How much of a kilogram is a gram?

Use the worked example above to help you solve this problem. An Alaskan rescue plane drops a package of emergency rations to str

Vlad [161]

Answer:

(a) The package lands 682 meters horizontally ahead from the point the package was dropped from the plane

(b) The horizontal component = 39.0 m/s

The vertical component = 171.55 m/s

(c) The angle of impact is 77.19°

Explanation:

The parameters given are;

Velocity of the plane, vₓ = 39.0 m/s

Height of the plane above the ground, h = 1.50 × 10² m = 1,500 m

(a) The time, t, before the package hits the ground when dropped from the plane is given by the relation;

h = u·t + 1/2×g×t²

Where:

g = Acceleration due to gravity = 9.81 m/s²

u = Initial vertical velocity = 0 m/s

Hence;

1500 = 0×t + 1/2 × 9.81 × t² = 4.905·t²

∴ t = √(1500/4.905) = 17.49 s

The horizontal distance the package travels before landing = 17.49 × 39 ≈ 682 m

The package lands 682 meters horizontally ahead from the point the package was dropped from the plane

(b) The vertical velocity, $v_y$ , of the package just before landing is given by the relation;

$v_y$ ² = u² + 2·g·h

u = 0 m/s

∴ $v_y$ ² = 0 + 2×9.81×1500 = 29430 m²/s²

$v_y$ = √29430 = 171.55 m/s

Hence the horizontal component = 39.0 m/s

The vertical component = 171.55 m/s

(c) The angle of impact, θ, is given as follows;

$tan \theta = \dfrac{v_y}{v_x} = \dfrac{171.55}{39.0 } = 4.4$

∴ θ = tan⁻¹(4.4) = 77.19°.

6 0

3 years ago

A 5.0-kg block of wood is placed on a 2.0-kg aluminum frying pan. How much heat is required to raise the temperature of both the

Shalnov [3]

Heat required to raise the temperature of a given system is

$Q = ms\Delta T$

here we know that

m = mass

s = specific heat capacity

$\Delta T$ = change in temperature

now as we know that

mass of wood = 5 kg

mass of aluminium pan = 2 kg

change in temperature = 45 - 20 = 25 degree C

specific heat capacity of wood = 1700 J/kg C

specific heat capacity of aluminium = 900 J/kg C

now here we will find the total heat to raise the temperature of both

$Q = m_1s_1\Delta T_1 + m_2s_2\Delta T_2$

$Q = 5 * 1700 * 25 + 2 * 900 * 25$

$Q = 212500 + 45000$

$Q = 257500 J$

So heat required to raise the temperature of the system is 257500 J

4 0

3 years ago

The y component of a vector is 36, and the angle between the vector and the x axis is 27 what is the magnitude of the vector

xz_007 [3.2K]

Answer:

Magnitude of Vector = 79.3

Explanation:

When a vector is resolved into its rectangular components, it forms two vector components. These components are named as x-component and y-component, they are calculated by the following formulae:

x-component of vector = (Magnitude of Vector)(Cos θ)

y-component of vector = (Magnitude of Vector)(Sin θ)

where,

θ = angle of the vector with x-axis = 27°

Therefore, using the values in the equation of y-component, we get:

36 = (Magnitude of Vector)(Sin 27°)

Magnitude of Vector = 36/Sin 27°

<u>Magnitude of Vector = 79.3</u>

3 0

3 years ago

Mary pushes a crate by applying force of 18 newtons. Unable to push it alone, she gets help from her friend, Anne. Together they

maw [93]

<span>This problem is relatively simple, in order to solve this problem the only formula you need to know is the formula for friction, which is:

Ff = UsN

where Us is the coefficient of static friction and N is the normal force.

In order to get the crate moving you must first apply enough force to overcome the static friction:

Fapplied = Ff

Since Fapplied = 43 Newtons:

Fapplied = Ff = 43 = UsN

and it was given that Us = 0.11, so all you have to do is isolate N by dividing both sides by 0.11

43/0.11 = N = 390.9 which is approximately 391 or C. 3.9x10^2</span>

7 0

3 years ago

Read 2 more answers

YAHOO ANSWERSA 1260 kg car moving south at 12.9 m/s collides with a 2430 kg car moving north. The cars stick together and move a

Fofino [41]

Explanation:

Below is an attachment containing the solution.

7 0

3 years ago