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cluponka [151]
2 years ago
8

Pls help! Fill in the blanks.

Physics
1 answer:
mart [117]2 years ago
5 0

Answer:

attracting iron and producing a magnetic field

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A 500-n parachutist opens his chute and experiences an air resistance force of 800 n. the net force on the parachutist is then
Tasya [4]

Force of 500 N is acting on the parachutist.

Parachutist applies 500 N force in downward direction.

Answer:

300 N upward

Solution:

Parachutist feels air resistance of 800 N.

Thus, 800 N of force is acting in upward direction.

Total force acting on the parachutist is given by,

F_{net} = air resistance force - force of parachutist

F_{net} = 800-500

F_{net} = 300 N

Direction of force is in upward direction because the air resistance force is more than force of parachutist.

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3 years ago
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How are objects in space able to “fall” into orbit?
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Answer:

MRCORRECT has answered the question

Explanation:

Newton realized that the reason the planets orbit the Sun is related to why objects fall to Earth when we drop them. The Sun's gravity pullson the planets, just as Earth's gravity pulls down anything that is not held up by some other force and keeps you and me on the ground.

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The observation deck of a skyscraper is 420 m above
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2 years ago
When looking through a fish tank, the mediums which light passes through are:
Alinara [238K]

The answer is : C ) air,water,and the tank glass.
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Una placa de cobre a 20°C tiene unas dimensiones de 65cm x 78 cm. Encuentra el área de la placa a 400°C; Coeficiente de dilataci
ValentinkaMS [17]

Answer:

El área de la placa es aproximadamente 5102.752 centímetros cuadrados.

Explanation:

Asumamos que el cambio dimensional como consecuencia de la temperatura es pequeña, entonces podemos estimar el área de la placa de cobre en función de la temperatura mediante la siguiente aproximación:

A_{f} = w\cdot l \cdot [1 + 2\cdot \alpha\cdot (T_{f}-T_{o})] (1)

Donde:

w - Ancho de la placa, en centímetros.

l - Longitud de la placa, en centímetros.

\alpha - Coeficiente de dilatación, en \frac{1}{^{\circ}C}.

T_{o} - Temperatura inicial, en grados Celsius.

T_{f} - Temperatura final, en grados Celsius.

Si sabemos que w = 65\,cm, l = 78\,cm, \alpha = 17\times 10^{-6}\,\frac{1}{^{\circ}C}, T_{o} = 20\,^{\circ}C and T_{f} = 400\,^{\circ}C, entonces el área de la placa a la temperatura final:

A_{f} = (65\,cm)\cdot (78\,cm)\cdot \left[1+\left(17\times 10^{-6}\,\frac{1}{^{\circ}C} \right)\cdot (400\,^{\circ}C-20\,^{\circ}C)\right]

A_{f} = 5102.752\,cm^{2}

El área de la placa es aproximadamente 5102.752 centímetros cuadrados.

4 0
3 years ago
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