Conservation of Energy Quiz<br> Please help! 5 points per question (50 total)

If the mass of a material is 87 grams and the volume of the material is 14 cm3, what would the density of the material be?

tresset_1 [31]

Answer:

6.214g/cm³

Explanation:

The question is on density of a material

Density=mass/volume

Given, mass=87grams and volume= 14 cm³ density=?

Density=m/v 87/14 =6.214g/cm³

8 0

2 years ago

A beryllium-9 ion has a positive charge that is double the charge of a proton, and a mass of 1.50 ✕ 10−26 kg. At a particular in

seropon [69]

Answer:

Magnetic force, $F = 3.52\times 10^{-13}\ N$

Explanation:

Given that,

A beryllium-9 ion has a positive charge that is double the charge of a proton, $q=2\times 1.6\times 10^{-19}\ C=3.2\times 10^{-19}\ C$

Speed of the ion in the magnetic field, $v=5\times 10^6\ m/s$

Its velocity makes an angle of 61° with the direction of the magnetic field at the ion's location.

The magnitude of the field is 0.220 T.

We need to find the magnitude of the magnetic force on the ion. It is given by :

$F=qvB\\\\F=3.2\times 10^{-19}\times 5\times 10^6\times 0.22\\\\F=3.52\times 10^{-13}\ N$

So, the magnitude of magnetic force on the ion is $3.52\times 10^{-13}\ N$ .

3 0

2 years ago

Read 2 more answers

A 1.120 kg car is traveling with a speed of 40 m/s. find its energy

Aleonysh [2.5K]

Answer:

896 kJ

Explanation:

KInetic Energy = 1/2 m v^2

= 1/2 (1120)(40^2) = 896 000 J or 896 kJ

4 0

1 year ago

What minimum speed does a 200 g puck need to make it to the top of a frictionless ramp that is 4.1 m long and inclined at 22 ∘?

myrzilka [38]

Answer:

5.5 m/ sec

Explanation:

Because the inclined surface is frictionless so we can assume that total change of energy is zero

i-e ΔE = 0

Or we can say that difference between final and initial energy is zero i-e

Ef- Ei =0

Where,

Ef= final energy at the top of the ramp= KEf+PEf

Ei= Initial energy at the bottom of the ramp=KEi+PEi

So we have

(KEf+PEf)-(KEi+PEi)=0

==>KEf-KEi+PEf-PEi=0 -------------(1)

KEf = mgh = 200×9.8×h

Where h= Sin 22 = h/d= h/4.1

or

0.375×4.1=h

or h= 1.54 m

So, PEf= 200×9.8×1.54=3018.4 j

and KEf= 1/2 m $Vf^{2}$ = 0.5×200×0=0 j

PEi= mgh = 200×9.8×0=0 j

KEi= 1/2 m $Vi^{2}$ =0.5×200× $Vi^{2}$ =100 $Vi^{2}$ j

Put these values in eq 1, we get;

0-100 $Vi^{2}$ +3018.4-0=0

-100 $Vi^{2}$ =-3018.4

==> $Vi^{2}= \frac{3018.4}{100}$ = 30.184

==> Vi = $\sqrt{30.184} = 5.5 m.sec$

7 0

3 years ago

calculate the mass of potassium chlorate (kcio3) required to obtain 10g of oxygen in the following reaction:kclO3-kcl+O2

igor_vitrenko [27]

First, balance the reaction:

_ KClO₃ ==> _ KCl + _ O₂

As is, there are 3 O's on the left and 2 O's on the right, so there needs to be a 2:3 ratio of KClO₃ to O₂. Then there are 2 K's and 2 Cl's among the reactants, so we have a 1:1 ratio of KClO₃ to KCl :

2 KClO₃ ==> 2 KCl + 3 O₂

Since we start with a known quantity of O₂, let's divide each coefficient by 3.

2/3 KClO₃ ==> 2/3 KCl + O₂

Next, look up the molar masses of each element involved:

• K: 39.0983 g/mol

• Cl: 35.453 g/mol

• O: 15.999 g/mol

Convert 10 g of O₂ to moles:

(10 g) / (31.998 g/mol) ≈ 0.31252 mol

The balanced reaction shows that we need 2/3 mol KClO₃ for every mole of O₂. So to produce 10 g of O₂, we need

(2/3 (mol KClO₃)/(mol O₂)) × (0.31252 mol O₂) ≈ 0.20835 mol KClO₃

KClO₃ has a total molar mass of about 122.549 g/mol. Then the reaction requires a mass of

(0.20835 mol) × (122.549 g/mol) ≈ 25.532 g

of KClO₃.

7 0

2 years ago

Conservation of Energy Quiz Please help! 5 points per question (50 total)