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2 years ago

12

A student drops an egg from the roof of their house onto a trampoline. The egg feels a change in momentum of 2.2 kg^ * m/s in 1.

1s How many Newtons of force did the egg feel?

1 answer:

geniusboy [140]2 years ago

8 0

Answer:

F = m a = m v / t where v is the change in velocity in time t

F = p / t since m v is equal to p

F = 2.2 (kg m / s) / 1.1 s = 2 kg-m / s^2 = 2 N

Or you can use the impulse equation

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When you put a pot of water on the stove, the stove transfers thermal energy to the water. As the water gains large

STALIN [3.7K]

Answer:

It releases some of the energy into the atmosphere as hot steam.

Explanation:

8 0

2 years ago

In this lab you will use a cart and track to explore various aspects of motion. You will measure and record the time it takes th

Keith_Richards [23]

It is given that by using track and cart we can record the time and the distance travelled and also the speed of the cart can be recorded. With all this data we can solve questions on the laws of motion.

Like using the first law of motion we can determine the force of gravity acting on the cart that has moved a certain distance and the velocity or the speed of card has already been registered and since time is known putting the values in formula would help us calculate the gravitational pull acting on cart.

3 0

3 years ago

Read 2 more answers

A 65.0-Ω resistor is connected to the terminals of a battery whose emf is 12.0 V and whose internal resistance is 0.5 Ω. Calcula

Luda [366]

Answer:

a) 0.1832 A

b) 11.91 Volts

c) 2.18 Watt , 0.0168 Watt

Explanation:

(a)

R = external resistor connected to the terminals of the battery = 65 Ω

E = Emf of the battery = 12.0 Volts

r = internal resistance of the battery = 0.5 Ω

i = current flowing in the circuit

Using ohm's law

E = i (R + r)

12 = i (65 + 0.5)

i = 0.1832 A

(b)

Terminal voltage is given as

$V_{ab}$ = i R

$V_{ab}$ = (0.1832) (65)

$V_{ab}$ = 11.91 Volts

(c)

Power dissipated in the resister R is given as

$P_{R}$ = i²R

$P_{R}$ = (0.1832)²(65)

$P_{R}$ = 2.18 Watt

Power dissipated in the internal resistance is given as

$P_{r}$ = i²r

$P_{r}$ = (0.1832)²(0.5)

$P_{r}$ = 0.0168 Watt

5 0

3 years ago

1) Si un mango cae a una velocidad de 75m/s y tarda 26 seg. en caer. ¿ Cuál habrá sido la velocidad con qué el mango llegó al su

Lyrx [107]

Answer:

El mango llega al suelo a una velocidad de 329.982 metros por segundo.

Explanation:

El mango experimenta un movimiento de caída libre, es decir, un movimiento uniformemente acelerado debido a la gravedad terrestre, despreciando los efectos de la viscosidad del aire y la rotación planetaria. Entonces, la velocidad final del mango, es decir, la velocidad con la que llega al suelo, se puede determinar mediante la siguiente fórmula cinemática:

$v = v_{o}+g\cdot t$ (1)

Donde:

$v_{o}$ - Velocidad inicial, en metros por segundo.

$v$ - Velocidad final, en metros por segundo.

$g$ - Aceleración gravitacional, en metros por segundo al cuadrado.

$t$ - Tiempo, en segundos.

Si sabemos que $v_{o} = -75\,\frac{m}{s}$ , $g = -9.807\,\frac{m}{s^{2}}$ y $t = 26\,s$ , entonces la velocidad final del mango es:

$v = v_{o}+g\cdot t$

$v = -75\,\frac{m}{s}+\left(-9.807\,\frac{m}{s} \right)\cdot (26\,s)$

$v = -329.982\,\frac{m}{s}$

El mango llega al suelo a una velocidad de 329.982 metros por segundo.

8 0

2 years ago

A 1.50 µF capacitor and a 3.50 µF capacitor are connected in series across a 2.50 V battery. How much charge (in µC) is stored o

Nataly_w [17]

Explanation:

The given data is as follows.

$C_{1} = 1.50 \times 10^{-6} F$

$C_{1} = 3.50 \times 10^{-6} F$

Voltage = 2.50 V

Hence, calculate the equivalence capacitor as follows.

$\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}}$

$\frac{1}{C} = \frac{1}{1.50 \times 10^{-6} F} + \frac{1}{3.50 \times 10^{-6} F}$

= $0.945 \times 10^{-6} F$

C = $1.06 \times 10^{-6} F$

Now, we will calculate the charge across each capacitance as follows.

Q = CV

= $1.06 \times 10^{-6} F \times 2.50 V$

= $2.65 \times 10^{-6} C$

= $2.65 \mu C$

Thus, we can conclude that $2.65 \mu C$ is the charge stored on each given capacitor.

5 0

3 years ago