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MArishka [77]
3 years ago
6

A person is trying to lift a crate that has a mass of 30 kg. The normal force of the floor is currently supplying 150 N of force

. How much force is the person currently exerting?
144

150

294

294
Physics
2 answers:
pochemuha3 years ago
8 0

5 meters per second

frez [133]3 years ago
6 0

Answer:

The exerting force by the person is 294 N.

(3) is correct option.

Explanation:

Given that,

Mass of crate = 30 kg

Normal force = 150 N

When a person is lifting a crate then the crate does not contact the floor

So, the normal force is zero.

We need to calculate the exerting force by the person

Using formula of force

F = ma

F=30\times9.8

F=294\ N

Hence, The exerting force by the person is 294 N.

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A solid conducting sphere of radius 2.00 cm has a charge of 6.77 μC. A conducting spherical shell of inner radius 4.00 cm and ou
madreJ [45]

Answer:

Part a)

E = 0

Part b)

E = 6.77 \times 10^7 N/C

Part c)

Electric field inside the conductor is again zero

E = 0

Part d)

E = 8.52 \times 10^6 N/C

Explanation:

Part a)

conducting sphere is of radius

R = 2 cm

so electric field inside any conductor is always zero

So electric field at r = 1 cm

E = 0

Part b)

Now at r = 3 cm

By Gauss law

E = \frac{kq}{r^2}

E = \frac{(9\times 10^9)(6.77 \muC)}{0.03^2}

E = 6.77 \times 10^7 N/C

Part c)

Again when we use r = 4.50 cm

then we will have

Electric field inside the conductor is again zero

E = 0

Part d)

Now at r = 7 cm

again by Gauss law

E = \frac{kQ}{r^2}

E = \frac{(9\times 10^9)(6.77\mu C - 2.13\mu C)}{0.07^2}

E = 8.52 \times 10^6 N/C

5 0
3 years ago
A single-phase transformer has a turns ratio of 20,000/5,000. If a DC voltage of 50 V is applied to the primary winding, what wi
Kipish [7]

Sneaky question.

The secondary voltage will be zero.

Secondary voltage is the result of CHANGES in the primary voltage. That means the primary is energized with AC. The transformer in this question is energized with DC, which doesn't change. So there is no secondary voltage, this transformer doesn't work, and what you get out of it is: Smoke !

6 0
3 years ago
The diagram shows a heat engine. In which area of the diagram is unusable thermal energy detected?
Marat540 [252]
Nope, I disagree with the former answer. The answer is definitely Z. <u>W area</u> (boxed with red outline) is represented as the hot reservoir while <u>Z area</u> is the cold reservoir (boxed with blue outline). X area is the heat engine itself and Y area is the work produced from thermal energy from hot reservoir. Typically, all heat engines lose some heat to the environment (based from the second law of thermodynamics) that is symbolically illustrated by the lost energy in the cold reservoir. This lost thermal energy is basically the unusable thermal energy. The higher thermal energy lost, the less efficient your heat engine is. 
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3 years ago
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The force that moving charged particles exert on one another is called
Irina-Kira [14]
It should be electromagnetic force
6 0
3 years ago
18. Un avión de rescate de animales que vuela hacia el este a 36.0 m/s deja caer una paca de
Alona [7]

Answer:

Definimos momento como el producto entre la masa y la velocidad

P = m*v

(tener en cuenta que la velocidad es un vector, por lo que el momento también será un vector)

Sabemos que el peso de la paca de heno es 175N, y el peso es masa por aceleración gravitatoria, entonces.

Peso = m*9.8m/s^2 = 175N

m = (175N)/(9.8m/s^2) = 17.9 kg

Ahora debemos calcular la velocidad de la paca justo antes de tocar el suelo.

Sabemos que la velocidad horizontal será la misma que tenía el avión, que es:

Vx = 36m/s

Mientras que para la velocidad vertical, usamos la conservación de la energía:

E = U + K

Apenas se suelta la caja, esta tiene velocidad cero, entonces su energía cinética será cero y la caja solo tendrá energía potencial (Si bien la caja tiene velocidad horizontal en este punto, por la superposición lineal podemos separar el problema en un caso horizontal y en un caso vertical, y en el caso vertical no hay velocidad inicial)

Entonces al principio solo hay energía potencial:

U = m*g*h

donde:

m = masa

g = aceleración gravitatoria

h = altura  

Sabemos que la altura inicial es 60m, entonces la energía potencial es:

U = 175N*60m = 10,500 N

Cuando la paca esta próxima a golpear el suelo, la altura h tiende a cero, por lo que la energía potencial se hace cero, y en este punto solo tendremos energía cinética, entonces:

10,500N = (m/2)*v^2

De acá podemos despejar la velocidad vertical justo antes de golpear el suelo.

√(10,500N*(2/ 17.9 kg)) = 34.25 m/s

La velocidad vertical es 34.25 m/s

Entonces el vector velocidad se podrá escribir como:

V = (36 m/s, -34.25 m/s)

Donde el signo menos en la velocidad vertical es porque la velocidad vertical es hacia abajo.

Reemplazando esto en la ecuación del momento obtenemos:

P = 17.9kg*(36 m/s, -34.25 m/s)  

P = (644.4 N, -613.075 N)

6 0
3 years ago
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