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3 years ago

A car speeds upu as it rolls down a hill. Which is this an example of?

Physics

2 answers:

olga2289 [7]3 years ago

8 0

<span>positive acceleration</span>

anygoal [31]3 years ago

4 0

Positive acceleration

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What is polarization?

worty [1.4K]

Answer:

if your glasses are polarized, you can see the fish in the water. also im pretty sure its d

Explanation:

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3 years ago

The amp is the unit for _________.

adell [148]

B.) <span>The amp is the unit for "Current"

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7 0

3 years ago

Read 2 more answers

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

An automobile engine delivers 47.4 hp. how much time will it take for the engine to do 6.82 × 105 j of work? one horsepower is e

kogti [31]

1 horsepower is equal to 746 W, so the power of the engine is
$P=47.4 hp \cdot 746 \frac{W}{hp}=35360 W$
The power is also defined as the energy E per unit of time t:
$P= \frac{E}{t}$
Where the energy corresponds to the work done by the engine, which is $E=6.82 \cdot 10^5 J$ . Re-arranging the formula, we can calculate the time t needed to do this amount of work:
$t = \frac{E}{P}= \frac{6.82 \cdot 10^5 J}{35360 W}=19.3 s$

8 0

3 years ago

how much power is needed to lift a box with a force of 780 newtons over a distance of 2 meters in 45 seconds

-BARSIC- [3]

Answer:

Explanation:

Power is the rate at which work is done and can be found by using the formula

$p = \frac{w}{t} \\$

p is the power in Watts (W)

w is the workdone in joules

t is time in s

but workdone = force × distance

From the question

force = 780 N

distance = 2 m

workdone = 780 × 2 = 1560 N

Since we now have the value of workdone we can find the power

We have

$p = \frac{1560}{45} = 34.6666666... \\$

We have the final answer as

Hope this helps you

3 0

2 years ago