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3 years ago

If the car goes exits a freeway and goes from 65 mph to 35 mph is it accelerating?

Physics

1 answer:

Zolol [24]3 years ago

8 0

Answer:

No, the car is decelerating

Explanation:

No the car is decelerating if it exits a freeway and goes from 65

mph to 35 mph since the change in velocity is negative.

change in velocity = final - initial

change in velocity = 35 - 65

change in velocity = -30mph

Since the change in velocity is negative, hence the car is decelerating. Deceleration is a negative acceleration

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When you must add three vectors together, what is not true this process? You must only give a magnitude of the resultant vector

vovangra [49]

The addition of any numbers of vector provide the magnitude as well as the direction of the resultant vector, hence the mentioned first option is not true.

The addition of vector required to connect the head of the one vector with the tail of the other vector and any vector can be moved in the plane parallet to the previous location, so, the mentioned second and third options are true.

4 0

3 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

What are the four system of measurement

Rudiy27

Answer:

ask google

Explanation:

5 0

3 years ago

Explain why not everyone can become a bodybuilder, even if they train hard enough.

Anni [7]

People have diffrent body builds and bone structure

5 0

4 years ago

25% part (c) assume that d is the distance the cheetah is away from the gazelle when it reaches full speed. Derive an expression

levacccp [35]

maximum speed of cheetah is

$v_1 = v_{max}$

speed of gazelle is given as

$v_2 = v_{g}$

Now the relative speed of Cheetah with respect to Gazelle

$v_{12} = v_1 - v_2$

$v_{12} = v_{max} - v_g$

now the relative distance between Cheetah and Gazelle is given initially as "d"

now the time taken by Cheetah to catch the Gazelle is given as

$d = v_{12}* t$

so by rearranging the terms we can say

$t = \frac{d}{v_{12}}$

$t = \frac{d}{v_{max} - v_g}$

so above is the relation between all given variable

6 0

3 years ago