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MrMuchimi
1 year ago
14

A small block of wood of inertia mb is released from rest a distance h above the ground, directly above your head. you decide to

shoot it with your pellet gun, which fires a pellet of inertia mp. after the block has fallen a distance d, the pellet hits it and becomes embedded in it, kicking it upward. at the instant of impact, the pellet is moving at speed vp.
Physics
1 answer:
Zinaida [17]1 year ago
7 0

Since I'm assuming that its perfectly elastic, considering there's not enough information given, so I think that no energy is dissipated in the collision


hmax = h - d + { [ mpvp - mb√(2gd) ] / (mp+mb) }² / (2g)

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If Earth were twice as far from the sun, the force of gravity attracting the Earth to the Sun would be
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If Earth was twice as far from the sun, the force of gravity attracting the Earth to the sun would be only one-quarter as strong. The correct answer will be C. 
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Rhea kicks a soccer ball at 13 km/h to Sean. After kicking the ball, the speed of the soccer ball from Rhea’s reference frame is
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Explanation: Your welcome :)

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What is the formula that can be used to find velocity if kinetic energy and mass are known?
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The formula for kinetic energy is \frac{1}{2}m\Delta v^2. Thus, the equation for velocity is v=   \sqrt{ \frac{2TotalKineticEnergy}{m} }. 
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The equilibrium fraction of lattice sites that are vacant in silver (Ag) at 600°C is 1 × 10-6. Calculate the number of vacancies
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Answer :

The number of vacancies (per meter cube) = 5.778 × 10^22/m^3.

Explanation:

Given,

Atomic mass of silver = 107.87 g/mol

Density of silver = 10.35 g/cm^3

Converting to g/m^3,

= 10.35 g/cm^3 × 10^6cm^3/m^3

= 10.35 × 10^6 g/m^3

Avogadro's number = 6.022 × 10^23 atoms/mol

Fraction of lattice sites that are vacant in silver = 1 × 10^-6

Nag = (Na * Da)/Aag

Where,

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Na = Avogadro's number

Da = Density of silver

Aag = Atomic weight of silver

= (6.022 × 10^23 × (10.35 × 10^6)/107.87

= 5.778 × 10^28 atoms/m^3

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6 0
2 years ago
Two fun-loving otters are sliding toward each other on a muddy (and hence frictionless) horizontal surface. One of them, of mass
zvonat [6]

Answer:

(a). The magnitude and direction of the velocity of the otters after collision is 1.35 m/s toward left.

(b). The mechanical energy dissipates during this play is 226.98 J.

Explanation:

Given that,

Mass of one otter = 8.50 kg

Speed = 6.00 m/s

Mass of other = 5.75 kg

Speed = 5.50 m/s

(a). We need to calculate the magnitude and direction of the velocity of these free-spirited otters right after they collide

Using conservation of momentum

m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v

Put the value into the formula

8.50\times(-6.00)+5.75\times5.50=(8.50+5.75)\times v

v=\dfrac{-19.375}{14.25}

v=-1.35\ m/s

Negative sign shows the direction of motion of the object after collision is toward left.

(b). We need to calculate the initial kinetic energy

Using formula of kinetic energy

K.E_{i}=\dfrac{1}{2}m_{1}v_{1}^2+\dfrac{1}{2}m_{2}v_{2}^2

Put the value into the formula

K.E_{i}=\dfrac{1}{2}\times8.50\times(6.00)^2+\dfrac{1}{2}\times5.75\times(5.50)^2

K.E_{i}=239.96\ J

We need to calculate the final kinetic energy

Using formula of kinetic energy

K.E_{f}=\dfrac{1}{2}(m_{1}+m_{2})v^2

Put the value into the formula

K.E_{f}=\dfrac{1}{2}\times(8.50+5.75)\times(-1.35)^2

K.E_{f}=12.98\ J

We need to calculate the mechanical energy dissipates during this play

Using formula of loss of mechanical energy

\Delta K.E=K.E_{f}-K.E_{i}

Put the value into the formula

\Delta K.E=12.98-239.96

\Delta K.E=-226.98\ J

Negative sign shows the loss of mechanical energy

Hence, (a). The magnitude and direction of the velocity of the otters after collision is 1.35 m/s toward left.

(b). The mechanical energy dissipates during this play is 226.98 J.

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