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icang [17]
3 years ago
7

A 150-kg object takes 1.5 minutes to travel a 2,500- meter straight path. It begins the trip traveling 120 m/s and decelerates t

o a velocity of 20 m/s . What is the acceleration?
Physics
1 answer:
lana [24]3 years ago
4 0
The answer is it is -1.11 m/s^2.
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PROBLEMA DI FISICA:il volume di un attizzatoio da camino in ferro, a temperatura ambiente (20°C), è 12,2 cm cubi. A contatto con
lesantik [10]

Responder:

20.3 ° C

Explicación:

<u>Según la ley de Charles</u>: <em>cuando la presión sobre una muestra de gas seco se mantiene constante, la temperatura y el volumen estarán en proporción directa. </em>

Paso uno:

datos dados

Temperatura T1 = 20 ° C

Temperatura T2 =?

Volumen V1 = 12.2 cm ^ 3

Volumen V2 = 12.4 cm ^ 3

Aplicar la relación temperatura y volumen

\frac {V_{1}}{T_{1}}=\frac {V_{2}}{T_{2}}

sustituyendo tenemos

\frac {12.2}{20}=\frac {12.4}{T_{2}}\\\\

Cruz multiplicar tenemos

T_2=\frac{20*12.4}{12.2} \\\\T_2=\frac{248}{12.2}\\\\T_2=20.3

Temperatura delle braci 20.3°C

4 0
3 years ago
A 1300-N crate rests on the floor. How much work is required to move it at constant speed (a)
kherson [118]

a) The work done is 920 J

b) The work done is 5200 J

Explanation:

a)

In this first part of the problem, the crate is moved horizontally at constant speed.

The work required in this case is given by

W=Fd cos \theta

where

F is the magnitude of the force applied

d is the displacement of the crate

\theta is the angle between the direction of the force and of the displacement

Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so

F = 230 N

The displacement is

d = 4.0 m

And the angle is \theta=0^{\circ}, since the force is applied horizontally. Therefore, the work done is

W=(230)(4.0)(cos 0^{\circ})=920 J

b)

In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore

F = W = 1300 N

The displacement this time is again

d = 4.0 m

And the angle is \theta=0^{\circ}, since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is

W=(1300)(4.0)(cos 0^{\circ})=5200 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0
3 years ago
How is thermal energy naturally transferred between objects?
Sladkaya [172]

The second law of thermodynamics establishes restrictions on the flow of thermal energy between two bodies. This law states that the energy does not flow spontaneously from a low temperature object T1, to another object that is at a high temperature T2.

For example. Suppose you place your cell phone on the table. Your phone is at a temperature of 40 ° C and the table is at 19 ° C. Then, it is impossible for the table to spontaneously transfer its thermal energy to the telephone, and so that the table gets colder and the telephone warmer.


Finally we can say that the correct option is B: From the hotter object to the cooler object

5 0
2 years ago
Scientific notation for 0.000468
Lubov Fominskaja [6]
I think that’s the correct answer

5 0
3 years ago
The current in some DC circuits decays according to the function I=I0e−t/τ, where I is the current at some point in time, I0 is
almond37 [142]

Answer: 1.95

Explanation:

You should start off from the decay formula and solve for τ:

I = I_{0}e^{\frac{t}{\tau\\  } }

\frac{I}{I_{0}} = e^{\frac{-t}{\tau} }

Apply inverse logarithmic function:

ln(\frac{0.2 A}{1.2 A} ) = \frac{-t}{\tau}

The final form will be:

\tau=\frac{-3.5s}{ln(\frac{0.2A}{1.2A} )}

Inputing values for I, IO, and t:

\tau=\frac{-3.5S}{ln(\frac{0.2 A}{1.2 A} )} = 1.95

3 0
3 years ago
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