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2 years ago

11

Professor blossom expects her students to be on time for class holds, them for the entire class period and has firm deadlines se

t for reports and other assignment. Sociologically which best describes professor blossom’s approach to teaching

1 answer:

defon2 years ago

8 0

Answer:

A dictatorship or tyranny.

Explanation:

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Ground reaction force acting on carter

mezya [45]

We don't know Carter, and we don't know where he is or what
he's doing, so I'm taking a big chance speculating on an answer.

I'm going to say that if Carter is pretty much just standing there,
or, let's say, lying on the ground taking a nap, then the force of
the ground acting on him is precisely exactly equal to his weight.

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2 years ago

The site from which an airplane takes off is the origin. The X axis points east, the y axis points straight up. The position and

Contact [7]

Answer:

d = 3.5*10^4 m

Explanation:

In order to calculate the displacement of the airplane you need only the information about the initial position and final position of the airplane. THe initial position is at the origin (0,0,0) and the final position is given by the following vector:

$\vec{r}=(1.21*10^3\hat{i}+3.45*10^4\hat{j})m$

The displacement of the airplane is obtained by using the general form of the Pythagoras theorem:

$d=\sqrt{(x-x_o)^2+(y-y_o)^2}$ (1)

where x any are the coordinates of the final position of the airplane and xo and yo the coordinates of the initial position. You replace the values of all variables in the equation (1):

$d=\sqrt{(1.12*10^3-0)^2+(3.45*10^4-0)^2}=3.45*10^4m$

hence, the displacement of the airplane is 3.45*10^4 m

6 0

2 years ago

How do you write a scale for distance-time graph in physics?

Studentka2010 [4]

Answer:

Time always is on X axis.

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2 years ago

Read 2 more answers

What is a difference in electrical charge from one point to another called?

spayn [35]

Answer:electrical potential

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2 years ago

61. A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A

Mademuasel [1]

Answer:

Part a)

$percentage = 21.3$ %

Part b)

$percentage = 2.13 \times 10^{-5}$ %

Explanation:

As we know that total power used in the room is given as

$P = P_1 + P_2 + P_3 + P_4$

here we have

$P_1 = (110)(3) = 330 W$

$P_2 = 100 W$

$P_3 = 60 W$

$P_4 = 3 W$

$P = 330 + 100 + 60 + 3$

$P = 493 W$

Part a)

Since power supply is at 110 Volt so the current obtained from this supply is given as

$110\times i = 493$

$i = 4.48 A$

now resistance of transmission line

$R = \frac{\rho L}{A}$

$R = \frac{(2.8 \times 10^{-8})(10\times 10^3)}{\pi(4.126\times 10^{-3})^2}$

$R = 5.23 \ohm$

now power loss in line is given as

$P = i^2 R$

$P = (4.48)^2(5.23)$

$P = 105 W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{105}{493} \times 100$

$percentage = 21.3$ %

Part b)

now same power must have been supplied from the supply station at 110 kV, so we have

$110 \times 10^3 (i ) = 493$

$i = 4.48\times 10^{-3} A$

now power loss in line is given as

$P = i^2 R$

$P = (4.48 \times 10^{-3})^2(5.23)$

$P = 1.05 \times 10^{-4} W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{1.05 \times 10^{-4}}{493} \times 100$

$percentage = 2.13 \times 10^{-5}$ %

6 0

2 years ago