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aleksandrvk [35]

2 years ago

13

A runner traveling with an initial velocity of 1.1 m/s accelerates at a constant rate of 0.8 m/s2fora time of 2.0 s.(a).What is

his velocity at the end of this time

1 answer:

pychu [463]2 years ago

7 0

Answer:

The final velocity of the runner at the end of the given time is 2.7 m/s.

Explanation:

Given;

initial velocity of the runner, u = 1.1 m/s

constant acceleration, a = 0.8 m/s²

time of motion, t = 2.0 s

The velocity of the runner at the end of the given time is calculate as;

$v = u + at$

where;

v is the final velocity of the runner at the end of the given time;

v = 1.1 + (0.8)(2)

v = 2.7 m/s

Therefore, the final velocity of the runner at the end of the given time is 2.7 m/s.

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Two ice skaters, each with a mass of 72.0 kg, are skating at 5.45 m/s when they collide and stick together. If the angle between

Rudiy27

Answer:

The skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.

Explanation:

To solve the problem it is necessary to go back to the theory of conservation of momentum, specifically in relation to the collision of bodies. In this case both have different addresses, consideration that will be understood later.

By definition it is known that the conservation of the moment is given by:

$m_1v_1+m_2v_2=(m_1+m_2)v_f$

Our values are given by,

$m_1=m_2=72Kg$

As the skater 1 run in x direction, there is not component in Y direction. Then,

Skate 1:

$v_{x1}=5.45m/s$

$v_{y1}=0$

Skate 2:

$v_{x2} = 5.45*cos105= -1.41m/s$

$v_{y2} = 5.45*sin105 = 5.26m/s$

Then, if we applying the formula in X direction:

m_1v_{x1}+m_2v_{x2}=(m_1+m_2)v_{fx}

75*5.45-75*1.41=(75+75)v_{fx}

Re-arrange and solving for v_{fx}

v_{fx}=\frac{4.04}{2}

v_{fx}=2.02m/s

Now applying the formula in Y direction:

$m_1v_{y1}+m_2v_{y2}=(m_1+m_2)v_{fy}$

$0+75*5.25=(75+75)v_{fy}$

$v_{fy}=\frac{5.25}{2}$

$v_{fy}=2.63m/s$

Therefore the skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.

6 0

2 years ago

Examples of drawing packages

Marat540 [252]

Answer:

The answer are given above in attachment.

5 0

3 years ago

Consider the model above. It represents the electrical force. As r increases, the attractive force decreases. How would this mod

aivan3 [116]

Answer:

As we keep on increasing the radius the value of the gravitation force of attraction decreases and as we decrease the radius the gravitation force increases.

Explanation:

Like the coulombs law of electrostatics, the law of gravitation also depends inversely on the square of the value of r. Therefore, as we keep on increasing the value of r the value of the gravitation force decreases and as we decrease the value of the r the value of gravitation force increases.

Gravitation Force= $\frac{Gm_{1}m_{2} }{r^{2}}$

Coulombs's Law= $\frac{Kq_{1}q_{2} }{r^{2}}$

6 0

3 years ago

Read 2 more answers

an object is placed 15.8 cm in front of a thin converging lens with an unknown focal length. if a real image forms behind the le

storchak [24]

<span>On what:

f (is the focal length of the lens) = ?
p (is the distance from the object to the lens) =15.8 cm
p' (is the distance from the image to the spherical lens) = 4.2 cm

</span><span>Using the Gaussian equation, to know where the object is situated (distance from the point).
</span>
$\frac{1}{f} = \frac{1}{p} + \frac{1}{p'}$
$\frac{1}{f} = \frac{1}{15.8} + \frac{1}{4.2}$
$\frac{1}{f} = \frac{2.1}{33.18} + \frac{7.9}{33.18}$
$\frac{1}{f} = \frac{10}{33.18}$
Product of extremes equals product of means:
$10*f = 1*33.18$
$10f = 33.18$
$f = \frac{33.18}{10}$
$\boxed{\boxed{f = 3.318\:cm}}\end{array}}\qquad\quad\checkmark$

6 0

3 years ago

A block slides down a slope inclined at 37 degrees. If the slope is frictionless and the block starts off at rest, how fast is i

lianna [129]

Answer:

Vf= 3.435 m/s

Explanation:

Given:

Initial velocity Vi =0 m/s (starting from Rest position)

θ = 37⁰

Distance S = 1 m

To find: Final Velocity Vf=?

fist we have to find the down slope net acceleration a = g sin θ

a= 9.81 sin 37⁰ = 5.9 m/s²

By 3rd equation of motion

2 a S= Vf² - Vi²

Vf = Square root ( 2 × 5.9 m/s² × 1 + 0 m/s)

Vf = Square root (11.8)

Vf= 3.435 m/s

3 0

3 years ago