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3 years ago

Please please please please can someone please help me with this. thank you

Physics

1 answer:

Svetllana [295]3 years ago

4 0

Explanation:

Answer 1

45kg×4m/s

180kg m/s

Answer 2

the momentum of each body never changes when two or more bodies collide...

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Express each of the following in ms -1 <br>a) 18kmh-1<br>

vovangra [49]

1km=1000m; 1hr=3600secs
1km/hr=1000/3600= 5/18m/sec
To convert km/hr into m/sec, multiply the number by 5 and then divide it by 18.

18kmh-1= 18•5=90
90/18=5
5ms-1

3 0

3 years ago

s=1/2(u+v)t check the dimensional consistency for this? where s is displacement, t is time, u is initial velocity, v is final ve

Anettt [7]

Answer:

See the explanation below.

Explanation:

The units of work are consistent since if we work in the international system of measures we have the following dimensional quantities for velocity, distance and time.

s = displacement [m]

v and u = velocity [m/s]

t = time [s]

Now using these units in the given equation.

$s = 0.5*([m/s]+[m/s])*[s]\\s=0.5*[m/s]*[s]\\s = 0.5*[m]$

So the expression is good, and dimensional has consistency.

8 0

3 years ago

How did thomson’s findings revise dalton’s atomic theory??

Burka [1]

Dalton thought that atoms were indivisible particles, and Thomson's discovery of the electron proved the existence of subatomic particles. ... The positive and negative charges cancel producing a neutral atom. images.tutorvista.com. Later discoveries by Rutherford and others lead to additional revisions to atomic theory.

8 0

3 years ago

Which one of the following temperatures (in °C) is equivalent to 294 K?

vaieri [72.5K]

Answer:

Explanation: its actually 20.85 but i guess they round to 21

5 0

3 years ago

A battery with an internal resistance ofrand an emf of 10.00 V is connected to a loadresistorR=r. As the battery ages, the inter

yanalaym [24]

Answer:

The current is reduced to half of its original value.

Explanation:

Assuming we can apply Ohm's Law to the circuit, as the internal resistance and the load resistor are in series, we can find the current I₁ as follows:

$I_{1} = \frac{V}{R_{int} +r_{L} }$

where Rint = r and RL = r
Replacing these values in I₁, we have:

$I_{1} = \frac{V}{R_{int} +r_{L} } = \frac{V}{2*r} (1)$

When the battery ages, if the internal resistance triples, the new current can be found using Ohm's Law again:

$I_{2} = \frac{V}{R_{int} +r_{L} } = \frac{V}{(3*r) +r} = \frac{V}{4*r} (2)$

We can find the relationship between I₂, and I₁, dividing both sides, as follows:

$\frac{I_{2} }{I_{1} } = \frac{V}{4*r} *\frac{2*r}{V} = \frac{1}{2}$

The current when the internal resistance triples, is half of the original value, when the internal resistance was r, equal to the resistance of the load.

7 0

3 years ago

Please please please please can someone please help me with this. thank you ​

Please please please please can someone please help me with this. thank you