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PtichkaEL [24]
3 years ago
14

Which statement(s) accurately describe the conditions immediately before and after the firecracker explodes:

Physics
1 answer:
lidiya [134]3 years ago
8 0

Answer:

Option C, The total momentum of the fragments is equal to the original momentum of the firecracker.

Explanation:

Kinetic energy of cracker cannot remain constant before and after explosion. It is so because in the process of burning and bursting some amount of kinetic energy is lost in the form of light and heat energy. While the total mass before and after the explosion remains constant due to which the momentum is conserved before and after the explosion

Hence, option C is correct

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Ml(d^2θ/dt^2) =-mgθ
Nata [24]

The equation of motion of a pendulum is:

\dfrac{\textrm{d}^2\theta}{\textrm{d}t^2} = -\dfrac{g}{\ell}\sin\theta,

where \ell it its length and g is the gravitational acceleration. Notice that the mass is absent from the equation! This is quite hard to solve, but for <em>small</em> angles (\theta \ll 1), we can use:

\sin\theta \simeq \theta.

Additionally, let us define:

\omega^2\equiv\dfrac{g}{\ell}.

We can now write:

\dfrac{\textrm{d}^2\theta}{\textrm{d}t^2} = -\omega^2\theta.

The solution to this differential equation is:

\theta(t) = A\sin(\omega t + \phi),

where A and \phi are constants to be determined using the initial conditions. Notice that they will not have any influence on the period, since it is given simply by:

T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{g}{\ell}}.

This justifies that the period depends only on the pendulum's length.

4 0
3 years ago
. A student times a car traveling a distance of 2 m. She finds that it takes the car 5 s to
AveGali [126]
No, the car travels 1 metre in 5s at the start which is 0.2m/s, while the second meter it travels one metre in 8 seconds which is 0.125 m/s, the speed changes therefore it is not constant during the two metres the car travels
3 0
3 years ago
A 2 kg rock is at the edge of a cliff 20 meters above a lake The rock becomes loose and falls toward the water below. Calculate
natima [27]

Answer:

The potential energy (P.E) at the top is 392 J

The kinetic energy (K.E) at the top is 0 J

The potential energy (P.E) at the halfway point is 196 J.

The kinetic energy (K.E) at the halfway point is 196 J.

Explanation:

Given;

mass of the rock, m = 2 kg

height of the cliff, h = 20 m

speed of the rock at the halfway point, v = 14 m/s

The potential energy (P.E) and kinetic energy (K.E) when its at the top;

P.E = mgh

P.E = (2)(9.8)(20)

P.E= 392 J

K.E = ¹/₂mv²

where;

v is velocity of the rock at the top of the cliff = 0

K.E = ¹/₂(2)(0)²

K.E = 0

The potential energy (P.E) and kinetic energy (K.E) at the halfway point;

P.E = mg(¹/₂h)

P.E = (2)(9.8)(¹/₂ x 20)

P.E = 196 J

K.E = ¹/₂mv²

where;

v is velocity of the rock at the halfway point = 14 m/s

K.E = ¹/₂(2)(14)²

K.E = 196 J.

4 0
3 years ago
Clarence is taking his dog for a walk around town. After wandering around for an hour, his mother calls him, telling him dinner
Anvisha [2.4K]

Answer:

he can subtract the distance and then divide by the time it takes him

Explanation:

8 0
3 years ago
An electric field of 1139 V/m is applied to a section of silver of uniform cross section. Find the resulting current density if
N76 [4]

Answer:

So, you're going to need the equation ρ = ρo [1 + α(T-To)]  

1.59x10^-8 ohms*m is your ρo because that is measured at your reference temperature (To), 20◦C. T is your 6◦C and α is 0.0038(◦C)−1. So, using that you solve for ρ. If you keep up with the units though, you notice it comes out to be ohms*m and that isn't what you want.  

So, the next equation you need is J=σE where E is your electric field (3026 V/m) and σ is the electrical conductivity which is the inverse of your answer you got in the previous equation. So find the inverse of that answer and multiply it by your electric field and that will give you the current density.  

I hope this helps!

Explanation:

8 0
3 years ago
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