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3 years ago

What distance does a car travel as its speed changes from 0 to 20 m/s in 17 s at constant acceleration?

Physics

1 answer:

love history [14]3 years ago

4 0

Since acceleration is constant

So the distance moved in constant acceleration will be given as

$d = \frac{v_f + v_i}{2} t$

as we know that

$v_f = 20 m/s$

$v_i = 0$

t = 17 s

so distance moved is given by

$d = \frac{0 + 20}{2}(17)$

$d = 170 m$

so distance moved will be 170 m

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An elevator together with its passengers weights 5000 N. At a certain instant, the tension in its supporting cable is 6000 N. De

gayaneshka [121]

We have to forces acting on the system (elevator+passengers):
1) The weight (W=5000 N), acting downward
2) The cable's tension (T=6000 N), acting upward
So, the two forces have opposite direction. The resultant (in upward direction) will be
$F=T-W$
And for Newton's second law, the resultant of the forces acting on the system causes an acceleration on the system itself, given by
$a= \frac{F}{m}$
where m is the mass of the system.

So, we need to find F and m.
The resultant of the forces is
$F=T-W=6000 N-5000 N=1000 N$
To find m, we can use the weight of the system. In fact, the weight of an object is given by
$W=mg$
where $g=9.81 m/s^2$ . Solving for m, and using W=5000 N, we find
$m= \frac{W}{g}= \frac{5000 N}{9.81 m/s^2}=510 kg$

and at this point, we can calculate the acceleration of the system (elevator+people):
$a= \frac{F}{m}= \frac{1000 N}{510 kg}=1.96 m/s^2$
and the acceleration has the same direction of the resultant force, so upward.

8 0

3 years ago

In the periodic table, similar elements are arranged in vertical columns called

tester [92]

Answer:

The vertical columns on the periodic table are called groups or families because of their similar chemical behavior.

Explanation:

3 0

3 years ago

Read 2 more answers

An unknown radioactive sample is observed to decrease in activity by a factor of two in a one hour period. What is its half-life

elena55 [62]

Answer:

The half-life is $t_{1/2} = 1.005 h$

Explanation:

Using the decay equation we have:

$A=A_{0}e^{-\lambda t}$

Where:

λ is the decay constant
A(0) the initial activity
A is the activity at time t

We know the activity decrease by a factor of two in a one hour period (t = 1 h), it means that $A = \frac{A_{0}}{2}$

$\frac{A_{0}}{2}=A_{0}e^{-\lambda*1 h}$

$0.5=e^{-\lambda*1 h}$

Taking the natural logarithm on each side we have:

$ln(0.5)=-\lambda$

$\lambda=0.69 h^{-1}$

Now, the relationship between the decay constant λ and the half-life t(1/2) is:

$\lambda = \frac{ln(2)}{t_{1/2}}$

$t_{1/2} = \frac{ln(2)}{\lambda}$

$t_{1/2} = \frac{ln(2)}{0.69}$

$t_{1/2} = 1.005 h$

I hope it helps you!

6 0

3 years ago

A gas station owner suspects that he is being overcharged for gasoline deliveries by a gasoline supplier. The overcharge seems p

Tems11 [23]

Answer:

Explanation:

delta V = v * alpha * delta T

= V * 0.00053 * (92.2 - 55.0)

= 0.019716 V

percentage that the owner

= [delta V / V] * 100

= [0.019716 V / V] * 100

= 1.9716 %

4 0

2 years ago

Suppose a car travels 108 km at a speed of 30.0 m/s, and uses 1.90 gallons of gasoline. Only 30% of the gasoline goes into usefu

Aleksandr [31]

Answer:

686.11 N

1.7733 gallons

Explanation:

$\eta$ = Efficiency = 30%

V = Volume of gasoline

E = Energy content of gasoline = $1.3\times 10^8\ J/gal$

F = Force

s = Displacement = 108000 m

v = Velocity

Work done is given by

$W=F\times s\\\Rightarrow F=\frac{W}{s}\\\Rightarrow F=\frac{VE\eta}{s}\\\Rightarrow F=\frac{1.9\times 0.3\times 1.3\times 10^8}{108000}\\\Rightarrow F=686.11\ N$

The force required to keep the car moving at a constant speed is 686.11 N

Here the force is directly proportional to speed

$\\\Rightarrow F=v$

$\\\Rightarrow \frac{F_1}{v_1}=\frac{F_2}{v_2}\\\Rightarrow F_2=\frac{F_1\times v_2}{v_1}\\\Rightarrow F_2=\frac{686.11\times 28}{30}\\\Rightarrow F_2=640.36\ N$

$W=F\times s\\\Rightarrow 0.3\times 1.3\times 10^8\times V=640.36\times 108000\\\Rightarrow V=\frac{640.36\times 108000}{0.3\times 1.3\times 10^8}\\\Rightarrow V=1.7733\ gal$

The gallons that will be used is 1.7733 gallons

7 0

3 years ago