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3 years ago

8

Match the measurement with the proper Sl unit Acceleration: Velocity Distance:

2 answers:

Ket [755]3 years ago

6 0

Answer:

Acceleration - ms^-2

Velocity- ms^-1

Distance-m

Xelga [282]3 years ago

3 0

Answer:

Acceleration:

$m/s^{2}$

Velocity:

$m/s$

Distance:

$m$

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The cockroach Periplaneta americana can detect a static electric field of magnitude 8.50 kN/C using their long antennae. If the

erica [24]

Answer:

0.235 nC

Explanation:

Given:

$E$ = the magnitude of electric field = $8.50\ kN/C =8.50\times 10^{3}\ N/C$

$F$ = the magnitude of electric force on each antenna = $2.00\ \mu N =2.00\times 10^{-6}\ N$

$q$ = The magnitude of charge on each antenna

Since the electric field is the electric force applied on a charged body of unit charge.

$\therefore E = \dfrac{F}{q}\\\Rightarrow q =\dfrac{F}{E}\\\Rightarrow q =\dfrac{2.00\times 10^{-6}\ N}{8.50\times 10^{3}\ N/C}\\\Rightarrow q =0.235\times 10^{-9}\ C\\\Rightarrow q =0.235\ nC$

Hence, the value of q is 0.235 nC.

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3 years ago

Which characteristic must a food have to receive a “natural” label from the FDA ?

adelina 88 [10]

No artificial ingredients
i only know that because my mom was vegan for two years

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3 years ago

Emeka carried out a reaction in which a gas was given off. He followed the progress of the reaction by measuring the mass of the

pickupchik [31]

Answer:

17.5

or

1.1 g/min

I know it's one of these, try getting a second opinion

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2 years ago

A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the

Maurinko [17]

Answer:

(a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity $R_{0}=10\ mCi$

Time $t_{1}=4\ hours$

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

$R=R_{0}e^{-\lambda t}$

$\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})$

Put the value into the formula

$\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})$

$\lambda=0.0000154\ s^{-1}$

$\lambda=1.55\times10^{-5}\ s^{-1}$

We need to calculate the half life

Using formula of half life

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}$

Put the value into the formula

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}$

$T_{\dfrac{1}{2}}=44.719\times10^{3}\ s$

$T_{\dfrac{1}{2}}=11.3\ hr$

(b). We need to calculate the value of N₀

Using formula of $N_{0}$

$N_{0}=\dfrac{3.70\times10^{6}}{\lambda}$

Put the value into the formula

$N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}$

$N_{0}=2.38\times10^{11}\ nuclei$

(c). We need to calculate the sample's activity

Using formula of activity

$R=R_{0}e^{-\lambda\times t}$

Put the value intyo the formula

$R=10e^{-(1.55\times10^{-5}\times30\times3600)}$

$R=1.87\ mCi$

Hence, (a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

4 0

3 years ago

Which velocity-time graph matches the position-time graph?

skelet666 [1.2K]

The answer is Graph C. To explain, this is because as we look at the position vs time graph, we see that after the first second, it was 30 meters from the start. That would mean that it took 1 second to get to 30 meters. That is shown in Graph c

7 0

3 years ago

Read 2 more answers