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3 years ago

A paper airplane moving 2.33 m/s has 0.161 J of KE. What is its mass? (Unit = kg)

Physics

1 answer:

LekaFEV [45]3 years ago

4 0

The kinetic energy of an object is given by:

KE = 0.5mv²

KE = kinetic energy, m = mass, v = speed

Given values:

KE = 0.161J, v = 2.33m/s

Plug in and solve for m:

0.161 = 0.5m(2.33)²

m = 0.059kg

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Why is an element considered a pure substance?

anastassius [24]

Answer:

An element is a pure substance and is made of only one type of atom.

Explanation:

It cannot be broken down into a simpler substance.

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3 years ago

Positive charge Q is distributed uniformly along the x-axis fromx=0 to x=a. A positive point charge q is located on the positive

dybincka [34]

Answer:

a. b- x= y

dx = -dy

b. F = $\frac{-kQqi}{r (a+r)}$

c. F = $\frac{-kQqi}{r^{2} }$

Explanation:

a. x components:

$dE = \frac{kdq}{(a+r-x^{2}) } \\$

= $\frac{kQdx}{(a(a+r-x)^2}$

Integrating and solving gives:

b- x= y

dx = -dy

b. the force is given by the equation derived from (a.):

F = $\frac{-kQqi}{r (a+r)}$

c. Given that r>>a, the expression becomes:

F = $\frac{-kQqi}{r^{2} }$

Explanation:

When the size of the charge distribution is less than the distance to the deviation point of the charge then the charge distribution would produce the same effect such as a linear charge.

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4 years ago

What does activation energy has to do with a chemical reaction.

asambeis [7]

Explanation:

Activation energy and reaction rate

The activation energy of a chemical reaction is closely related to its rate. Specifically, the higher the activation energy, the slower the chemical reaction will be. ... The released energy helps other fuel molecules get over the energy barrier as well, leading to a chain reaction.

6 0

3 years ago

Question Part Points Submissions Used A car is stopped for a traffic signal. When the light turns green, the car accelerates, in

olya-2409 [2.1K]

(a) 328.6 kg m/s

The linear impulse experienced by the passenger in the car is equal to the change in momentum of the passenger:

$I=\Delta p = m\Delta v$

where

m = 62.0 kg is the mass of the passenger

$\Delta v$ is the change in velocity of the car (and the passenger), which is

$\Delta v = 5.30 m/s - 0 = 5.30 m/s$

So, the linear impulse experienced by the passenger is

$I=(62.0 kg)(5.30 m/s)=328.6 kg m/s$

(b) 404.7 N

The linear impulse experienced by the passenger is also equal to the product between the average force and the time interval:

$I=F \Delta t$

where in this case

$I=328.6 kg m/s$ is the linear impulse

$\Delta t = 0.812 s$ is the time during which the force is applied

Solving the equation for F, we find the magnitude of the average force experienced by the passenger:

$F=\frac{I}{\Delta t}=\frac{328.6 kg m/s}{0.812 s}=404.7 N$

7 0

4 years ago

Initially sliding with a speed of 1.9 m/s, a 1.8 kg block collides with a spring and compresses it 0.35 m before coming to rest.

Alika [10]

Let k = the force constant of the spring (N/m).

The strain energy (SE) stored in the spring when it is compressed by a distance x=0.35 m is
SE = (1/2)*k*x²
= 0.5*(k N/m)*(0.35 m)²
= 0.06125k J

The KE (kinetic energy) of the sliding block is
KE = (1/2)*mass*velocity²
= 0.5*(1.8 kg)*(1.9 m/s)²
= 3.249 J

Assume that negligible energy is lost when KE is converted into SE.
Therefore
0.06125k = 3.249
k = 53.04 N/m

Answer: 53 N/m (nearest integer)

3 0

3 years ago