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AnnyKZ [126]
3 years ago
13

An engineer in a locomotive sees a car stuck on the track at a railroad crossing in front of the train . When the engineer first

sees the car, the locomotive is 110 m from the crossing and its speed is 21 m/s . If the engineer's reaction time is 0.6 s, what should be the magnitude of the minimum de celeration to avoid an accident ? Answer in units of m/s^ 2
Physics
1 answer:
8_murik_8 [283]3 years ago
4 0

Answer:

Explanation:

distance traveled during reaction time

d = vt = 21(0.6) = 12.6 m

distance remaining 110 - 12.6 = 97.4 m

Assuming we want zero speed in 97.4 m using constant acceleration.

a = (0² - 21²) / 2(97.4) = - 2.2638603...

the negative value indicates acceleration opposes initial velocity.

magnitude is

a = 2.3 m/s²

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The Storm Surge of a hurricane

Explanation:

A hurricane can ultimately cause a tsunami and is just as deadly.

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What 2 layers have temperature increasing with elevation?
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Stratosphere and Thermosphere

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Read 2 more answers
Two fish swimming in a river have the following equations of motion:
IRINA_888 [86]

Answer:

The second fish, X2, is moving faster than the first fish, X1

Explanation:

The given parameters for the equation of motion of the fishes are;

X1 = -6.4 m + (-1.2 m/s)×t

X2 = 1.3 m + (-2.7 m/s)×t

The given equation are straight line equations in the slope and intercept form, where the slope is the speed and in m/s and the intercept is the starting point of swimming of the fishes

For the first fish, the intercept = -6.4 m, the slope = the  speed = -1.2 m/s

For the second fish, the intercept = 1.3 m, the slope = the  speed = -2.7 m/s

Whereby the fishes are swimming in the opposite direction of the measurement of length, we have;

The magnitude of the speed of the second fish \left | -2.7 \ m/s \right | = 2.7 \ m/s, is larger than the magnitude of the speed of the first fish \left | -1.2 \ m/s \right | = 1.2 \ m/s

Therefore, the second fish, X2, is moving faster than the first fish, X1.

8 0
3 years ago
Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m
netineya [11]

Answer: m \frac{d}{dt}v_{(t)}

Explanation:

In the image  attached with this answer are shown the given options from which only one is correct.

The correct expression is:

m \frac{d}{dt}v_{(t)}

Because, if we derive velocity v_{t} with respect to time t we will have acceleration a, hence:

m \frac{d}{dt}v_{(t)}=m.a

Where m is the mass with units of kilograms (kg) and a with units of meter per square seconds \frac{m}{s}^{2}, having as a result kg\frac{m}{s}^{2}

The other expressions are incorrect, let’s prove it:

\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)} This result has units of kg\frac{m}{s}

m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m This result has units of kg

m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C This result has units of kgm^{2} and C is a constant

m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m This result has units of kg

m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m This result has units of kg

\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C This result has units of kg \frac{m^{3}}{s^{3}} and C is a constant

m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{4}} and C is a constant

\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0 because v_{(x)} is a constant in this derivation respect to t

m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{2}} and C is a constant

6 0
3 years ago
An experimental analysis of the natural oscillations in a particular structure shows the dominant frequencies of interest appear
Irina18 [472]

Answer:

In order to prevent the aliased frequency from we need to sample at  with a sampling frequency that is minimum 2 times the highest frequency where information exit i.e 2 × 750 = 1500 Hz

Explanation:

The explanation is shown on the first and second uploaded image

6 0
3 years ago
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