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Oksanka [162]
3 years ago
6

When Marie sees the red traffic light, Marie brakes to a halt from a speed of 70 m/s in just 2 seconds. What is her deceleration

?
Physics
1 answer:
SashulF [63]3 years ago
7 0

Answer:

-35 m/s^2

Explanation:

<u>Deceleration is the complete opposite of acceleration and is calculated by adding a negative sign to the formula for finding acceleration.</u>

Since acceleration is calculated by diving the change in velocity with time, deceleration will become:

 Deceleration =   -Δv/t, where Δv = change in velocity and t = time

In this case, Δv = 70 m/s and t = 2 s, hence;

Deceleration = -70/2 = -35 m/s^2

<em>The deceleration is 35 </em>m/s^2<em>.</em>

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Answer:

(1)p(x)\geq 0\\(2)\int_{0}^{\infty} p(x) dx=0

Explanation:

A function f(x) is a Probability Density Function if it satisfies the following conditions:

(1)f(x)\geq 0\\(2)\int_{0}^{\infty} f(x) dx=0

Given the function:

p(x)=\dfrac{1}{r}e^{-x/r} \: on\: [0,\infty), where\:r=\dfrac{20}{ln(2)}

(1)p(x) is greater than zero since the range of exponents of the Euler's number will lie in [0,\infty).

(2)

\int_{0}^{\infty} p(x)=\int_{0}^{\infty} \dfrac{1}{r}e^{-x/r}\\=\dfrac{1}{r} \int_{0}^{\infty} e^{-x/r}\\=-\dfrac{r}{r}\left[e^{-x/r}\right]_{0}^{\infty}\\=-\left[e^{-\infty/r}-e^{-0/r}\right]\\=-e^{-\infty}+e^{-0}\\SInce \: e^{-\infty} \rightarrow 0\\e^{-0}=1\\\int_{0}^{\infty} p(x)=1

The function p(x) satisfies the conditions for a probability density function.

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3 years ago
If a ball is thrown horizontally with a speed of 65 mph, how far will it fall while traveling 90 ft of horizontal distance?
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6. A girl pushes her little brother on his sled with a force of 300 N for 750 m.
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w=225000 J  

E=225000 J  

P=9000 W  

P=22500 W

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2 years ago
A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for
stealth61 [152]

Answer:

(a) a_s=0.62\frac{m}{s^2}

(b) a_s=0.13\frac{m}{s^2}

(c) x_f=2.6m

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}

(c) Using the kinematics equation:

x_f=x_0+v_0t \pm  \frac{at^2}{2}

For the girl, we have x_0=0 and v_0=0. So:

x_f_g=\frac{a_gt^2}{2}(1)

For the sled, we have v_0=0. So:

x_f_s=x_0_s-\frac{a_st^2}{2}(2)

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s

Now, we solve (1) for x_f_g

x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m

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