All categories

3 years ago

The answer would be ?

Physics

1 answer:

Cerrena [4.2K]3 years ago

4 0

Answer:

D. demand; increased

Explanation:

Demand is how much people want it.

You might be interested in

A flying saucer moving initially at 20 m/s[E] accelerates to 50 m/s[W] in 3.8 s. Find the saucer's average

Rashid [163]

Answer:A flying saucer moving initially at 20 m/s[E] accelerates to 50 m/s[W] in 3.8 s. Find the saucer's average

acceleration during the time interval.

22. Two arrows are launched at the same time with the same speed of 10 m/s. Arrow A is launched at an angle of 65 degrees, and arrow B at an angle of 25 degrees. Both land at the same spot on the horizontal ground. Ignore air resistance and the height of the archet How long does it take for both arrows to land on the ground?

*Our experts’ time to respond varies by subject and question (we average 46 mins)

**Content not available for all Textbook Solutions or subjects

Explanation:

8 0

2 years ago

This diagram shows a process the power stars. This process is called

Damm [24]

I think its oxidation.

6 0

3 years ago

Read 2 more answers

complete the following equation that represents the complete combustion of hexane in a car engine.C6H14 + O2

mina [271]

The balanced combustion reaction for butane is;

2C4H10 + 13O2 → 8CO2 + 10H2O

Molar mass of water = 18.02 g/mol

Molar mass of oxygen = 32 g/mol

2.46g H2O(1 mol H2O/18.02g)(13 mol O2/10 mol H2O)(32g O2/1 mol O2) = 5.68g O2

5 0

3 years ago

A leaky 10-kg bucket is lifted from the ground to a height of 10 m at a constant speed with a rope that weighs 0.9 kg/m. Initial

sladkih [1.3K]

Answer:

The work done shall be 14715 Joules

Explanation:

The work done by a force 'F' in a displacement 'dy' is given by

$W=m(y)g\times dy$

At any position 'y' the weight shall be sum of weft of water and weight of string

$\therefore m(y)=m_{water}(y)+m_{string}(y)\\\\m(y)=30(1-\frac{y}{10})+0.9y$

Thus applying values we get

$W=\int m(y)g\times dy\\\\W=\int_{0}^{10}(30(1-\frac{y}{10}+0.9y)\times g)dy\\\\Solving\\W=294.3\int_{0}^{10}(1-\frac{y}{10}+0.9y)dy\\\\W=14715J$

8 0

2 years ago

Q.Solve the following circuit find total resistance RT. Also find value of voltage across resister RC.

vagabundo [1.1K]

Answer:

R_total = 14.57 Ω , V_C = 1.176 V

Explanation:

To solve this circuit we are going to find the equivalent resistance of each branch, let's remember

* Serial resistance

$R_{eq}$ = ∑ $R_{i}$

* For resistance in parallel

1 / R_{eq} = ∑ 1/R_{i}

We solve the two branches of the wheatstone bridge

Series resistors

Branch B

R_B = Rb + R4

R_B = 2 + 18

R_B = 20 Ω

Branch C

R_C5 = Rc + R5

R_C5 = 3 + 12

R_C5 = 15 Ω

Resistance in parallel R_B and R_C5

1 / R_BC = 1 / R_B + 1 / R_C5

1 / R_BC = 1/20 + 1/15 = 0.116666

R_BC = 8.57 Ω

Now we have a single branch, we solve the series resistance

R_total = R_A + R_BC

R_total = 6 + 8.57

R_total = 14.57 Ω

b) they ask us for the voltage in the resistance R_C

Let's remember that the voltage in a series circuit is the sum of the voltages

10 = V_a + V_BC

10 = i R_a + i R_BC = i (R_a + R_BC)

i = 10 / (R_a + R_BC)

i = 10 / (14.57)

i = 0.6863 A

The current in the series circuit is constant

V_BC = i R_BC

V_BC = 0.6863 8.57

V_BC = 5.8819 V

This voltage is divided in the bridge, for the two branches in parallel it is the same, but the resistance is different in each branch.

Branch C

V_BC = i R_C5

i = V_BC / R_C5

i = 5.8819 / 15

i = 0.39213 A

In this branch we have two resistors in series, let's remember that the current in a series circuit is constant

V_C = i R_C

V_C = 0.39213 3

V_C = 1.176 V

3 0

3 years ago