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Marizza181 [45]
3 years ago
10

What is the Volcanic Intrusion "rule"-

Physics
1 answer:
poizon [28]3 years ago
7 0
The volcanic intrusion rule is relative dating 
HOPE I HELPED!!!! =D
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A 70-kg astronaut is space walking outside the space capsuleand is stationary when the tether line breaks. As a means of returni
Bogdan [553]

Answer:

0.4 m/s

Explanation:

Law of conservation of momentum tell us that the change in momentum of the hammer will be equal to the change in momentum of the astronaut

change in momentum of hammer = change in momentum of astronaut

2 kg (14 m/s - 0 m/s) = 70 kg * (v-0)

                                v  = 0.4 m/s

7 0
3 years ago
Read 2 more answers
a ball of mass 100g moving at a velocity of 100m/s collides with another ball of mass 400g moving at 50m/s in same direction, if
klio [65]

Answer:

Velocity of the two balls after collision: 60\; \rm m \cdot s^{-1}.

100\; \rm J of kinetic energy would be lost.

Explanation:

<h3>Velocity</h3>

Because the question asked about energy, convert all units to standard units to keep the calculation simple:

  • Mass of the first ball: 100\; \rm g = 0.1\; \rm kg.
  • Mass of the second ball: 400\; \rm g = 0.4 \; \rm kg.

The two balls stick to each other after the collision. In other words, this collision is a perfectly inelastic collision. Kinetic energy will not be conserved. The velocity of the two balls after the collision can only be found using the conservation of momentum.

Assume that the system of the two balls is isolated. Thus, the sum of the momentum of the two balls will stay the same before and after the collision.

The momentum of an object of mass m and velocity v is: p = m \cdot v.

Momentum of the two balls before collision:

  • First ball: p = m \cdot v = 0.1\; \rm kg \times 100\; \rm m \cdot s^{-1} = 10\; \rm kg \cdot m \cdot s^{-1}.
  • Second ball: p = m \cdot v = 0.4\; \rm kg \times 50\; \rm m \cdot s^{-1} = 20\; \rm kg \cdot m \cdot s^{-1}.
  • Sum: 10\; \rm kg \cdot m \cdot s^{-1} + 20 \; \rm kg \cdot m \cdot s^{-1} = 30 \; \rm kg \cdot m \cdot s^{-1} given that the two balls are moving in the same direction.

Based on the assumptions, the sum of the momentum of the two balls after collision should also be 30\; \rm kg \cdot m \cdot s^{-1}. The mass of the two balls, combined, is 0.1\; \rm kg + 0.4\; \rm kg = 0.5\; \rm kg. Let the velocity of the two balls after the collision v\; \rm m \cdot s^{-1}. (There's only one velocity because the collision had sticked the two balls to each other.)

  • Momentum after the collision from p = m \cdot v: (0.5\, v)\; \rm kg \cdot m \cdot s^{-1.
  • Momentum after the collision from the conservation of momentum: 30\; \rm kg \cdot m \cdot s^{-1}.

These two values are supposed to describe the same quantity: the sum of the momentum of the two balls after the collision. They should be equal to each other. That gives the equation about v:

0.5\, v = 30.

v = 60.

In other words, the velocity of the two balls right after the collision should be 60\; \rm m \cdot s^{-1}.

<h3>Kinetic Energy</h3>

The kinetic energy of an object of mass m and velocity v is \displaystyle \frac{1}{2}\, m \cdot v^{2}.

Kinetic energy before the collision:

  • First ball: \displaystyle \frac{1}{2} \, m \cdot v^2 = \frac{1}{2}\times 0.1\; \rm kg \times \left(100\; \rm m \cdot s^{-1}\right)^{2} = 500\; \rm J.
  • Second ball: \displaystyle \frac{1}{2} \, m \cdot v^2 = \frac{1}{2}\times 0.4\; \rm kg \times \left(50\; \rm m \cdot s^{-1}\right)^{2} = 500\; \rm J.
  • Sum: 500\; \rm J + 500\; \rm J = 1000\; \rm J.

The two balls stick to each other after the collision. Therefore, consider them as a single object when calculating the sum of their kinetic energies.

  • Mass of the two balls, combined: 0.5\; \rm kg.
  • Velocity of the two balls right after the collision: 60\; \rm m\cdot s^{-1}.

Sum of the kinetic energies of the two balls right after the collision:

\displaystyle \frac{1}{2} \, m \cdot v^{2} = \frac{1}{2}\times 0.5\; \rm kg \times \left(60\; \rm m \cdot s^{-1}\right)^2 = 900\; \rm J.

Therefore, 1000\; \rm J - 900\; \rm J = 100\; \rm J of kinetic energy would be lost during this collision.

7 0
3 years ago
Explain two scenarios where a large truck can have the same momentum as a small car.
KengaRu [80]

The momentum, p, of any object having mass m and the velocity v is

p=mv\cdots(i)

Let M_L and M_S be the masses of the large truck and the car respectively, and V_L and V_S be the velocities of the large truck and the car respectively.

So, by using equation (i),

the momentum of the large truck = M_LV_L

and the momentum of the small car = M_SV_S.

If the large truck has the same momentum as a small car, then the condition is

M_LV_L=M_SV_S\cdots(ii)

The equation (ii) can be rearranged as

\frac {M_L}{M_S}=\frac {V_S}{V_L} \; or \; \frac{M_L}{V_S}=\frac{M_S}{V_L}

So, the first scenario:

\frac {M_L}{M_S}=\frac {V_S}{V_L}

\Rhghtarrow M_L:M_S=V_S:V_L

So, to have the same momentum, the ratio of mass of truck to the mass of the car must be equal to the ratio of velocity of the car to the velocity of the truck.

The other scenario:

\frac{M_L}{V_S}=\frac{M_S}{V_L}

\Rhghtarrow M_L:V_S= M_S:V_L

So, to have the same momentum, the ratio of mass of truck to the velocity of the car must be equal to the ratio of mass of the car to the velocity of the truck.

5 0
3 years ago
Para abrir una puerta se comprime un resorte y, por defecto de este, la puerta se cierra automáticamente. ¿Qué forma de energía
harkovskaia [24]

Answer:

El resorte al comprimirse adquiere energía potencial elástica.

Explanation:

Por conservación de la energía, y si no hay agentes externos (como pérdida de energía por rozamiento), la energía cinética que proviene del movimiento de la puerta al abrirse se transfiere al resorte en forma de energía potencial elástica cuando el resorte se comprime.

En el proceso de descompresión ocurre lo contrario, es decir, la energía potencial elástica del resorte se transforma en energía cinética que es transferida a la puerta para cerrarse.      

Por lo tanto, el resorte al comprimirse adquiere energía potencial elástica.

Espero que te sea de utilidad!            

4 0
2 years ago
an object starts at rest and accelerates until it is moving at 247 km/h. if it takes 1.7minutes for this change to occur what is
d1i1m1o1n [39]

The average acceleration in 0.67 m/s^2.

The average acceleration is calculated as change in velocity/time

a=(247*5/18)/(1.7*60)

=68.61/102=0.67 m/s^2

Therefor the average acceleration of the object is 0.67 m/s^2

5 0
3 years ago
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