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3 years ago

10

A 50.0 g projectile is launched with a horizontal velocity of 657 m/s from a 4.65 kg launcher moving in the same direction at 2.

00 m/s. What is the launcher’s velocity after the launch?

1 answer:

Anni [7]3 years ago

4 0

Backwards with the speed 5.24m/s

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garik1379 [7]

It would take about 2 thirds of a second or .66666666 repeating of a second. please give brainliest?

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3 years ago

Romashka-Z-Leto [24]

Answer:

A car traveling south with its cruise control set

Explanation:

Turning could cause anything to slow down but going forward could mean that anything could go faster.

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2 years ago

The range of all electromagnetic radiation is known as the

Readme [11.4K]

B.) <span>The range of all electromagnetic radiation is known as the "Electromagnetic Spectrum"

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2 years ago

You are in the frozen food section of the grocery store and you notice that your hand gets cold when you place it on the glass w

Simora [160]

Answer:

I think my friend got it all wrong, as coolness can not be transferred but heat was actually transferred between my hand and the glass windows

Explanation:

In thermodynamics, coolness can not be transferred, only heat can be transferred

Here is how the mechanism of why i felt cold works, my body gave out heat, hence there was heat transfer from a region of high to a low heat region, equilibrium was reached and I started feeling the coolness in my hands.

4 0

2 years ago

a nonrelativistic proton is confined to a length of 2.0 pm on the x-axis. what is the kinetic energy of the proton if its speed

Elina [12.6K]

Answer:

K = 1.29eV

Explanation:

In order to calculate the kinetic energy of the proton you first take into account the uncertainty principle, which is given by:

$\Delta x \Delta p\geq \frac{h}{4\pi}$ (1)

Δx : uncertainty of position = 2.0pm = 2.0*10^-12m

Δp: uncertainty of momentum = ?

h: Planck's constant = 6.626*10^-34 J.s

You calculate the minimum possible value of Δp from the equation (1):

$\Delta p=\frac{h}{4\pi \Delta x}=\frac{6.626*10^{-34}J.s}{4\pi(2.0*10^{-12}m)}\\\\\Delta p=2.63*10^{-23}kg.\frac{m}{s}$

The minimum kinetic energy is calculated by using the following formula:

$k=\frac{(\Delta p)^2}{2m}$ (2)

m: mass of the proton = 1.67*10^{-27}kg

$k=\frac{(2.63*10^{-23}kgm/s)^2}{2(1.67*10^{-27}kg)}=2.08*10^{-19}J$

in eV you have:

$2.08*10^{-19}J*\frac{6.242*10^{18}eV}{1J}=1.29eV$

The kinetic energy of the proton is 1.29eV

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2 years ago