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nignag [31]
3 years ago
10

A block of mass m1 = 4 kg is moving at 4 m/s and collides with a block of mass m2 = 2 kg, which is moving at 5 m/s in the opposi

te direction on a frictionless floor. After the collision, the blocks stick together and move as a single unit. a) [5 points] Determine the total kinetic energy of the system before the collision.
Physics
1 answer:
vladimir1956 [14]3 years ago
4 0

Answer:

The total kinetic energy of the system before the collision is 57 Joules.                  

Explanation:

Given that,

Mass of first block, m_1=4\ kg

Speed of first block, v_1=4\ m/s

Mass of second block, m_2=2\ kg

Speed of first block, v_2=-5\ m/s

We need to find the total kinetic energy of the system before the collision. It is equal to the kinetic energies of both the blocks. It is given by :

K_i=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_2^2\\\\K_i=\dfrac{1}{2}(m_1v_1^2+m_2v_2^2)\\\\K_i=\dfrac{1}{2}(4\times (4)^2+2\times (-5)^2)\\\\K_i=57\ J

So, the total kinetic energy of the system before the collision is 57 Joules.

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The energy an object has due to its motion is called kinetic energy.

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malfutka [58]

Acceleration is the rate of change in an object's velocity

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Suppose a wheel with a tire mounted on it is rotating at the constant rate of 3.37 times a second. A tack is stuck in the tire a
zhannawk [14.2K]

Answer:

The tangential speed of the tack is 8.19 m/s.

Explanation:

The wheel rotates 3.37 times a second that means wheel complete 3.37 revolutions in a second. Therefore, the angular speed ω of the wheel is given as follows:

\omega =3.37rev/s \times(\frac{2\pi rad}{1s} )\\\\=21.174rad/s

Use the relation of angular speed with tangential speed to find the tangential speed of the tack.

The tangential speed v of the tack is given by following expression

v = ω r

Here, r is the distance to the tack from axis of rotation.

Substitute 21.174 rad/s for ω, and 0.387 m for r in the above equation to solve for v.

v = 21.174 × 0.387

v = 8.19m/s

Thus, The tangential speed of the tack is 8.19 m/s.

4 0
3 years ago
Read 2 more answers
An owl weighing 40N is sitting in a tree waiting to dive down to catch a mouse. If the owl's potential energy is 800 J with resp
Natali5045456 [20]

Answer:

<em>h = 20 m</em>

Explanation:

<u>Gravitational Potential Energy</u>

Gravitational potential energy (GPE) is the energy stored in an object due to its vertical position or height in a gravitational field.

It can be calculated with the equation:

U=m.g.h

Where m is the mass of the object, h is the height with respect to a fixed reference, and g is the acceleration of gravity or 9.8 m/s^2.

The weight of an object of mass m is:

W = m.g

Thus, the GPE is:

U=W.h

Solving for h:

\displaystyle h=\frac{U}{W}

The weight of the owl is W=40 N and its GPE is U=800 J.

\displaystyle h=\frac{800}{40}=20

h = 20 m

3 0
3 years ago
An experimental tungsten light bulb filament has a length of 5 cm and a diameter of 0.074 cm. The filament is basically just a w
adell [148]

Answer:

power emitted is 1.75 W

Explanation:

given data

length l = 5 cm = 5 ×10^{-2} m

diameter d = 0.074 cm = 74 ×10^{-5} m

total filament emissivity = 0.300

temperature = 3068 K

to find out

power emitted

solution

we find first area that is π×d×L

area = π×d×L

area = π×74 ×10^{-5}×5 ×10^{-2}

area = 1162.3892  ×10^{-5} m²

so here power emitted  is express as

power emitted  = E × σ × area × (temperature)^4

put here all value

power emitted  = 0.300× 5.67 × 1162.3892  ×10^{-5}  × (3068)^4

power emitted = 1.75 W

5 0
3 years ago
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