PLZ HELP Lab: Circuit Design

Write whether true or false...

elixir [45]

Answer:

a --> true, b --> false, c --> true, d -->false

Explanation:

a) since it stays floating the gravity force and the upqards push is the same

b) if it's balanced the rocket won't move from the ground, the force of the rocket, has to exceed the force of gravity

c) since it's going in a diretion the force of gravity is exceeding the force pushing it up

d) since that are speeding up at a rate, meaning growing, the force is unbalanced.

8 0

2 years ago

A train at a constant 60.0 km/h moves east for 40.0 min, then in a direction 50.0 degrees east of due north for 20.0 min, and th

son4ous [18]

Answer:

Part a)

$v = 7.57 km/h$

Part b)

$\theta = 67.5 degree$ North of East

Explanation:

Speed of train towards East = 60 km/h

displacement towards East is given as

$d_1 = 40 km$

now it turns towards 50 degree East of North

so its distance is given as

$d_2 = 20 km(sin50 \hat i + cos50\hat j)$

$d_2 = 15.3 \hat i + 12.8 \hat j$

then finally it moves towards west for 50 min

$d_3 = -50 \hat i$

Now the total displacement of the train is given as

$d = d_1 + d_2 + d_3$

$d = (40 + 15.3 - 50)\hat i + 12.8 \hat j$

$d = 5.3\hat i + 12.8 \hat j$

now total time duration of the motion is given as

$T = 40 min + 20 min + 50 min$

$T = 1.83 h$

now average velocity is given as

$v_{avg} = \frac{5.3\hat i + 12.8\hat j}{1.83}$

$v_{avg} = 2.89\hat i + 6.99\hat j$

Part a)

magnitude of the average velocity is given as

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{2.89^2 + 6.99^2}$

$v = 7.57 km/h$

Part b)

Direction of the velocity is given as

$tan\theta = \frac{v_y}{v_x}$

$tan\theta = \frac{6.99}{2.89}$

$\theta = 67.5 degree$ North of East

6 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: $n=1-\frac{423.15}{723.15} =0.415=41.5%$