A student conducted an experiment to measure the acceleration due to gravity 'g' of a simple pendulum.

Mass, size and color are examples of ___

BigorU [14]

The short answer (and the long one for that matter) is physical properties of chemicals. If you are being marked by a machine, likely the answer is going to be physical properties.

6 0

3 years ago

A cannon fires a 0.2 kg shell with initial velocity vi = 9.2 m/s in the direction θ = 46 ◦ above the horizontal. The shell’s tra

Sedbober [7]

Answer:

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

Vy(t) and y(t)

We start by decomposing the speed in the direction ''y''

$sin(\alpha) = \frac{Vyi}{Vi}$

$Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}$

Vy in this problem will follow this equation =

$Vy(t) = Vyi -g.t$

where g is the gravity acceleration

$Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t$

This is equation (1)

For Y(t) :

$Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}$

We suppose yi = 0

$Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}$

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

$Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s$

So in t = 0.675 s → Vy = 0. Now we calculate the y in which this happen using (2)

$Y(0.675s) = 6.62\frac{m}{s}.(0.675s)-4.9 \frac{m}{s^{2} } .(0.675s)^{2} \\Y(0.675s) =2.236 m$

2.236 m is the maximum height from the shell (in which Vy=0 m/s)

Let's calculate now the height for t = 0.555 s

$Y(0.555s)= 6.62 \frac{m}{s} .(0.555s)-4.9\frac{m}{s^{2} } .(0.555s)^{2} \\Y(0.555s) = 2.165m$

The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

6 0

3 years ago

Let's say that we have a pilot that is dropping a package from a plane that is flying horizontally at a constant speed. If we ne

antoniya [11.8K]

Answer:

The package will be directly below the location of the plane.

Explanation:

Look up projectile motion for more information. The horizontal speed of the package is separate from the vertical speed of the package. The vertical speed of the falling package will be based on the rate of acceleration and the height of the package when dropped. The horizontal speed of the package will be the same as the plane so the package will remain directly below the plane the entire time until the package hits the ground.

6 0

2 years ago

Can we fly if we were falling for two days

nevsk [136]

...................no

4 0

2 years ago

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A Boeing 737 airliner has a mass of 20,000 kg and the total area of both wings (top or bottom) is 100 m2. What is the pressure d

kvv77 [185]

Answer:

The correct option is A = 1960 N/m²

Explanation:

Given that,

Mass m= 20,000kg

Area A = 100m²

Pressure different between top and bottom

Assume the plane has reached a cruising altitude and is not changing elevation. Then sum the forces in the vertical direction is given as

∑Fy = Wp + FL = 0

where

Wp = is the weight of the plane, and

FL is the lift pushing up on the plane.

Let solve for FL since the mass of the plane is given:

Wp + FL = 0

FL = -Wp

FL = -mg

FL = -20,000× -9.81

FL = 196,200N

FL should be positive since it is opposing the weight of the plane.

Let Equate FL to the pressure differential multiplied by the area of the wings:

FL = (Pb −Pt)⋅A

where Pb and Pt are the static pressures on bottom and top of the wings, respectively

FL = ∆P • A

∆P = FL/A

∆P = 196,200 / 100

∆P = 1962 N/m²

∆P ≈ 1960 N/m²

The pressure difference between the top and bottom surface of each wing when the airplane is in flight at a constant altitude is approximately 1960 N/m². Option A is correct

5 0

3 years ago

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