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2 years ago

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Newtons first law stateS that object will move With a constant velocity if nothing acts on it. Does our every day experience con

tradict one of newtons law ? Explain

1 answer:

bazaltina [42]2 years ago

4 0

Answer:

hi

Explanation:

hi

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a bus is moving with the velociity of 36 km/hr . after seeing a boy at 20 m ahead on the road, the driver applies the brake and

Klio2033 [76]

Answer:

Assumption: the acceleration of this bus is constant while the brake was applied.

Acceleration of this bus: approximately $\left(-6.0\; \rm m \cdot s^{-2}\right)$ .

It took the bus approximately $1.7\;\rm s$ to come to a stop.

Explanation:

Quantities:

Displacement of the bus: $x = 10\; \rm m$ .
Initial velocity of the bus: $\displaystyle u = 36\; \rm km \cdot hr^{-1} = 36\; \rm km \cdot hr^{-1}\times \frac{1\; \rm m \cdot s^{-1}}{3.6\; \rm km\cdot hr^{-1}} = 10\; \rm m \cdot s^{-1}$ .
Final velocity of the bus: $v = 0\; \rm m\cdot s^{-1}$ because the bus has come to a stop.
Acceleration, $a$ : unknown, but assumed to be a constant.
Time taken, $t$ : unknown.

Consider the following SUVAT equation:

$\displaystyle x = \frac{1}{2}\, \left(a\, t^2\right) + u\, t$ .

On the other hand, assume that the acceleration of this bus is indeed constant. Given the initial and final velocity, the time it took for the bus to stop would be inversely proportional to the acceleration of this bus. That is:

$\displaystyle t = \frac{v - u}{a}$ .

Therefore, replace the quantity $t$ with the expression $\displaystyle \left(\frac{v - u}{a}\right)$ in that SUVAT equation:

$\displaystyle x = \frac{1}{2}\, \left(a\, \left(\frac{v -u}{a}\right)^2\right) + u\, \left(\frac{v - u}{a}\right)$ .

Simplify this equation:

$\begin{aligned}x &= \frac{1}{2}\, \left(a\, {\left(\frac{v -u}{a}\right)}^2\right) + u\, \left(\frac{v - u}{a}\right) \\ &= \frac{1}{2}\left(\frac{{(v - u)}^2}{a}\right) + \frac{u\, (v - u)}{a} =\frac{1}{a}\, \left(\frac{{(v - u)}^2}{2} + u\, (v - u)\right)\end{aligned}$ .

Therefore, $\displaystyle a= \frac{1}{x}\, \left(\frac{{(v - u)}^2}{2} + u\, (v - u)\right)$ .

In this question, the value of $x$ , $u$ , and $v$ are already known:

$x = 10\; \rm m$ .
$\displaystyle u =10\; \rm m \cdot s^{-1}$ .
$v = 0\; \rm m\cdot s^{-1}$ .

Substitute these quantities into this equation to find the value of $a$ :

$\begin{aligned} a &= \frac{1}{x}\, \left(\frac{{(v - u)}^2}{2} + u\, (v - u)\right) \\ &= \frac{1}{10\; \rm m}\times \left(\frac{{\left(0\; \rm m \cdot s^{-1} - 10\; \rm m \cdot s^{-1}\right)}^2}{2} + \left(0\; \rm m \cdot s^{-1} - 10\; \rm m \cdot s^{-1}\right)\times 10\; \rm m \cdot s^{-1}\right)\\ &\approx -6.0\; \rm m \cdot s^{-2}\end{aligned}$ .

(The value of acceleration $a$ is less than zero because the velocity of the bus was getting smaller.)

Substitute $a \approx -6.0\; \rm m \cdot s^{-2}$ (alongside $u = 10\; \rm m \cdot s^{-1}$ and $v = 0\; \rm m \cdot s^{-1}$ ) to estimate the time required for the bus to come to a stop:

$\begin{aligned}t &= \frac{v - u}{a} \\ &\approx \frac{0\; \rm m \cdot s^{-1} - 10\; \rm m \cdot s^{-1}}{-6.0\; \rm m \cdot s^{-2}} \approx 1.7\; \rm s\end{aligned}$ .

8 0

3 years ago

A particle travels along the x-axis in such a way that its acceleration at time t is a(t) = t + t2. if it starts at the origin w

Olegator [25]

The acceleration of the particle as a function of time t is
$a(t) = t + t^2$
The velocity of the particle at time t is the integral of the acceleration:
$v(t) = \int {a(t)} \, dt = \frac{t^2}{2} + \frac{t^3}{3} + C$
where the constant C can be found by requiring that the velocity at time t=0 is v=3:
$v(0) = 3$
and we find $C=v_0=3$
so the velocity is
$v(t)=3+ \frac{t^2}{2}+ \frac{t^3}{3}$

The position of the particle at time t is the integral of the velocity:
$x(t)=\int {v(t) } \,dt = 3t + \frac{t^3}{6}+ \frac{t^4}{12} +D$
where D can be found by requiring that the initial position at time t=0 is zero:
x(0)=0
from which we find D=0, so
$x(t)=3t + \frac{t^3}{6}+ \frac{t^4}{12}$

To solve the problem, now we just have to substitute t=5 into x(t) and v(t) to find the position and the velocity of the particle at t=5.

The position is:
$x(5)=3(5) + \frac{5^3}{6}+ \frac{5^4}{12}=87.92$
and the velocity is:
$v(5) = 3+ \frac{5^2}{2}+ \frac{5^3}{3}=57.17$

6 0

3 years ago

Two 5000-kg passenger cars roll without friction (one at 1 m/s, the other at 2 m/s) toward one another on a level track. They co

balu736 [363]

The combined momentum of the passengers is 5000 kgm/s.

<h3>Combined momentum of the passenger</h3>

The combined momentum of the passengers is calculated as follows;

P = mv1 + mv2

where;

m is mass of the passengers
v1 is velocity of the first passenger
v2 is velocity of the second passenger

P = m(v1 + v2)

P = 5000(-1 + 2)

P = 5000 kgm/s

Thus, the combined momentum of the passengers is 5000 kgm/s.

Learn more about momentum here: brainly.com/question/7538238

#SPJ1

5 0

2 years ago

What is 18 degrees celsius in fahrenheit ?

Butoxors [25]

18 degree is equal to 64.4 fahrenheit

4 0

3 years ago

How much force must be applied on a blade of length 4cm and thickness of 0.1mm to exert a pressure of 4000000pa?

Viktor [21]

Answer:

F= 403429 kpa

Explanation:

Pressure is the product of force and area

Mathematically,

P=F*A -------where F is force and A is area.

A= 40 *0.1 = 4mm² -----convert to m²

A= 4e⁻⁶ m²

P= 4000000 pa

F= P/A = 4000000/4e⁻⁶

F= 403428793.493 pa

F= 403429 kpa

7 0

3 years ago