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3 years ago

A child slides down a hill on a toboggan with acceleration of 1.8 m/s2. if she starts from rest, how far has she traveled in:

1 answer:

4 0

A child slides down a hill on a toboggan with acceleration of 1.8 m/s2. if she starts from rest, how far has she traveled in: 2 seconds

Answer:

2.4 m

Explanation:

From the question above,

Applying equation of motion,

s = ut+at²/2....................... Equation 1

Where t = time, u = initial velocity, a = acceleration, s = distance.

make s the subject of the equation,

Given: a = 1.8 m/s², t = 2 seconds, u = 0 m/s (from rest)

Substitute these value into equation 1

s = 0(2)+1.8(2²)/2

s = 1.2(4)/2

s = 2.4 m.

Hence she has traveled 2.4 m

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In the physics lab, a block of mass M slides down a frictionless incline from a height of 35cm. At the bottom of the incline it

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Solution :

Given :

M = 0.35 kg

$m=\frac{M}{2}=0.175 \ kg$

Total mechanical energy = constant

or $K.E._{top}+P.E._{top} = K.E._{bottom}+P.E._{bottom}$

But $K.E._{top} = 0$ and $P.E._{bottom} = 0$

Therefore, potential energy at the top = kinetic energy at the bottom

$\Rightarrow mgh = \frac{1}{2}mv^2$

$\Rightarrow v = \sqrt{2gh}$

$=\sqrt{2 \times 9.8 \times 0.35}$ (h = 35 cm = 0.35 m)

= 2.62 m/s

It is the velocity of M just before collision of 'm' at the bottom.

We know that in elastic collision velocity after collision is given by :

$v_1=\frac{m_1-m_2}{m_1+m_2}v_1+ \frac{2m_2v_2}{m_1+m_2}$

here, $m_1=M, m_2 = m, v_1 = 2.62 m/s, v_2 = 0$

∴ $$v_1=\frac{0.35-0.175}{0.5250}+\frac{2 \times 0.175 \times 0}{0.525}$

$=\frac{0.175}{0.525}+0$

= 0.33 m/s

Therefore, velocity after the collision of mass M = 0.33 m/s

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3 years ago

Which most likely indicates a chemical change has occurred?

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green liquid becoming a red liquid

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2 years ago

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An applied force of 50 N is used to accelerate an object, that weighs 73 N, to the right across a frictional surface. The object

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Answer:

5.38 m/s^2

Explanation:

NET force causing the object to accelerate = 50 -10 = 40 N

Mass of the object = 73 N / 9.81 m/s^2 = 7.44 kg

F = ma

40 = 7.44 * a a = 5.38 m/s^2

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1 year ago

A .5 kg toy train car moving forward at 3 m/s collides with and sticks to a .8 kg toy car that is traveling at 2 m/s what is the

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Here we have perfectly inelastic collision. Perfectly inelastic collision is type of collision during which two objects collide, stay connected and momentum is conserved. Formula used for conservation of momentum is:
$m_{1} * v_{1} + m_{2} * v_{2} = m_{1} * v'_{1}+ m_{2} * v'_{2}$

In case of perfectly inelastic collision v'1 and v'2 are same.

We are given information:
m₁=0.5kg
m₂=0.8kg
v₁=3m/s
v₂=2m/s
v'₁=v'₂=x

0.5*3 + 0.8*2 = 0.5*x + 0.8*x
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3 years ago

Which of the following is an example of rolling friction?

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Answer:

O bike tires on the road as you ride

Explanation:

is the rolling friction

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