Take into account that conduction is the process at which heat is propagated by the termical motion of the molecules, without a real desplacement of the molecules.

Based on the previous definition, the following is an example of heat transfer by condu

7 0

1 year ago

A 10 lb. cat needs 31⁄2 tablets that each contain 6 mg of a drug. what's the equivalent dose in mg/kg?

mina [271]

Mass of cat = 10 lb

1 lb = 0.454 kg

So mass of cat is 4.54 kg

Number of tablets given to cat

$N = 3\frac{1}{2}$

Each tablet has 6 mg of drug

Total drug given to cat

$Drug = 3\frac{1}{2} * 6mg$

$Drug = 21 mg$

Now dose given to cat in mg/kg form

$Dose = \frac{mass of drug in mg}{mass of cat in kg}$

$Dose = \frac{21mg}{4.54kg}$

$Dose = 4.63mg/kg$

4 0

4 years ago

A boy shoves his stuffed toy zebra down a frictionless chute. It starts at a height of 1.69 m above the bottom of the chute with

Veseljchak [2.6K]

Answer:

x = 6.94 m

Explanation:

For this exercise we can find the speed at the bottom of the ramp using energy conservation

Starting point. Higher

Em₀ = K + U = ½ m v₀² + m g h

Final point. Lower

$Em_{f}$ = K = ½ m v²

Em₀ = Em_{f}

½ m v₀² + m g h = ½ m v²

v² = v₀² + 2 g h

Let's calculate

v = √(1.23² + 2 9.8 1.69)

v = 5.89 m / s

In the horizontal part we can use the relationship between work and the variation of kinetic energy

W = ΔK

-fr x = 0- ½ m v²

Newton's second law

N- W = 0

The equation for the friction is

fr = μ N

fr = μ m g

We replace

μ m g x = ½ m v²

x = v² / 2μ g

Let's calculate

x = 5.89² / (2 0.255 9.8)

x = 6.94 m

6 0

4 years ago

Read 2 more answers

A block of mass 200g is oscillating on the end of a horizontal spring of spring constant 100 N/m and natural length 12 cm. When

malfutka [58]

In order to determine the acceleration of the block, use the following formula:

$F=ma$

Moreover, remind that for an object attached to a spring the magnitude of the force acting over a mass is given by:

$F=kx$

Then, you have:

$ma=kx$

by solving for a, you obtain:

$a=\frac{kx}{m}$

In this case, you have:

k: spring constant = 100N/m

m: mass of the block = 200g = 0.2kg

x: distance related to the equilibrium position = 14cm - 12cm = 2cm = 0.02m

Replace the previous values of the parameters into the expression for a:

$a=\frac{(\frac{100N}{m})(0.02m)}{0.2\operatorname{kg}}=10\frac{m}{s^2}$

Hence, the acceleration of the block is 10 m/s^2

8 0

1 year ago