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3 years ago

If two objects collide and one is initially at rest, is itpossible for both to be at rest after the collision?

Physics

1 answer:

Kruka [31]3 years ago

7 0

Answer:

Not possible

Explanation:

Suppose there's no external force, then the momentum must be conserved before and after the collision.

Since momentum is defined as the product of object mass and speed. This means that if after the collision, for both object to be at rest then the total momentum is 0.

But before the collision, only 1 is at rest, the other is not -> the total momentum is non 0 before the collision.

Therefor this is not possible with the law of momentum conservation.

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A lamp can work on 50V mains taking 2 amps. What value of resistance must be connected in series with it, so that it can be oper

wolverine [178]

The resistance of the lamp is apparently 50V/2A = 25 ohms.

When the circuit is fed with more than 50V, we want to add
another resistor in series with the 25-ohm lamp so that the
current through the combination will be 2A.

In order for 200V to cause 2A of current, the total resistance
must be 200V/2A = 100 ohms.

The lamp provides 25 ohms, so we want to add another 75 ohms
in series with the lamp. Then the total resistance of the circuit is
(75 + 25) = 100 ohms, and the current is 200V/100 ohms = 2 Amps.

The power delivered by the 200V mains is (200V) x (2A) = 400 watts.

The lamp dissipates ( I² · R ) = (2² · 25 ohms) = 100 watts.

The extra resistor dissipates ( I² · R) = (2² · 75 ohms) = 300 watts.

Together, they add up to the 400 watts delivered by the mains.

CAUTION:
300 watts is an awful lot of power for a resistor to dissipate !
Those little striped jobbies can't do it.
It has to be a special 'power resistor'.
300 watts is even an unusually big power resistor.
If this story actually happened, it would be cheaper, easier,
and safer to get three more of the same kind of lamp, and
connect THOSE in series for 100 ohms. Then at least the
power would all be going to provide some light, and not just
wasted to heat the room with a big moose resistor that's too
hot to touch.

3 0

3 years ago

A child on a high dive has a mass of 40 kilograms. If the high dive is 10 meters in the air, what is the potential energy? GPE=m

saw5 [17]

Answer:

Ep = 3924 [J]

Explanation:

To calculate this value we must use the definition of potential energy which tells us that it is the product of mass by the acceleration of gravity by height.

$E_{p}=m*g*h\\$

where:

Ep = potential energy [J] (units of Joules)

m = mass = 40 [kg]

g = gravity acceleration = 9.81 [m/s²]

h = elevation = 10 [m]

$E_{p} =40*9.81*10\\E_{p} = 3924 [J]$

7 0

3 years ago

skateboarder, starting from rest, rolls down a 13.5 m ramp. When she arrives at the bottom of the ramp her speed is 7.37 m/s. If

scZoUnD [109]

Answer:

1.7 m/s²

Explanation:

d = length of the ramp = 13.5 m

v₀ = initial speed of the skateboarder = 0 m/s

v = final speed of the skateboarder = 7.37 m/s

a = acceleration

Using the equation

v² = v₀² + 2 a d

7.37² = 0² + 2 a (13.5)

a = 2.01 m/s²

θ = angle of the incline relative to ground = 29.9

a' = Component of acceleration parallel to the ground

Component of acceleration parallel to the ground is given as

a' = a Cosθ

a' = 2.01 Cos29.9

a' = 1.7 m/s²

7 0

3 years ago

मारवतन गनुहास (What Is MKS Syste<br>Convert 5 solar days into second.)

mr_godi [17]

Answer:

5 Days to Seconds = 432000

Explanation:

7 0

3 years ago

How do i find stretch? The problem in questioning has already given me the elastic energy and k-value, but I have no idea how to

finlep [7]

Answer:

Stretch can be obtained using the Elastic potential energy formula.

The expression to find the stretch (x) is $x=\sqrt{\frac{2\times EPE}{k}}$

Explanation:

Given:

Elastic potential energy (EPE) of the spring mass system and the spring constant (k) are given.

To find: Elongation in the spring (x).

We can find the elongation or stretch of the spring using the formula for Elastic Potential Energy (EPE).

The formula to find EPE is given as:

$EPE=\frac{1}{2}kx^2$

Rewriting the above expression in terms of 'x', we get:

$x=\sqrt{\frac{2\times EPE}{k}}$

Example:

If EPE = 100 J and spring constant, k = 2 N/m.

Elongation or stretch is given as:

$x=\sqrt{\frac{2\times EPE}{k}}\\\\x=\sqrt{\frac{2\times 100}{2}}\\\\x=\sqrt{100}=10\ m$

Therefore, the stretch in the spring is 10 m.

So, stretch in the spring can be calculated using the formula for Elastic Potential Energy.

6 0

3 years ago