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Pie
2 years ago
7

A 150 g sample of brass at 100 °C is placed in a Styrofoam cup of water containing 120 mL of water at 10 °C. No heat is lost to

the cup or surroundings. What is the final temperature of the mixture? answer in celsius​
Physics
1 answer:
fredd [130]2 years ago
4 0

Answer:

≈19.144°C.

Explanation:

all the details are in the attachment.

Note, that c₁, m₁, t₁ are the parameters of the sample of brass; c₂, m₂ and t₂ are  the parameters of the sample of water.

P.S. change the provided design according Your requirements.

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For this quiz, we shall return to the radio control car track that we visited briefly on the last quiz. The track is 10 meters l
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Answer:

v=1.54\,\,metres\,\,per\,\,second

Explanation:

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Put x_0=2,\,,\,x=8\,,\,t=3.9 in equation x=x_0+vt

8=2+3.9t\\8-2=3.9t\\6=3.9v\\v=\frac{6}{3.9}\\ v=1.54\,\,metres\,\,per\,\,second

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3 years ago
If Earth's mass was cut in half, what would happen to your mass? Group of answer choices
Alex777 [14]

Answer:

nothing, mass is not affected by gravitational force

Explanation:

Weight is the gravitational force a planet exerts on a mass on the surface.

It is the product of the mass of an object with the gravitational acceleration that the planet produces.

The weight is the gravitational force

W=mg

where,

m = Mass of the object

g = Acceleration due to gravity = 9.81 m/s²

Mass is the property that matter has which opposes the force being applied to it. It is intrinsic to the object itself and does not change according to the gravitational force. But, the weight changes.

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4 years ago
A Cessna 150 aircraft has a lift-off speed of approximately 125 km/h. What minimum constant acceleration does this require if th
Marysya12 [62]

Answer:

Mininmum acceleration required = 2.74 m/s^{2}

Explanation:

Given,

Lift off speed = 125km/hr.

We know km/hr conversion factor to m/s is \frac{5}{18},

Therefore,

Lift off speed required = 125 \times\frac{5}{18} m/s

Lift off speed required = 34.72222 m/s

Length of Takeoff Run given = 220 m.

So,

the flight needs to achieve its takeoff speed at a constant acceleration before it reaches the end of the 220 m runway.

Consider the formula from kinematics,

v^{2}-u^{2}=2as,

where,

v - final velocity of particle,

u - initial velocity of particle

a - the constant acceleration at which the particle is moving with

s - distance travelled by particle

So at minimum acceleration,it reaches the takeoff speed at the end of the runway.

Therefore in the formula,

we use, v = Lift off speed = 34.72222 m/s;

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s = 220 m;

Substituting these values in the formula, we get,

(34.72222)^{2}-0 = 2\times a \times 220

a = \frac{(34.72222)^{2} }{2\times220}

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Therefore,the minimum acceleration of the plane required = 2.74 m/s^{2}.

7 0
3 years ago
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