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2 years ago

14

You push against a wall and the wall pushes back on you.

1 answer:

emmainna [20.7K]2 years ago

4 0

Answer:

Its the third law of motion.

Explanation:

Newton's third law of motion states that every action has equal and opposite reaction.

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What is newton first law

andrew11 [14]

Answer:

Newtons first law states that:

<em>If</em><em> </em><em>a</em><em> </em><em>body</em><em> </em><em>i</em><em>s</em><em> </em><em>in</em><em> </em><em>rest</em><em> </em><em>or</em><em> </em><em>motion</em><em> </em><em>in</em><em> </em><em>a</em><em> </em><em>straight</em><em> </em><em>line</em><em>,</em><em> </em><em>it</em><em> </em><em>remains</em><em> </em><em>at</em><em> </em><em>rest</em><em> </em><em>or</em><em> </em><em>at</em><em> </em><em>motion</em><em> </em><em>in</em><em> </em><em>a</em><em> </em><em>straight</em><em> </em><em>line</em><em> </em><em>with</em><em> </em><em>constant</em><em> </em><em>speed</em><em> </em><em>until</em><em> </em><em>and</em><em> </em><em>unless</em><em> </em><em>and</em><em> </em><em>external</em><em> </em><em>unbalanced</em><em> </em><em>force</em><em> </em><em>acts</em><em> </em><em>on</em><em> </em><em>it</em><em>.</em>

<em>'</em><em>This</em><em> </em><em>law</em><em> </em><em>i</em><em>s</em><em> </em><em>also</em><em> </em><em>known</em><em> </em><em>as</em><em> </em><em>the</em><em> </em><em>law</em><em> </em><em>of</em><em> </em><em>Inertia</em><em>.</em><em>'</em>

5 0

2 years ago

Two workers are sliding 290 kg crate across the floor. One worker pushes forward on the crate with a force of 430 N while the ot

m_a_m_a [10]

Answer:0.27

Explanation:

Given

One worker Pushes with force $F_1=430 N$

other Pulls it with a rope of rope $F_2=360 N$

mass of crate $m=290 kg$

both forces are horizontal and crate slides with a constant speed

Both forces are in the same direction so Friction will oppose the forces and will be equal in magnitude of sum of two forces because crate is moving with constant speed i.e. net force is zero on it

$f_r=F_1+F_2$

where $f_r$ is the friction force

$f_r=430+360$

$f_r=790 N$

$f_r=\mu N$

where $\mu$ is the coefficient of static friction

$N=mg$

$790=\mu 29\cdot 9.8$

$\mu =0.27$

6 0

3 years ago

A car is initially travelling at 12.1m/s It accelerates at a rate of -3.9m/s^2 for 7 seconds. What is it’s final velocity?

rosijanka [135]

It will be traveling in the reverse direction it was originally going at 15.2 m/s

8 0

2 years ago

A proton has_charge.

ladessa [460]

Answer:

a positive charge of 1

Explanation:

electrons have negative charges

8 0

2 years ago

Read 2 more answers

A 2 eV electron encounters a barrier 5.0 eV high and width a. What is the probability b) 0.5 that it will tunnel through the bar

denis23 [38]

Answer:

The tunnel probability for 0.5 nm and 1.00 nm are $5.45\times10^{-4}$ and $7.74\times10^{-8}$ respectively.

Explanation:

Given that,

Energy E = 2 eV

Barrier V₀= 5.0 eV

Width = 1.00 nm

We need to calculate the value of $\beta$

Using formula of $\beta$

$\beta=\sqrt{\dfrac{2m}{\dfrac{h}{2\pi}}(v_{0}-E)}$

Put the value into the formula

$\beta = \sqrt{\dfrac{2\times9.1\times10^{-31}}{(1.055\times10^{-34})^2}(5.0-2)\times1.6\times10^{-19}}$

$\beta=8.86\times10^{9}$

(a). We need to calculate the tunnel probability for width 0.5 nm

Using formula of tunnel barrier

$T=\dfrac{16E(V_{0}-E)}{V_{0}^2}e^{-2\beta a}$

Put the value into the formula

$T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times0.5\times10^{-9}}$

$T=5.45\times10^{-4}$

(b). We need to calculate the tunnel probability for width 1.00 nm

$T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times1.00\times10^{-9}}$

$T=7.74\times10^{-8}$

Hence, The tunnel probability for 0.5 nm and 1.00 nm are $5.45\times10^{-4}$ and $7.74\times10^{-8}$ respectively.

6 0

3 years ago