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Alchen [17]
4 years ago
12

50 g of liquid Y at 10 Celcius and 200 g of liquid Y at 40Celcius are mixed. Final temperature of the mixture is measured as 15

celcius. Find the ratio of cx/cy.

Physics
1 answer:
Sholpan [36]4 years ago
5 0
50 g of liquid X at 10 Celcius and 200 g of liquid Y

mx*cx*(t-tx)+my*cy*(t-ty)=0
cx/cy = - my*(t-ty) : mx*(t-tx) = (my/mx)  * (ty - t) / (t-tx)
cx/cy  = 200/50*(40-15)/(15-10) = 20
cx/cy   = 20



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(a) Calculate the magnitude of the gravitational force exerted on a 445-kg satellite that is a distance of 1.77 earth radii from
victus00 [196]

Answer:

a)1396.52 N

b)1396.52 N

c)a_{satellite}= 3.13 m/sec^2

d)a_{earth}=2.32\times10^{-22} m/s^2

Explanation:

The force experienced by the satelite is giveb by

F= \frac{Gm_{satellite}m_{earth}}{r^2} \\

m_{satellite}= 445 Kg

m_{earth}= 6×10^24 Kg

radius r= 1.77Re= 1.77×6.38×10^6 m

now putting values we get

F= \frac{6.67\times10^{-11}(445)(6\times10^24)}{(1.77\times6.38\times10^6)^2}

⇒F= 1396.52 N

now,

a_{satellite}= \frac{F}{m_{satellite}}

a_{satellite}= \frac{1396.52}{445}

a_{satellite}= 3.13 m/sec^2

also,

a_{earth}= \frac{F}{m_{earth}}

a_{earth}= \frac{1396.52}{(6\times10^24)}

a_{earth}=2.32\times10^{-22} m/s^2

3 0
3 years ago
How much force is needed to accelerate a 2500 kg car at a rate of 3.5 m/s^2?
Sati [7]

F = ma  

F = applied force in newtons = to be determined  

m = mass of the car = 2,500 kg  

a = acceleration of the car = 3.5 m/s²  

F = (2,500 kg)(3.5 m/s²)  

F =8750

4 0
4 years ago
Please help me with this question someone
Gekata [30.6K]
From the answers provided, I believe the possible answer would be the last option, silicon, oxygen, and one or more metals. Here's my reasoning: the most abundant mineral group found in the Earth's crust is the silicate group. The silicate materials contain both oxygen and silicon. Silicates are the most common minerals in the rock-formation process, and it has, in fact, been estimated that they make up 75 to 90 percent of the Earth's crust. From this piece of evidence, I can guess that the answer will possibly be D, silicon, oxygen, and one or more metals.
It should also be noted that the additional elements that combine with the silicon-oxygen tetrahedron are involved with the other elements commonly found in the Earth's crust and mantle. They are aluminum, calcium, iron, magnesium, potassium and sodium.
8 0
3 years ago
Read 2 more answers
A volleyball player’s hand applies a 39 N force while in contact with a volleyball for 2 seconds. What is the impulse on the bal
nikklg [1K]
Given:
F = 39 N, the force applied
t = 2 s, the time interval in which the force is applied.

By definition, the impulse is
I = \int_{0}^{2} F(t) dt = \int _{0}^{2} 39dt = 39*2=78 \, N-s

Answer: 78 N-s
3 0
3 years ago
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A flat, circular, steel loop of radius 75 cm is at rest in a uniform magnetic field, as shown in an edge-on view in the figure (
SIZIF [17.4K]

The solution to the questions are given as

  • t=40.39 \mathrm{sec}
  • \varepsilon &=(0.12v)e^{0.057t}
  • the direction of induced current will be Counterclock vise.

<h3>What is the direction of the current induced in the loop, as viewed from above the loop.?</h3>

Given, $B(t)=(1.4 T) e^{-0.057 t}$

$\varepsilon m f(\varepsilon)=-\frac{d \phi_{B}}{d t}

\quad$ and, $\phi_{B}=\int B \cdot d A=\int B \cdot d A \cdot \cos \theta$

\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}

\varepsilon &=(0.12v)e^{0.057t}

(b) Here, $\varepsilon_{0}=0.12 \mathrm{~V} \quad\left(a t_{2} t=0 \mathrm{sec}\right)$

\begin{aligned}&\therefore 1 . \varepsilon_{0}=\varepsilon_{0} \cdot e^{-e .057 t} \\&\therefore e^{0.057 t}=10 \quad \text { (taking log both thesides) } \\&\therefore 0.057 t=\ln (10)=2.303 \\&\therefore t=40.39 \mathrm{sec}\end{aligned}

c)

In conclusion, the direction of the induced current will be Counterclockwise.

Read more about current

brainly.com/question/13076734

#SPJ1

4 0
2 years ago
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