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3 years ago

A 1200kg car moving at 25m/s has the brakes applied with a deceeration of 8.0m/s^2. How far does the car travel before it stops?

1 answer:

5 0

V=vnaut+at 0(want it to stop)=25+-8t(negative accel cuz its deceling)
-25/-8=t t=3.125 seconds or with sig figs 3.1 seconds
x=xnaut+vnaut*time+1/2*a*time^2
x(where it ends up)=0(we start distance from when it started to brake)+25*3.1+1/2*-8*3.1^2=39.06 meters or 39.1 meters
x=0+25*3.1-.5*8*3.1^2 is the same equation as above with nothing written in. Are you in physics ap or reg? either way friend me lol(just joined website tonight)

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An aluminum bar 600mm long, with diameter 40mm long has a hole drilled in the center of the bar.The hole is 30mm in diameter and

Svetradugi [14.3K]

Answer:

Total contraction on the Bar = 1.22786 mm

Explanation:

Given that:

Total Length for aluminum bar = 600 mm

Diameter for aluminum bar = 40 mm

Hole diameter = 30 mm

Hole length = 100 mm

elasticity for the aluminum is 85GN/m² = 85 × 10³ N/mm²

compressive load P = 180 KN = 180 × 10³ N

Calculate the total contraction on the bar = ???

The relation used in calculating the contraction on the bar is:

$\delta L = \dfrac{P *L }{A*E}$

The relation used in calculating the total contraction on the bar can be expressed as :

Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)

i.e

Total contraction on the Bar = $\dfrac{P *L_1 }{A_1*E} + \dfrac{P *L_2 }{A_2 *E}$

Let's find the area of cross section without the hole and with the hole

Area of cross section without the hole is :

Using A = πd²/4

A = π (40)²/4

A = 1256.64 mm²

Area of cross section with the hole is :

A = π (40²-30²)/4

A = 549.78 mm²

Total contraction on the Bar = $\dfrac{P *L_1 }{A_1*E} + \dfrac{P *L_2 }{A_2 *E}$

Total contraction on the Bar = $\dfrac{180 *10^3 \N }{85*10^3 \ N/mm^2} [\dfrac{500}{1256.64}+ \dfrac{100}{549.78}]$

Total contraction on the Bar = 2.117( 0.398 + 0.182)

Total contraction on the Bar = 2.117*(0.58)

Total contraction on the Bar = 1.22786 mm

5 0

3 years ago

A motorbike reaches a speed of 20 m/s over 60m, whilst

Fynjy0 [20]

Initial speed = 2√10 m/s

<h3>Further explanation </h3>

Linear motion consists of 2: constant velocity motion with constant velocity and uniformly accelerated motion with constant acceleration

An equation of uniformly accelerated motion

V = vo + at

Vt² = vo² + 2a (x-xo)

x = distance on t

vo / vi = initial speed

vt / vf = speed on t / final speed

a = acceleration

vf=20 m/s

d = 60 m

a = 3 m/s²

$\tt vf^2=vi^2+2.ad\\\\20^2=vi^2+2\times 3\times 60\\\\400=vi^2+360\\\\40=vi^2\\\\vi=\sqrt{40}=2\sqrt{10}~m/s$

7 0

3 years ago

Some neodymium glass lasers can provide 100TW of power in 1.0 ns pulses at a wavelength of 0.26 micrometers. how much energy is

Aleksandr-060686 [28]

Answer:

E = 10⁵ J

Explanation:

given,

Power, P = 100 TW

= 100 x 10¹² W

time, t = 1 ns

= 1 x 10⁻⁹ s

The energy of a single pulse is:-

Energy = Power x time

E = P t

E = 100 x 10¹² x 1 x 10⁻⁹

E = 10⁵ J

The energy contained in a single pulse is equal to 10⁵ J

7 0

3 years ago

The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio

Sveta_85 [38]

Answer:

temperature on left side is 1.48 times the temperature on right

Explanation:

GIVEN DATA:

$\gamma = 5/3$

T1 = 525 K

T2 = 275 K

We know that

$P_1 = \frac{nRT_1}{v}$

$P_2 = \frac{nrT_2}{v}$

n and v remain same at both side. so we have

$\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}$

$P_1 = \frac{21}{11} P_2$ ..............1

let final pressure is P and temp $T_1 {f} and T_2 {f}$

$P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}$

$P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3}$ ..................2

similarly

$P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3}$ .............3

divide 2 equation by 3rd equation

$\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}$

$T_1 {f} = 1.48 T_2 {f}$

thus, temperature on left side is 1.48 times the temperature on right

6 0

3 years ago

A river flows due south at 5 mi/h. A swimmer attempting to cross the river heads due east swimming at 3 mi/h relative to the wat

Dafna11 [192]

Answer:

<u>velocity of swimmer relative to ground = 3 i -5 j</u>

Explanation:

To cross a river the swimmer swims relative to river in perpendicular direction.

Velocity of river = -5 j (south)

Velocity of swimmer relative to river = 3 i(north)

<h2>Velocity of swimmer relative to ground = Velocity of swimmer relative to river + Velocity of river</h2>

Velocity of swimmer relative to ground = 3 i -5 j

So magnitude of total velocity is $\sqrt{3^2+(-5)^2}$ = $\sqrt{9+25}$ = $\sqrt{34}$

3 0

3 years ago