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VashaNatasha [74]
4 years ago
8

To find force you multiply mass by accelerations; if I have a 5.0 gram Matchbox car accelerating at 7.0 m/sec2 How many Newtons

(N) of force will it hit the wall with?
Physics
2 answers:
cestrela7 [59]4 years ago
6 0

Answer:

The answer to your question is:  F = 0.035 N

Explanation:

Data

mass = 5 gram

acceleration = 7 m/s²

Force = ?

Formula

               F = ma

Process

Convert mass

                                1000 g ----------------------- 1 kg

                                      5 g  ----------------------   x

                                x = (5 x 1) / 1000

                                x = 0.005 kg

              F = (0.005)(7)

              F = 0.035 N                

MakcuM [25]4 years ago
5 0

You don't understand.

If you WANT the little car to accelerate at 7 m/s^2, then YOU have to push it with (0.005 kg)x(7m/s^2) = 0.035 Newton of force. If you ever stop pushing, the car will stop accelerating.

Does it hit a wall ? Well then, the average force it hits the wall with is

(the car's speed when it hits) times (0.005) divided by (the length of time the hit lasts) .

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Determine the accelerations that result when a 45 N net force is applied to 3kg object
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Radon-222 ( 222/86 Rn) is a radioactive gas with a half-life of 3.82 days. A gas sample contains 4.1 e 8 radon atoms initially.
kow [346]

Answer :

(a) The number of radon atoms will remain after 12 days is, 4.67\times 10^7

(b) The number of radon nuclei have decayed by this time will be, 3.6\times 10^8

Explanation :

<u>For part (a) :</u>

Half-life = 3.82 days

First we have to calculate the rate constant, we use the formula :

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{3.82\text{ days}}

k=1.81\times 10^{-1}\text{ days}^{-1}

Now we have to calculate the number of radon atoms will remain after 12 days.

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 1.81\times 10^{-1}\text{ days}^{-1}

t = time passed by the sample  = 12 days

a = initially number of radon atoms  = 4.1\times 10^8

a - x = number of radon atoms left = ?

Now put all the given values in above equation, we get

12=\frac{2.303}{1.81\times 10^{-1}}\log\frac{4.1\times 10^8}{a-x}

a-x=4.67\times 10^7

Thus, the number of radon atoms will remain after 12 days is, 4.67\times 10^7

<u>For part (b) :</u>

Now we have to calculate the number of radon nuclei will have decayed by this time.

The number of radon nuclei have decayed = Initial number of radon atoms - Number of radon atoms left

The number of radon nuclei have decayed = (4.1\times 10^8)-(4.67\times 10^7)

The number of radon nuclei have decayed = 3.6\times 10^8

Thus, the number of radon nuclei have decayed by this time will be, 3.6\times 10^8

5 0
3 years ago
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