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NNADVOKAT [17]
2 years ago
12

An object has a mass of 785 g and a volume of 15 cm³. What is its density? (Give your answer in g/cm³ to 2 decimal places).

Physics
1 answer:
Anvisha [2.4K]2 years ago
4 0

Answer:

denisity = 52.33 g/cm^{3}

Explanation:

Density:

d = \frac{m}{v}

We have that m = 785 and that v = 15 cm^{3}.

d = \frac{785}{15}

d = 52.33 m^{3}

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3.00 textbook rests on a frictionless, horizontal tabletop surface. A cord attached to the book passes over a pulley whose diame
sammy [17]

Answer:

a1 = 3.56 m/s²

Explanation:

We are given;

Mass of book on horizontal surface; m1 = 3 kg

Mass of hanging book; m2 = 4 kg

Diameter of pulley; D = 0.15 m

Radius of pulley; r = D/2 = 0.15/2 = 0.075 m

Change in displacement; Δx = Δy = 1 m

Time; t = 0.75

I've drawn a free body diagram to depict this question.

Since we want to find the tension of the cord on 3.00 kg book, it means we are looking for T1 as depicted in the FBD attached. T1 is calculated from taking moments about the x-axis to give;

ΣF_x = T1 = m1 × a1

a1 is acceleration and can be calculated from Newton's 2nd equation of motion.

s = ut + ½at²

our s is now Δx and a1 is a.

Thus;

Δx = ut + ½a1(t²)

u is initial velocity and equal to zero because the 3 kg book was at rest initially.

Thus, plugging in the relevant values;

1 = 0 + ½a1(0.75²)

Multiply through by 2;

2 = 0.75²a1

a1 = 2/0.75²

a1 = 3.56 m/s²

6 0
2 years ago
Find the total electric charge of 1.7 kg of electrons. me=9.11×10−31kg, e=1.60×10−19C.
Gelneren [198K]

Answer:

2.99\cdot 10^{11}C

Explanation:

The mass of one electron is

m_e = 9.11\cdot 10^{-31}kg

So the number of electrons contained in M=1.7 kg of mass is

N=\frac{M}{m_e}=\frac{1.7 kg}{9.11\cdot 10^{-31}kg}=1.87\cdot 10^{30}

The charge of one electron is

e=1.60\cdot 10^{-19} C

So, the total charge of these electrons is equal to the charge of one electron times the number of electrons:

Q=Ne=(1.87\cdot 10^{30})(1.6\cdot 10^{-19}C)=2.99\cdot 10^{11}C

8 0
2 years ago
During a circus act, an elderly performer thrills the crowd by catching a cannon ball shot at him. The cannon ball has a mass of
Ipatiy [6.2K]

Answer:

m v1 = (m + M) v2

v2 = m v1 / (m + M)

v2 = 7 * 74 / (74 + 65)

3.73 m/s

74 kg is too heavy for the cannonball (over 150 lbs)

4 0
2 years ago
Consider a short time span just before and after the spark plug in a gasoline engine ignites the fuel-air mixture and releases 1
Tju [1.3M]

Answer:

Temperature after ignition=7883.205 K

Explanation:

The number of moles is,

n=PV/RT

=(1.18x10^6)(47.9x10^-6)/8.314(325)

= 0.0209 moles

a) In this process volume is constant

Q=U

=nCv.dT

dT= Q/nCv

=1970/(1.5x8.314)(0.0209)

= 7558.205 K

The final temperature is,

= 7558.205+325

= 7883.205 K

5 0
3 years ago
Read 2 more answers
This whole worksheet. I am not good at this and I am completely lost
Ket [755]
1.) potential energy
2.)potential and kinetic
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4.)potential energy
5.)kinetic energy
6.) potential and kinetic energy
3 0
3 years ago
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