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ioda
3 years ago
10

1. What average force is exerted on a 25 g egg by a bed sheet if the egg hits the sheet at 4 m/s and takes

Physics
1 answer:
attashe74 [19]3 years ago
5 0

Answer:

F = -0.5 N

Explanation:

Given that,

Mass of an egg, m = 25 g = 0.025 kg

Initial speed, u = 4 m/s

Final speed, v = 0 (it stops)

Time, t = 0.2 s

We need to find the average force exerted on the egg. The force is given by :

F = ma

F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.025\times (0-4)}{0.2}\\\\F=-0.5\ N

So, the average force exerted on the egg is (-0.5 N).

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Please help!How is constant or uniform acceleration used to explain free fall?
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Free fall is a special case of motion with constant acceleration, because acceleration due to gravity is always constant and downward. For example, when a ball is thrown up in the air, the ball's velocity is initially upward.
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3 years ago
What factor about the planets caused you to weigh more or less?
ad-work [718]

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7 0
3 years ago
(a) A load of coal is dropped (straight down) from a bunker into a railroad hopper car of inertia 3.0 × 104 kg coasting at 0.50
Firlakuza [10]

Answer:

a) m=20000Kg

b) v=0.214m/s

Explanation:

We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.

For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, p=m_Av_A=m_Bv_B=m_Cv_C, were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as m_h, and the mass of the first and second coals as m_1 and m_2 respectively

We start with the transition between parts A and B, so we have:

m_Av_A=m_Bv_B

Which means

m_hv_A=(m_h+m_1)v_B

And since we want the mass of the first coal thrown (m_1) we do:

m_hv_A=m_hv_B+m_1v_B

m_hv_A-m_hv_B=m_1v_B

m_1=\frac{m_hv_A-m_hv_B}{v_B}=\frac{m_h(v_A-v_B)}{v_B}

Substituting values we obtain

m_1=\frac{(3\times10^4Kg)(0.5m/s-0.3m/s)}{0.3m/s}=20000Kg=2\times10^4Kg

For the transition between parts B and C, we can write:

m_Bv_B=m_Cv_C

Which means

(m_h+m_1)v_B=(m_h+m_1+m_2)v_C

Since we want the new final speed of the car (v_C) we do:

v_C=\frac{(m_h+m_1)v_B}{(m_h+m_1+m_2)}

Substituting values we obtain

v_C=\frac{(3\times10^4Kg+2\times10^4Kg)(0.3m/s)}{(3\times10^4Kg+2\times10^4Kg+2\times10^4Kg)}=0.214m/s

5 0
3 years ago
Por favor ayuda que ejercicio debo hacer para enflacar en una semana xd
Serggg [28]

Answer:

Un excelente ejercicio para perder barriga rápidamente es correr, ya que el cuerpo gasta una mayor cantidad de calorías en un corto período de tiempo, en tan sólo 30 minutos corriendo se queman unas 300 calorías.

...

Qué ejercicios realizar

Correr. ...

Clase aeróbica. ...

Saltar cuerda. ...

Bicicleta. ...

Caminar rápido. ...

Natación.

Explanation:

dame coronita por favor

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3 years ago
a 0.4 kg block rests on a desk. the coefficient of static friction is 0.2. You push the side of the block but do not have a spri
ladessa [460]
Static friction is greater than Applied force
7 0
3 years ago
Read 2 more answers
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