Need help in science <br><br> Please help me pass my exam hw

A cat chase a mouse across a 1.0 m high table. The mouse steps out of the way, and the car slides off the table and strikes the

Tema [17]

The cat fell 1.0 m from the ground.

Using the formula

$h = v_{oy}t - \frac{gt^2}{2}$

Here, $v_{oy} = 0, \; h = 1.0 \; m.$

Solving for t, the time it spent in the air is

$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2(1.0 \; m)}{9.8 \; m/s^2}} = 0.451753951 \; s$

The cat does not accelerate along the horizontal, so it has constant horizontal velocity. Since it strikes the floor 2.2 m from the table, then

$v_x = \Delta x/t = \frac{2.2\; m}{0.451753951 s} = 4.869907597 \; m/s \Rightarrow 4.9 \; m/s$

5 0

3 years ago

When being unloaded from a moving truck, a 16.0- kilogram suitcase is placed on a flat ramp inclined at 40.0 o. When released fr

9966 [12]

Answer:

0.65812

Explanation:

m = Mass of suitcase = 16 kg

$\theta$ = Incline angle = 40°

g = Acceleration due to gravity = 9.81 m/s²

a = Acceleration of the block = 1.36 m/s²

f = Frictional force

The normal force is given by

$N=mgcos\theta$

In x direction

$mgsin\theta-f=ma\\\Rightarrow f=mgsin\theta-ma\\\Rightarrow f=16\times 9.81\times sin40-16\times 1.36\\\Rightarrow f=79.13194\ N$

Frictional force is given by

$f=\mu N=\mu mgcos\theta\\\Rightarrow \mu=\dfrac{f}{mgcos\theta}\\\Rightarrow \mu=\dfrac{79.13194}{16\times 9.81\times cos40}\\\Rightarrow \mu=0.65812$

The coefficient of kinetic friction between the suitcase and the ramp is 0.65812

8 0

3 years ago

A 2.0-mm-diameter copper ball is charged to 40 nC . What fraction of its electrons have been removed? The density of copper is 8

murzikaleks [220]

Answer:

0.02442 × 10⁻⁹

Explanation:

Given:

Diameter of copper ball = 2.00 mm = 0.002 m

Charge on ball = 40 nC = 40 × 10⁻⁹ C

Density of copper = 8900 Kg/m³

Now,

The number of electrons removed, n = $\frac{\textup{Charge on ball}}{\textup{Charge of an electron}}$

also, charge on electron = 1.6 × 10⁻¹⁹ C

Thus,

n = $\frac{40\times10^{-9}}{1.6\times10^{-19}}$

or

n = 25 × 10¹⁰ Electrons

Now,

Mass of copper ball = volume × density

Or

Mass of copper ball = $\frac{4}{3}\pi(\frac{d}{2})^3$ × 8900

or

Mass of copper ball = $\frac{4}{3}\pi(\frac{0.002}{2})^3$ × 8900

or

Mass of copper ball = 0.03726 grams

Also,

molar mass of copper = 63.546 g/mol

Therefore,

Number of mol of copper in 0.03726 grams = $\frac{ 0.03726}{63.546}$

or

Number of mol of copper in 0.03726 grams = 5.86 × 10⁻⁴ mol

and,

1 mol of a substance contains = 6.022 × 10²³ atoms

Therefore,

5.86 × 10⁻⁴ mol of copper contains = 5.86 × 10⁻⁴ × 6.022 × 10²³ atoms.

or

5.86 × 10⁻⁴ mol of copper contains = 35.88 × 10¹⁹ atoms

Now,

A neutral copper atom has 29 electrons.

Therefore,

Number of electrons in ball = 29 × 35.88 × 10¹⁹ = 1023.37 × 10¹⁹ electrons.

Hence,

The fraction of electrons removed = $\frac{25\times10^{10}}{1023.37\times10^{19}}$

or

The fraction of electrons removed = 0.02442 × 10⁻⁹

3 0

4 years ago

The star Sirius has an apparent magnitude of -1.46 and appears 95-times brighter compared to the more distant star Tau Ceti, whi

dimaraw [331]

Answer:

(a) Apparent magnitude is the perceived brightness of an astronomical object

Absolute magnitude is the luminosity based on viewing an object from a 32.6 light-years distance

Bolometric magnitude is the total emitted radiation of a star

(b) The apparent magnitude of the star Tau Ceti = 3.51

(c) The distance between the Earth and Tau Ceti is 1.13 × 10¹⁴ km

Explanation:

(a) Apparent magnitude is an estimate of an astronomical objects' brightness as the object is perceived from the Earth

The absolute magnitude is the magnitude an object appears to have when viewed from a 32.6 light-years distance while having constant transfer of its luminosity that is not affected by cosmic dust and objects present in the line of sight

The bolometric magnitude of a star is the sum total of the star's radiation released over all electromagnetic spectrum wavelengths

(b) The apparent magnitude of the star Tau Ceti is found using the following equation;

$m_{2}-m_{1} = -2.512\times log\left (\dfrac{B_{2}}{B_{1}} \right )$