What is the force in Newtons when the mass of a truck is 1000Kg and the acceleration of the truck is 3m/s2?

__________ is when a person becomes less sensitive to a drug. A. Tolerance B. Craving C. Addiction D. Dependence

Luden [163]

A) Tolerance

Tolerance is developed after using a drug repeatedly, so the body adapts to it. Because of that, people who develop a tolerance would then need to use more of that drug to get the same effect.

4 0

3 years ago

An instrument is thrown upward with a speed of 15 m/s on the surface of planet X where the acceleration due to gravity is 2.5 m/

Katen [24]

<h2>Answer: 12 s</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told <u>the instrument is thrown upward</u> from the surface, we will only use the equations related to the Y axis.

In this sense, the main movement equation in the Y axis is:

$y-y_{o}=V_{o}.t-\frac{1}{2}g.t^{2}$ (1)

Where:

$y$ is the instrument's final position

$y_{o}=0$ is the instrument's initial position

$V_{o}=15m/s$ is the instrument's initial velocity

$t$ is the time the parabolic movement lasts

$g=2.5\frac{m}{s^{2}}$ is the acceleration due to gravity at the surface of planet X.

As we know $y_{o}=0$ and $y=0$ when the object hits the ground, equation (1) is rewritten as:

$0=V_{o}.t-\frac{1}{2}g.t^{2}$ (2)

Finding $t$ :

$0=t(V_{o}-\frac{1}{2}g.t^{2})$ (3)

$t=\frac{2V_{o}}{g}$ (4)

$t=\frac{2(15m/s)}{2.5\frac{m}{s^{2}}}$ (5)

Finally:

$t=12s$

3 0

3 years ago

A cyclical heat engine, operating between temperatures of 450º C and 150º C produces 4.00 MJ of work on a heat transfer of 5.00

gogolik [260]

Answer:

(a) Heat transfer to the environment is: 1 MJ and (b) The efficiency of the engine is: 41.5%

Explanation:

Using the formula that relate heat and work from the thermodynamic theory as: $W=Q=Q_{in}-Q_{out}$ solving to Q_out we get: $Q_{out}=Q_{in}-W=5(MJ)-4(MJ)=1(MJ)$ this is the heat out of the cycle or engine, so it will be heat transfer to the environment. The thermal efficiency of a Carnot cycle gives us: $n=1-\frac{T_{Low} }{T_{High}}$ where T_Low is the lowest cycle temperature and T_High the highest, we need to remember that a Carnot cycle depends only on the absolute temperatures, if you remember the convertion of K=°C+273.15 so T_Low=150+273.15=423.15 K and T_High=450+273.15=723.15K and replacing the values in the equation we get: $n=1-\frac{423.15}{723.15} =0.415=41.5%$