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Mumz [18]
3 years ago
15

what is the specific heat of a substance that requires 99,100J of thermal energy to heat 3.47 kg of this substance from 15 C to

41C​
Physics
1 answer:
Nady [450]3 years ago
7 0

Answer:

1098.42 J/kg.c°

Explanation:

Q=mcT

you have : Q = 99100

m=3.47kg

and the change in temperature T = 41-15=26c°

Solve for specific heat c :

c = Q/mT === 99100/3.47(26) = 1098.42J/kg . c°

I hope that it's a clear solution and explanation.

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How do prokaryotic cells replicate?
Alika [10]

Answer:

The usual method of prokaryote cell division is termed binary fission. The prokaryotic chromosome is a single DNA molecule that first replicates, then attaches each copy to a different part of the cell membrane. When the cell begins to pull apart, the replicate and original chromosomes are separated.

6 0
3 years ago
the base of a rectangular vessel measure 10m by 18cm. water is poured into a depth of 4cm. (a) what is the pressure on the base?
Alex787 [66]

Answer:

a) P =392.4[Pa]; b) F = 706.32[N]

Explanation:

With the input data of the problem we can calculate the area of the tank base

L = length = 10[m]

W = width = 18[cm] = 0.18[m]

A = W * L = 0.18*10

A = 1.8[m^2]

a)

Pressure can be calculated by knowing the density of the water and the height of the water column within the tank which is equal to h:

P = density * g *h

where:

density = 1000[kg/m^3]

g = gravity = 9.81[m/s^2]

h = heigth = 4[cm] = 0.04[m]

P = 1000*9.81*0.04

P = 392.4[Pa]

The force can be easily calculated knowing the relationship between pressure and force:

P = F/A

F = P*A

F = 392.4*1.8

F = 706.32[N]

4 0
2 years ago
A kangaroo can jump straight up to a height of 2.0 m. What is its takeoff speed
12345 [234]
7.17m/s glad I could help
5 0
3 years ago
an object with a mass of 5kg is moving with an initial velocity of 20m/s.the object accelerates at a rate of 15m/s/s for 8s.what
Alinara [238K]

It doesn't matter what the object's initial velocity is, or how long
the acceleration lasts.  All that matters is the object's mass and
acceleration.

Force = (mass) x (acceleration) =

                (5kg) x (15 m/s²) =

                         75 kg-m/s² = <em>75 newtons .</em>


5 0
2 years ago
Three point charges are arranged along the x axis. Charge q1=-4.00nC is located at x= .250 m and q2= 2.40 nC is at the x= -.300m
Umnica [9.8K]

Answer:

q₃=5.3nC

Explanation:

First, we have to calculate the force exerted by the charges q₁ and q₂. To do this, we use the Coulomb's Law:

F= k\frac{|q_aq_b|}{r^{2} } \\\\\\F_{13}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(-4.00*10^{-9}C)q_3|}{(.250m)^{2} } =576q_3N/C\\\\F_{23}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(2.40*10^{-9}C)q_3|}{(.300m)^{2} } =240q_3N/C\\

Since we know the net force, we can use this to calculate q₃. As q₁ is at the right side of q₃ and q₁ and q₃ have opposite signs, the force F₁₃ points to the right. In a similar way, as q₂ is at the left side of q₃, and q₂ and q₃ have equal signs, the force F₂₃ points to the right. That means that the resultant net force is the sum of these two forces:

F_{Net}=F_{13}+F_{23}\\\\4.40*10^{-9} N=576q_3N/C+240q_3N/C\\\\4.40*10^{-6} N=816q_3N/C\\\\\implies q_3=5.3*10^{-9}C=5.3nC

In words, the value of q₃ must be 5.3nC.

7 0
3 years ago
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