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timofeeve [1]
3 years ago
5

We run a distance of 1000 m at a speed of 4.3 m/s. Calculate the time elapsed to cover this distance

Physics
1 answer:
pashok25 [27]3 years ago
6 0

Answer:

So if we need to cover 1000 meters. And we go at a speed of 4.3 m/s. That means that every 4.3 meters we cover is 1 second. So we divide both amd get

1000/4.3 = 232.56 is approx the answer. Also the meters cancel out because

m/(m/s) = m*s/m, cancels out giving s as a unit.

<h2><u>Therefore the answer is 232.56 seconds</u></h2>

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Well most of the particles did pass through and a few were deflected. but i think the answer is A

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3 years ago
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One ring of radius a is uniformly charged with charge +Q and is placed so its axis is the x-axis. A second ring with charge –Q i
kati45 [8]

Answer:

The force exerted on an electron is 7.2\times10^{-18}\ N

Explanation:

Given that,

Charge = 3 μC

Radius a=1 m

Distance  = 5 m

We need to calculate the electric field at any point on the axis of a charged ring

Using formula of electric field

E=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}

E_{1}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}

Put the value into the formula

E_{1}=\dfrac{9\times10^{9}\times3\times10^{-6}\times5}{(1^2+5^2)^{\frac{3}{2}}}

E_{1}=1.0183\times10^{3}\ N/C

Using formula of electric field again

E_{2}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}

Put the value into the formula

E_{2}=\dfrac{9\times10^{9}\times(-3\times10^{-6})\times5}{((0.5)^2+5^2)^{\frac{3}{2}}}

E_{2}=-1.064\times10^{3}\ N/C

We need to calculate the resultant electric field

Using formula of electric field

E=E_{1}+E_{2}

Put the value into the formula

E=1.0183\times10^{3}-1.064\times10^{3}

E=-0.045\times10^{3}\ N/C

We need to calculate the force exerted on an electron

Using formula of electric field

E = \dfrac{F}{q}

F=E\times q

Put the value into the formula

F=-0.045\times10^{3}\times(-1.6\times10^{-19})

F=7.2\times10^{-18}\ N

Hence, The force exerted on an electron is 7.2\times10^{-18}\ N

8 0
3 years ago
A 85 kg lineman tackles a 90 kg receiver. The receiver is running 5.8 m/s, and the lineman is moving 4.1 m/s, at a right angle t
weeeeeb [17]

Answer:

3.59 m/s

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We are given that

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Mass of receiver,m'=90 kg

Speed of receiver,v'=5.8 m/s

Speed of lineman,v=4.1 m/s

\theta=90^{\circ}

We have to find the their velocity immediately after the tackle.

Initial momentum,P_i=\sqrt{p^2_1+p^2_2}=\sqrt{(85\times 4.1)^2+(90\times 5.8)^2}=627.6 kgm/s

According to law of conservation of momentum

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3 0
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Answer:

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Answer:

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Based on the description, H2O2 will be 2H2O2 as it is said to contain four atoms of oxygen. This means that, in order to have a balanced equation, we must place coefficient 2 in front of water and coefficient 1 in front of oxygen. That is;

2H2O2 (l) ----> 2H2O (l) + O2(g)

6 0
3 years ago
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