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4 years ago

Explain how you know the car is accelerating when it reaches a curve on the road

2 answers:

6 0

Acceleration is a Vector, so speed and direction. If you change your direction you are also changing your acceleration because it's a vector.

ankoles [38]4 years ago

4 0

Answer:

When car reaches to the end of the curve then due to change in the direction of the motion the car its velocity is changing and hence the motion of the car is accelerated motion.

Explanation:

As we know that the acceleration of the car is defined as rate of change in velocity

$a = \frac{dv}{dt}$

here we know that velocity is a vector quantity and the change in velocity is defined in two ways

1) change in magnitude

2) change in direction

So here when car enters the curved path then due to change in the direction of the motion of the car its velocity will change and hence we can say that its motion is accelerated motion

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After your school's team wins the regional championship, students go to the dorm roof and start setting off fireworks rockets. T

oksian1 [2.3K]

Answer:

required distance is 233.35 m

Explanation:

Given the data in the question;

Sound intensity $I$ = 1.62 × 10⁻⁶ W/m²

distance r = 165 m

at what distance from the explosion is the sound intensity half this value?

we know that;

Sound intensity $I$ is proportional to 1/(distance)²

i.e

$I$ ∝ 1/r²

Now, let r² be the distance where sound intensity is half, i.e $I$ ₂ = $I$ ₁/2

Hence,

$I$ ₂/ $I$ ₁ = r₁²/r₂²

1/2 = (165)²/ r₂²

r₂² = 2 × (165)²

r₂² = 2 × 27225

r₂² = 54450

r₂ = √54450

r₂ = 233.35 m

Therefore, required distance is 233.35 m

6 0

3 years ago

A state patrol officer saw a car start from rest at a highway on-ramp. She radioed ahead to another officer 20 mi along the hig

Vikentia [17]

Answer:

We know from the basic speed distance relation that

$Speed=\frac{Distance}{Time}$

Since the car started from rest and it covered the distance between the 2 officer's in 19 minutes we have speed of the car

$Speed=\frac{Distance}{Time}\\\\Speed=\frac{20}{\frac{19}{60}}=63.16mph$

Which clearly exceeds the limit of $60\frac{mi}{hr}$

5 0

3 years ago

3. Neha travels 4 m towards the north, 3m towards the east, and then 7 m towards south, find the displacement and total distance

Digiron [165]

Answer:

Explanation:

Explanation: total displacement =3√2m. and total distance covered=14m. I hope this is right and helps u.

7 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

Why do nonmetals have a negative oxidation number

galina1969 [7]

Oxygen has a higher electro negativity that then Sulfur, so Sulfur will " lose" electrons to Oxygen and that is the electrons will be pulled closer to the Oxygen causing, for oxygen to have a negative charge and the Sulfur to have a positive charge

8 0

4 years ago