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aleksandr82 [10.1K]
3 years ago
6

Please help with these questions as well! I need urgent help! I will give brainliest! God bless!

Physics
1 answer:
Radda [10]3 years ago
5 0

6.  Since we are not sure if the person in the question is actively lifting the crate, we have to determine the downwards force of the crate due to gravity and compare it to the normal force.  

F = ma

F = (15.3)(-9.8)

F = -150N

Since the downwards force of the crate is equivalent to the normal force, it means the person is applying no force in picking up the object.  So to pick up a 150N object from scratch, you would have to exert more force than the weight of the object, so the answer is 294N.


7.  Same idea as question 2.  

First determine the weight of the object:

F = ma

F = (30)(-9.8)

F = -294N

The crate in question is not moving, so the magnitudes of the forces in the upwards and downwards direction has to equal to 0.

-294 + 150N + x = 0

x = 144N  

So the person is exerting 144 N.


10.  First find the force of block B to the right due to its acceleration:

F = ma

F = (24)(0.5)

F = 12N

So block B is moving 12N to the right relative to block A due to block A's movement to the left.  However, block A is being applied a much greater force and is moving quicker to the left than block B is moving to the right of bock A.  The force that is causing block B to experience the lower relative force to the right is because of the friction.  To find the friction:

The sum of the forces in the leftward and rightward direction for block B must equal 12N.

75 - x = 12

x = 63N

So the force of friction of block A on block B is 63N to the left.


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dmitriy555 [2]

Answer:

-270.321012\ J

Explanation:

C_v = Heat capacity of Hg = 28 J/mol

\Delta T = Change in temperature = (234.32-(273.15+25))

\Delta H_{f} = Enthalpy of fusion = 2.29 kJ/mol

The number of moles is given by

n=13.3\times \dfrac{1}{200.59}\\\Rightarrow n=0.0663\ molHg

Heat is given by

Q_1=nC_v\Delta T\\\Rightarrow Q_1=0.0663\times 28\times (234.32-(273.15+25))\\\Rightarrow Q_1=-118.494012\ J

Heat released is given by

Q_2=-n\Delta H_{f}\\\Rightarrow Q_2=-0.0663\times 2.29\times 10^3\\\Rightarrow Q_2=-151.827\ J

Total heat is given by

Q=Q_1+Q_2\\\Rightarrow Q=-118.494012+(-151.827)\\\Rightarrow Q=-270.321012\ J

The total heat released is -270.321012\ J

4 0
3 years ago
The concentration of an isotope is 1/500 of its initial value after one day. What is the rate of decay?
mel-nik [20]
The exponential growth/decay formula is given by A = Pe^(rt), where A is the final amount, P is the initial amount, r is the rate of growth/decay and t is time. 

We are given A = (1/500)P and t=1, so substituting:
(1/500)P = Pe^(r*1)
Cancelling out P on both sides:
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Rate of decay is -6.215 per day.
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5j-Tj=6j+3 Tj<br> Determine the value of T from the given vector
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3 years ago
A sperical ballon with a diameter of 6 m filled with helium at 20 degree centigrade and 200kpa determine mole number and the mas
rodikova [14]

Answer:

A. 9280.78 moles.

B. 37123.12 g.

Explanation:

We'll begin by calculating the volume of the spherical balloon. This can be obtained as follow:

Diameter (d) = 6 m

Radius (r) = d/2 = 6/2 = 3 m

Pi (π) =3.14

Volume (V) =?

V = 4/3πr³

V = 4/3 × 3.14 × 3³

V = 4/3 × 3.14 × 27

V = 113.04 m³

Next, we shall convert 20°C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

T(°C) = 20°C

T(K) = 20°C + 273

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1 KPa = 1000 Pa

Therefore,

200 KPa = 200 KPa × 1000 Pa / 1 KPa

200 KPa = 2×10⁵ Pa

A. Determination of the number of mole of He in the spherical balloon.

Volume (V) = 113.04 m³

Temperature (T) = 293 K

Pressure (P) = 2×10⁵ Pa

Gas constant (R) = 8.314 m³Pa/Kmol

Number of mole (n) =?

PV = nRT

2×10⁵ × 113.04 = n × 8.314 × 293

22608000 = n × 2436.002

Divide both side by 2436.002

n = 22608000 / 2436.002

n = 9280.78 moles

B. Determination of the mass of He.

Mole of He (n) = 9280.78 moles

Molar mass of He = 4 g/mol

Mass of He =?

Mass = mole × molar mass

Mass of He = 9280.78 × 4

Mass of He = 37123.12 g

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What is Kinetic energy
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Answer:energy which a body possesses by virtue of being in motion.

Explanation:

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