A member of the cross country team is training for the big race. She runs 7.85 km one afternoon. How many miles is this?

The density of aluminum is 2.70 g/ml. A piece of aluminum foil has a mass of 44 g. What is the volume of this piece of aluminum

anyanavicka [17]

Answer:

C) 16.3 ml

Explanation:

Density is equal to the ratio between the mass of an object and its volume:

$d=\frac{m}{V}$

where

m is the mass

V is the volume

In our problem, we know:

- density of aluminium: $d=2.70 g/mL$

- mass of the aluminium foil: $m=44 g$

So we can re-arrange the equation above and use these data to find the volume of the piece of aluminium foil:

$V=\frac{m}{d}=\frac{44 g}{2.70 g/mL}=16.3 mL$

7 0

3 years ago

What is the minimal mass of helium (density 0.18 kg/m3) needed to lift a balloon carrying two people in a basket, if the total m

Sergio039 [100]

Answer:
M[min] = M[basket+people+ balloon, not gas] * Î”R/R[b]
Î”R is the difference in density between the gas inside and surrounding the balloon.
R[b] is the density of gas inside the baloon.
====================================
Let V be the volume of helium required.
Upthrust on helium = Weight of the volume of air displaced = Density of air * g * Volume of helium = 1.225 * g * V
U = 1.225gV newtons
----
Weight of Helium = Volume of Helium * Density of Helium * g
W[h] = 0.18gV N
Net Upward force produced by helium, F = Upthrust - Weight = (1.225-0.18) gV = 1.045gV N  -----

Weight of 260kg = 2549.7 N
Then to lift the whole thing, F > 2549.7
So minimal F would be 2549.7
----
1.045gV = 2549.7
V = 248.8 m^3
Mass of helium required = V * Density of Helium = 248.8 * 0.18 = 44.8kg (3sf)
=====
Let the density of the surroundings be R
Then U-W = (1-0.9)RgV = 0.1RgV
So 0.1RgV = 2549.7 N
V = 2549.7 / 0.1Rg
Assuming that R is again 1.255, V = 2071.7 m^3
Then mass of hot air required = 230.2 * 0.9R = 2340 kg
Notice from this that M = 2549.7/0.9Rg * 0.1R so
M[min] = Weight of basket * (difference in density between balloon's gas and surroundings / density of gas in balloon)
M[min] = M[basket] * Î”R/R[b]

3 0

3 years ago

A 60-kg rollerblader rolls 10 m down a 30? incline. When she reaches the level floor at the bottom, she applies the brakes. The

VARVARA [1.3K]

Answer:

s = 20 m

Explanation:

given,

mass of the roller blader = 60 Kg

length = 10 m

inclines at = 30°

coefficient of friction = 0.25