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tigry1 [53]
3 years ago
10

A car initially going 61 ft/sec brakes at a constant rate (constant negative acceleration), coming to a stop in 7 seconds.

Physics
1 answer:
Bond [772]3 years ago
4 0

Answer:

See the attachment below for the graphics in part (a)

The initial velocity for this time interval is u = 61ft/sec and the final velocity is 0m/s because the car comes to a stop.

This a constant acceleration motion considering the given time interview over which the brakes are applied. So the equals for constant acceleration motion apply here.

Explanation:

The full solution can be found in the attachment below.

Thank you for reading. I hope this post is helpful to you.

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Can someone help??
Paul [167]

Answer:

F = 2520 N

Explanation:

We have,

The maximum acceleration of a fist in a karate punch is 4200 m/s². The mass of the fist is 0.6 kg.

It is required to find the force that the wood place on the fist. Force is given by the product of mass and its acceleration such that,

F = ma

F=0.6\times 4200\\\\F=2520\ N

So, the force of 2520 N is acting on the wood.

8 0
3 years ago
Starting from rest, a 2-m-long pendulum swings from an angleof
Andrews [41]

Answer:

D.) 1m/s

Explanation:

Assume the initial angle of the swing is 12.8 degree with respect to the vertical. We can calculate the vertical distance from this initial point to the lowest point by first calculate the vertical distance from this point the the pivot point:

L_1 = L*cos(12.8) = 2*0.975 = 1.95 m

where L is the pendulum length

The vertical distance from the lowest point to the pivot point L_2 is the pendulum length 2m

this means the vertical distance from this initial point to the lowest point is simply:

L_3 = L_2 - L_1 = 2 - 1.95 = 0.05 m

As the pendulum travel (vertically) from the initial point to the bottom point, its potential energy is converted to kinetic energy:

E_p = E_k

mgh = mv^2/2

where m is the mass of the pendulum, g  = 10 m/s2 is the constant gravitational acceleration, h = 0.05 is the vertical it travels, v is the pendulum velocity at the bottom, which we are trying to solve for.

The m on both sides of the equation cancel out

v^2 = 2gh = 2*10*0.05 = 1

v = \sqrt{1} = 1 m/s

so D is the correct answer

5 0
3 years ago
Is an atom with one valence electron more reactive than an atom with two electrone? ​
andrey2020 [161]

Answer:

An atom with a closed shell of valence electrons (corresponding to an electron configuration s2p6) tends to be chemically inert. An atom with one or two valence electrons more than a closed shell is highly reactive, because the extra valence electrons are easily removed to form a positive ion.Explanation:

7 0
3 years ago
X rays of wavelength 0.0169 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr
mamaluj [8]

Answer:

a) 4.04*10^-12m

b) 0.0209nm

c) 0.253MeV

Explanation:

The formula for Compton's scattering is given by:

\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)

where h is the Planck's constant, m is the mass of the electron and c is the speed of light.

a) by replacing in the formula you obtain the Compton shift:

\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m

b) The change in photon energy is given by:

\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm

c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.

P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\

E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV

5 0
3 years ago
On a popular amusement park ride, the riders stand along the inside wall of a large cylinder with a diameter of 8.27 m. The cyli
____ [38]

Answer:

Explanation:

D = 8.27 m   ⇒  R = D / 2 = 8.27 m / 2 = 4.135 m

ω = 0.66 rev/sec = (0.66 rev/sec)*(2π rad/1 rev) = 4.1469 rad/s

We can apply the equation

Ff = W  ⇒  μ*N = m*g   <em>(I)</em>

then we have

N = Fc = m*ac = m*(ω²*R)

Returning to the equation <em>I</em>

<em />

μ*N = m*g    ⇒    μ*m*ω²*R = m*g   ⇒ μ = g / (ω²*R)

Finally

μ = (9.81 m/s²) / ((4.1469 rad/s)²*4.135 m) = 0.1379

3 0
3 years ago
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