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frez [133]
3 years ago
14

A 27.0 g marble sliding to the right at 56.8 cm/s overtakes and collides elastically with a 13.5 g marble moving in the same dir

ection at 14.2 cm/s. After the collision, the 13.5 g marble moves to the right at 71 cm/s. Find the velocity of the 27.0 g marble after the collision.
Physics
1 answer:
Nataly [62]3 years ago
7 0

Answer:v_1=28.4 cm/s

Explanation:

Given

mass of marble m_1=27 gm

velocity of marble u_1=56.8 cm/s \approx 0.568 m/s

mass of second marble m_2=13.5 gm

Velocity of second marble u_2=14.2 cm/s \approx 0.142 m/s

After collision 13.5 gm marble moves to the right  at i.e. v_2=71 cm/s

Conserving momentum

m_1u_1+m_2u_2=m_1v_1+m_2v_2

27\times 56.8+13.5\times 14.2=27\times v_1+13.5\times 71

1533.6+191.7=27\cdot v_1+958.5

27\cdot v_1=766.8

v_1=\frac{766.8}{27}=28.4 cm/s

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A 20 kg wagon is rolling to the right across a floor. A person attempts to catch and stop the crate and applies a force of 70 N,
Serggg [28]

Answer:

1.736m/s²

Explanation:

According to Newton's second law;

\sum F_x = ma_x\\

Fm - Ff = ma_x\\ where;

Fm is the moving force = 70.0N

Ff is the frictional force acting on the body

Ff = \mu R\\Ff = \mu mg\\

\mu is the coefficient of friction

m is the mass of the object

g is the acceleration due to gravity

a is the acceleration/deceleration

The equation becomes;

Fm - Ff = ma_x\\Fm - \mu mg = ma\\

Substitute the given parameters

Fm - \mu mg = ma\\70 - 0.18(20)(9.8) = 20a\\70-35.28 = 20a\\34.72 = 20a\\a = \frac{34.72}{20}\\a =  1.736m/s^2\\

Hence the deceleration rate of the wagon as it is caught is 1.736m/s²

7 0
3 years ago
What does an element’s molar mass tell you about the element?
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Answer:

The molar mass of a compound tells you the mass of one mole of that substance. In other words, it tells you the number of grams per mole of a compound.

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2 years ago
. Imagine that you are standing at the center of a giant bowl of gelatin. What type of wave will you make across the top of the
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What happens when bromine reacts with carbon?
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3 years ago
Read 2 more answers
A car is running at the speed of 45km/hr see a child 25 meter ahead and suddenly apllies a brakes. If the retradation of the car
kenny6666 [7]

Answer:

The car stops in 7.78s and does not spare the child.

Explanation:

In order to know if the car stops before the distance to the child, you take into account the following equation:

x=x_o+v_ot-\frac{1}{2}at^2        (1)

vo: initial speed of the car = 45km/h

a: deceleration of the car = 2 m/s^2

t: time

xo: initial distance to the child = 25m

x: final distance to the child = 0m

It is necessary that the solution of the equation (1) for time t are real.

You first convert the initial speed to m/s, then replace the values of the parameters and solve the quadratic polynomial for t:

45\frac{km}{h}*\frac{1h}{3600s}*\frac{1000m}{1km}=12.5\frac{m}{s}

0=25+12.5t-2t^2\\\\2t^2-12.5t-25=0\\\\t_{1,2}=\frac{-(-12.5)\pm \sqrt{(-12.5)^2-4(2)(-25)}}{2(2)}\\\\t_{1,2}=\frac{12.25\pm 18.87}{4}\\\\t_1=7.78s\\\\t_2=-1.65s

You take the first value t1 because it has physical meaning.

The solution for t is real, then, the car stops in 7.78s and does not spare the child.

4 0
3 years ago
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