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ira [324]
3 years ago
12

Danny is competing in the high jump. When he is in the air, his body has _______ energy due to its height, and it has _______ en

ergy due to its motion.
Physics
1 answer:
shusha [124]3 years ago
5 0

Answer:

Gravitational potential energy

kinetic energy

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A pebble sinks in water . A twig floats on top of the water . Compare the densities of the two objects
algol [13]
If the pebble sinks it has a greater density an the twig has a lighter density
just like water and oil. water sits on top of oil because of their density
4 0
3 years ago
BABY BABY BABY OOOHHHH I THOUGHT THAT U WOULD ALWAYS BE MINE
kenny6666 [7]

Answer:

Don't you worry, 'cause everything's gonna be alright, ai-a'ight

Be alright, ai-a'ight

Explanation:

6 0
3 years ago
Read 2 more answers
Millikan measured the electron's charge by observing tiny charged oil drops in an electric field. Each drop had a charge imbalan
IceJOKER [234]

Answer:

4.8\cdot 10^{-19} C

Explanation:

For a drop in equilibrium, the weight is equal to the electric force (in magnitude):

W = F_e

where here we have

W=1.00\cdot 10^{-14}N is the weight of the drop

F_e is the magnitude of the electric force, which can be rewritten as

F_e = qE

where

q is the charge of the oil drop

E=2.08 \cdot 10^4 N/C is the magnitude of the electric field

Substituting into the equation and solving for q, we find the charge of the oil drop:

q=\frac{W}{F_e}=\frac{1.00\cdot 10^{-14}N}{2.08\cdot 10^4 N/C}=4.8\cdot 10^{-19} C

7 0
3 years ago
The tendency for an object to sink or float has to do with the object's density.<br> True<br> False
Stella [2.4K]

Answer:

TRUE

Explanation:

BECAUSE AN OBJECTS DENSITY  SHOWS HOW MUCH WEIGHT IS PER CUBIC UNIT. LIKE SAD HAS MORE DENSITY THAN WATER SO IT SINKS WHILE WOOD HAS LESS DENSITY THAN WATER SO IT FLOATS

4 0
3 years ago
You have 10 ohm and a 100 ohm resistor in parallel. You place this equivalent resistance in series with an LED, which is rated t
Nataly [62]

Answer:

Approximately \rm 2.0\; V.

Approximately \rm 30 \; mA. (assumption: the LED here is an Ohmic resistor.)

Explanation:

The two resistors here R_1= 10\; \Omega and R_2= 100\; \Omega are connected in parallel. Their effective resistance would be equal to

\displaystyle \frac{1}{\dfrac{1}{R_1} + \dfrac{1}{R_2}} = \frac{1}{\dfrac{1}{10} + \dfrac{1}{100}} = \frac{10}{11} \; \Omega.

The current in a serial circuit is supposed to be the same everywhere. In this case, the current through the LED should be 20\; \rm mA = 0.020\; \rm A. That should also be the current through the effective \displaystyle \rm \frac{10}{11} \; \Omega resistor. Make sure all values are in standard units. The voltage drop across that resistor would be

V = I \cdot R = 0.020 \times \dfrac{10}{11} \approx 0.182\; \rm V.

The voltage drop across the entire circuit would equal to

  • the voltage drop across the resistors, plus
  • the voltage drop across the LED.

In this case, that value would be equal to 1.83 + 0.182 \approx 2.0\; \rm V. That's the voltage that needs to be supplied to the circuit to achieve a current of 20\; \rm mA through the LED.

Assuming that the LED is an Ohmic resistor. In other words, assume that its resistance is the same for all currents. Calculate its resistance:

\displaystyle R(\text{LED}) = \frac{V(\text{LED})}{I(\text{LED})}= \frac{1.83}{0.020} \approx 91.5\; \Omega.

The resistance of a serial circuit is equal to the resistance of its parts. In this case,

\displaystyle R = R(\text{LED}) + R(\text{Resistors}) = 91.5 + \frac{10}{11} \approx 100\; \Omega.

Again, the current in a serial circuit is the same in all appliances.

\displaystyle I = \frac{V}{R} = \frac{3}{100} \approx 0.030\; \rm A = 30\; mA.

7 0
3 years ago
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