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3 years ago

Suppose an automobile has a kinetic energy

Physics

1 answer:

Maksim231197 [3]3 years ago

3 0

Answer:

$KE_{new}=186300 J$

Explanation:

Given:

$KE=2300J$ $n=9$

$v_{new}=n*v_{old}$

$\frac{KE_{new}}{KE_{old}}=\frac{v^2_{new}}{v^2_{old}}=\frac{(n*v_{old})^2}{v^2_{old}}=n^2$

$KE_{new}=n^2KE_{old}$

$KE_{new}=(9)^2(2300)$

$KE_{new}=186300 J$

Scientific notation

$KE_{new}= 1.863*10^5 J$

Hope it helps

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Estimate the boiling point of water at a pressure of 1.1 atmosphere

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3 years ago

suppose a car manufacturer tested its cars for front end collsion by hauling them up on a crane and dropping them from a certain

IRINA_888 [86]

Initial height: 66.5 m

Explanation:

The problem can be solved by using the principle of conservation of energy.

If we neglect air resistance, the total mechanical energy of the car is conserved during the fall, therefore we can write:

$K_i + U_i = K_f + U_f$

where :

$K_i = 0$ is the kinetic energy of the car at the top (it starts from rest)

$U_i = mgh$ is the gravitational potential energy of the car at the top, with:

m = the mass of the car

g = the acceleration of gravity

h = the heigth of the car

$K_f = \frac{1}{2}mv^2$ is the kinetic energy of the car just before hitting the ground, with

v = 130 km/h final speed of the car

$U_f = 0$ is the gravitational potential energy of the car at the bottom

Re-arranging the equation, we find

$mgh=\frac{1}{2}mv^2$

and we have:

$g=9.8 m/s^2\\v = 130 km/h = 36.1 m/s$

Solving for h, we find the initial height of the car:

$h=\frac{v^2}{2g}=\frac{36.1^2}{2(9.8)}=66.5 m$

Learn more about kinetic energy and potential energy:

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3 years ago

A planned high-speed train between Houston and Dallas will travel a distance of 386 kilometers in 5.40 × 10^3 seconds. What is t

Mazyrski [523]

¡Hellow!

For this problem, first, lets convert the seconds in hours:

5,4x10³ $\rightarrow$ 5400

h = sec / 3600

h = 5400 s / 3600

h = 1,5

Let's recabe information:

d (Distance) = 386 km

t (Time) = 1,5 h

v (Velocity) = ?

For calculate velocity, let's applicate formula:

$\boxed{\boxed{\textbf{d = v * t} } }$

Reeplace according we information:

386 km = v * 1,5 h

v = 386 km / 1,5 h

v = 257,33 km/h

The velocity of the train is of <u>257,33 kilometers for hour.</u>

<u></u>

Extra:

For convert km/h to m/s, we divide the velocity of km/h for 3,6:

m/s = km/h / 3,6

Let's reeplace:

m/s = 257,33 km/h / 3,6

m/s = 71,48

¿Good Luck?

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4 years ago

A 115 g hockey puck sent sliding over ice is stopped in 15.1 m by the frictional force on it from the ice.

Hoochie [10]

Answer:

(a) Ff = 0.128 N

(b μk = 0.1135

Explanation:

kinematic analysis

Because the hockey puck moves with uniformly accelerated movement we apply the following formulas:

vf=v₀+a*t Formula (1)

d= v₀t+ (1/2)*a*t² Formula (2)

Where:

d:displacement in meters (m)

t : time in seconds (s)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s

Calculation of the acceleration of the hockey puck

We apply the Formula (1)

vf=v₀+a*t v₀=5.8 m/s , vf=0

0=5.8+a*t

-5.8 = a*t

a= -5.8/t Equation (1)

We replace a= -5.8/t in the Formula (2)

d= v₀*t+ (1/2)*a*t² , d=15.1 m , v₀=5.8 m/s

15.1 = 5.8*t+ (1/2)*(-5.8/t)*t²

15.1= 5.8*t-2.9*t

15.1= 2.9*t

t = 15.1 / 2.9

t= 5.2 s

We replace t= 5.2 s in the equation (1)

a= -5.8/5.2

a= -1.115 m/s²

(a) Calculation of the frictional force (Ff)

We apply Newton's second law

∑F = m*a Formula (3)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Look at the free body diagram of the hockey puck in the attached graphic

∑Fx = m*a m= 115g * 10⁻³ Kg/g = 0.115g , a= -1.12 m/s²

-Ff = 0.115*(-1.115) We multiply by (-1 ) on both sides of the equation

Ff = 0.128 N

(b) Calculation of the coefficient of friction (μk)

N: Normal Force (N)

W=m*g= 0.115*9.8= 1.127 N : hockey puck Weight

g: acceleration due to gravity =9.8 m/s²

∑Fy = 0

N-W=0

N = W

N = 1.127 N

μk = Ff/N

μk = 0.128/1.127

μk = 0.1135

8 0

3 years ago