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3 years ago

A net force of 500 N acts on a 100 kg cart. What is the acceleration of the cart if the mass is doubled?

Physics

1 answer:

oksian1 [2.3K]3 years ago

8 0

Answer:

2.5 m/s²

Explanation:

You can solve the following equation: F=ma for acceleration.

You'll be left with this:

a=F/m

And then you substitute the force and the doubled mass

a=500N/200kg

a=2.5 m/s²

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Einstein is probably most known for his equation E = mc2, where E = the energy of a particle of mass = m, and c = the speed of l

Zepler [3.9K]

Answer:

speed of light

Explanation:

it is speed because they are talking about speed in the equation

6 0

3 years ago

HELP!! WILL MARK BRAINLIEST!!!!!

krek1111 [17]

Answer:

Create

Broken

Explanation:

Bond formation or creation requires the use of energy. Energy is used during bond formation between chemical species. The energy is required for the reaction to occur.

When bonds are broken, energy is released
Bond breaking process is a procedure that releases energy.
This energy makes them able to react.

5 0

3 years ago

A long cylindrical capacitor is made of a central wire of radius a = 2.50 mm surrounded by a conducting shell of radius b = 7.50

asambeis [7]

Answer:

The capacitance per unit length is $5.06\times10^{-11}\ F/m$

(b) is correct option.

Explanation:

Given that,

Radius a= 2.50 mm

Radius b=7.50 mm

Dielectric constant = 3.68

Potential difference = 120 V

We need to calculate charge per length for the capacitance

Using formula of charge per length

$\lambda=\dfrac{4\pi\epsilon_{0}\Delta V}{2 ln(\dfrac{r_{2}}{r_{1}})}$

Put the value into the formula

$\lambda=\dfrac{120}{9\times10^{9}\times2 ln(\dfrac{7.50\times10^{-3}}{2.50\times10^{-3}})}$

$\lambda=6.068\times10^{-9}\ C/m$

We know that,

$\lambda=\dfrac{Q}{L}$

We need to calculate the capacitance per unit length

Using formula of capacitance per unit length

$C=\dfrac{\dfrac{Q}{L}}{\Delta V}$

$C=\dfrac{6.068\times10^{-9}}{120}$

$C=5.06\times10^{-11}\ F/m$

Hence, The capacitance per unit length is $5.06\times10^{-11}\ F/m$

7 0

4 years ago

A lady bug is sitting on the bottom of a can while you twirl it overhead on a string that is 65.0

MA_775_DIABLO [31]

The linear speed of the ladybug is 4.1 m/s

Explanation:

First of all, we need to find the angular speed of the lady bug. This is given by:

$\omega=\frac{2\pi}{T}$

where

T is the period of revolution

The period of revolution is the time taken by the ladybug to complete one revolution: in this case, since it does 1 revolution every second, the period is 1 second:

T = 1 s

Therefore, the angular speed is

$\omega=\frac{2\pi}{1 s}=6.28 rad/s$