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3 years ago

A man pushed the box with a force of 20 N. The mass of the box doubled, but he still pushed with a force of 20 N.

Physics

1 answer:

Nuetrik [128]3 years ago

7 0

Answer:

Option (c) Acceleration is 1/2x the original acceleration

Explanation:

To know the the correct answer to the question, we shall determine the acceleration of the box in each case.

Case 1:

Mass (m) = m

Force (F) = 20 N

Acceleration 1 (a₁) =?

F = ma₁

20 = m × a₁

Divide both side by m

a₁ = 20 / m

Case 2:

Mass (m) = 2m

Force (F) = 20 N

Acceleration 2 (a₂) =?

F = ma₂

20 = 2m × a₂

Divide both side by 2m

a₂ = 20 / 2m

a₂ = 10 / m

Finally, we shall determine the acceleration of the box after the mass was doubled. This can be obtained as illustrated below:

Acceleration 1 (a₁) = 20 / m

Acceleration 2 (a₂) = 10 / m

a₂ : a₁ = 10 / m : 20 / m

a₂ / a₁ = 10 / m ÷ 20 / m

a₂ / a₁ = (10 / m) × (m / 20)

a₂ / a₁ = ½

Cross multiply

a₂ = ½a₁

Thus, the acceleration of the box after the mass was doubled is ½ times the original acceleration.

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How many lenses with different focal lengths can be obtained by combining two surfaces whose radii of curvature are 4.00 cm and

Dovator [93]

Answer:

The lenses with different focal length are four.

Explanation:

Given that,

Radius of curvature R₁= 4

Radius of curvature R₂ = 8

We know ,

Refractive index of glass = 1.6

When, R₁= 4, R₂ = 8

We need to calculate the focal length of the lens

Using formula of focal length

$\dfrac{1}{f}=(n-1)(\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}})$

Put the value into the formula

$\dfrac{1}{f}=(1.6-1)(\dfrac{1}{4}+\dfrac{1}{8})$

$\dfrac{1}{f}=\dfrac{9}{40}$

$f=4.44\ cm$

When , R₁= -4, R₂ = 8

Put the value into the formula

$\dfrac{1}{f}=(1.6-1)(\dfrac{1}{-4}+\dfrac{1}{8})$

$\dfrac{1}{f}=-\dfrac{3}{40}$

$f=-13.33\ cm$

When , R₁= 4, R₂ = -8

Put the value into the formula

$\dfrac{1}{f}=(1.6-1)(\dfrac{1}{4}-\dfrac{1}{8})$

$\dfrac{1}{f}=\dfrac{3}{40}$

$f=13.33\ cm$

When , R₁= -4, R₂ = -8

Put the value into the formula

$\dfrac{1}{f}=(1.6-1)(\dfrac{1}{-4}-\dfrac{1}{8})$

$\dfrac{1}{f}=-\dfrac{9}{40}$

$f=-4.44\ cm$

Hence, The lenses with different focal length are four.

8 0

3 years ago

Debido al desorden en el laboratorio un científico tiene 2 termómetros diferentes pero no sabe en qué escalas están por lo que d

just olya [345]

Answer:

La escala del termómetro ''A'' es grados Celsius.

La escala del termómetro ''B'' es grados Fahrenheit.

Explanation:

Para hallar en qué escalas están los termómetros partimos de que la mezcla a la cuál se midió su temperatura mantuvo su temperatura constante.

Esto quiere decir que los termómetros están expresando la misma temperatura pero en una escala distinta.

Sabemos que dada una temperatura en grados Celsius ''C'' si la queremos convertir a grados Fahrenheit ''F'' debemos utilizar la siguiente ecuación :

$F=(\frac{9}{5})C+32$ (I)

Ahora, si reemplazamos y asumimos que la temperatura de 18° es en grados Celsius, entonces si reemplazamos $C=18$ en la ecuación (I) deberíamos obtener $F=64.4$ ⇒

$F=(\frac{9}{5}).(18)+32=32.4+32=64.4$

Efectivamente obtenemos el valor esperado. Finalmente, corroboramos que la temperatura del termómetro ''A'' está medida en grados Celsius y la temperatura del termómetro ''B'' en grados Fahrenheit.

6 0

3 years ago

Squids and octopuses propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracti

Alex

Answer:

A) The speed of the water must be 8.30 m/s.

B) Total kinetic energy created by this maneuver is 70.12 Joules.

Explanation:

A) Mass of squid with water = 6.50 kg

Mass of water in squid cavuty = 1.55 kg

Mass of squid = $m_1=6.50 kg- 1.55 kg=4.95 kg$

Velocity achieved by squid = $v_1=2.60 m/s$

Momentum gained by squid = $P=m_1v_1$

Mass of water = $m_2=1.55 kg$

Velocity by which water was released by squid = $v_2$

Momentum gained by water but in opposite direction = $P'=m_2v_2$

P = P'

$m_1v_1=m_2v_2$

$v_2=\frac{m_1v_1}{m_2}=\frac{4.95 kg\times 2.60 m/s}{1.55 kg}=8.30 m/s$

B) Kinetic energy does the squid create by this maneuver:

Kinetic energy of squid = K.E = $\frac{1}{2}m_1v_1^{2}$

Kinetic energy of water = K.E' = $\frac{1}{2}m_2v_2^{2}$

Total kinetic energy created by this maneuver:

$K.E+K.E'=\frac{1}{2}m_1v_1^{2}+\frac{1}{2}m_2v_2^{2}$

$=\frac{1}{2}\times 4.95 kg\times (2.60 m/s)^2+\frac{1}{2}\times 1.55 kg\times (8.30 m/s)^2=70.12 Joules$

4 0

3 years ago

Sam moves a box with a with a force of 400N a distance of 5 meters. How long did it take him to move the box if 20 Watts of powe

Vlad1618 [11]

Answer:

100s

Explanation:

there are many student how can not get answer on time and step by step. so there are a group of trusted physics experts who provide step by step answer. just join this wats up group.

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3 years ago

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Firlakuza [10]

Answer:

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