as

$T = 3 + 3 = 6 s$

so range is given as

$R = v_x T$

$R = (17.72)(6)$

$R = 106.4 m$

so the distance from the fence is given as

$d = 106.4 - 97.5$

$d = 8.86 m$

7 0

4 years ago

when approaching the front of an idling jet engine, the hazard area extends forward of the engine approximately

Dominik [7]

when approaching the front of an idling jet engine, the hazard area extends forward of the engine approximately 25 feet.

<h3>What impact, if any, would jet fuel and aviation gasoline have on a turbine engine?</h3>

Tetraethyl lead, which is present in gasoline, deposits itself on the turbine blades. Because jet fuel has a higher viscosity than aviation gasoline, it may retain impurities with greater ease.

Once the gasoline charge has been cleared, start the engine manually or with an electric starter while cutting the ignition and using the maximum throttle.

On the final approach, the aeroplane needs to be re-trimmed to account for the altered aerodynamic forces. A substantial nose-down tendency results from the airflow producing less lift on the wings and less downward force on the horizontal stabiliser due to the reduced power and slower velocity.

Learn more about turbine engine refer

brainly.com/question/807662

#SPJ4

7 0

2 years ago

A particle makes 800 revolution in 4 minutes of a circle of 5cm. Find

vladimir1956 [14]

Answer:

i) The period of the particle is 0.3 seconds

ii) The angular velocity is approximately 20.94 rad/s

iii) The linear velocity is approximately 1.047 m/s

iv) The centripetal acceleration is approximately 6.98 m/s²

Explanation:

The given parameters are;

The number of revolution of the particle, n = 800 revolution

The time it takes the particle to make 800 revolutions = 4 minutes

The dimension of the circle = 5 cm = 0.05 m

Given that the dimension of the circle is the radius of the circle, we have;

i) The period of the particle, T = The time to complete one revolution

T = 1/(The number of revolutions per second)

∴ T = 1/(800 rev/(4 min × 60 s/min)) = 3/10 s

The period, T = 3/10 seconds = 0.3 seconds

ii) The angular velocity, ω = Angle covered/(Time)

800 revolutions in 4 minutes = Angle of (800 × 2·π) in 4 minutes

∴ ω = (800 × 2·π)/(4 × 60) = 20·π/3

The angular velocity, ω = 20·π/3 rad/s ≈ 20.94 rad/s

iii) The linear velocity, v = r × ω

∴ The linear velocity, v = 0.05 m × 20·π/3 rad/s = π/3 m/s ≈ 1.047 m/s

iv) The centripetal acceleration, $a_c$ = v²/r

∴ The centripetal acceleration, $a_c$ = (π/3)²/(0.05) = 20·π/9

The centripetal acceleration, $a_c$ = 20·π/9 m/s² ≈ 6.98 m/s²

4 0

3 years ago