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gtnhenbr [62]
2 years ago
10

A car having a mass of 500 kg is initially at rest. A constant 1,000 N net force acts on the car over a distance of 50 m, causin

g the car to speed up. After it travels 50 m, the car moves with constant velocity.
a.) What is the total work done on the car over the 50 m distance it travels while speeding up?
b.) How fast is the car moving after 50 m?
c.) What is the net force on the car while its moving with constant velocity?
d.) What is the total work done on the car while its moving with constant velocity?
Physics
1 answer:
nirvana33 [79]2 years ago
6 0

Answer:

(a) 50 kJ

(b) 14.14 m/s

(c) 0 N

(d) 0 J

Solution:

As per the question:

Mass of car, m = 500 kg

Force, F = 1000 N

Distance traveled, s = 50 m

Now,

(a) The total work done can be given as:

W = F\times s = 1000\times 50 = 50000\ J = 50\ kJ

(b) To calculate the speed of the car:

Using work-energy relation, we can say that the work done equals the change in the kinetic energy of the car:

W = \Delta KE

\frac{1}{2}m(v'^{2} - v^{2}) = 50000

where

v' = final velocity

v = initial velocity = 0

Now,

\frac{1}{2}\times 500v'^{2} = 50000

v' = \sqrt{200} = 14.14\ m/s

(c) To calculate the net force on the car when the velocity is constant:

The net force is given as the product of mass and acceleration:

F = ma

<em>Acceleration can be defined as the rate with which the velocity of the car changes and the since the velocity is constant, the car will not have any acceleration, a = 0</em>

Thus the net force is also zero.

F_{net} = 500\times 0 = 0\ N

(d) The total work done when the car moves with constant velocity:

W = F_{net}.s = 0\ J

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Find the kinetic energy of an electron whose de broglie wavelength is 34.0 nm.
dezoksy [38]
The De Broglie wavelength of the electron is
\lambda=34.0 nm=34 \cdot 10^{-9} m
And we can use De Broglie's relationship to find its momentum:
p= \frac{h}{\lambda}= \frac{6.6 \cdot 10^{-34} Js}{34 \cdot 10^{-9} m}=1.94 \cdot 10^{-26} kg m/s

Given p=mv, with m being the electron mass and v its velocity, we can find the electron's velocity:
v= \frac{p}{m}= \frac{1.94 \cdot 10^{-26} kgm/s}{9.1 \cdot 10^{-31} kg}=  2.13 \cdot 10^4 m/s

This velocity is quite small compared to the speed of light, so the electron is non-relativistic and we can find its kinetic energy by using the non-relativistic formula:
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3 years ago
A golfer is on the edge of a 12.5 m bluff overlooking the 18th hole which is located 60 m from the base of the bluff. She launch
Lina20 [59]

Answer:

The ball impact velocity i.e(velocity right before landing) is 6.359 m/s

Explanation:

This problem is related to parabolic motion and can be solved by the following equations:

x=V_{o}cos \theta t----------------------(1)

y=y_{o}+V_{o} sin \theta t - \frac{1}{2}gt^{2}---------(2)

V=V_{o}-gt ----------------------- (3)

Where:

x = m is the horizontal distance travelled by the golf ball

V_{o} is the golf ball's initial velocity

\theta=0\° is the angle (it was  a horizontal shot)

t is the time

y is the final height of the ball

y_{o} is the initial height of the ball

g is the acceleration due gravity

V is the final velocity of the ball

Step 1: finding t

Let use the equation(2)

t=\sqrt{\frac{2 y_{o}}{g}}

t=\sqrt{\frac{2 (12.5 m)}{9.8 m/s^{2}}}

t=1.597s

Substituting (6) in (1):

67.1 =V_{o} cos(0\°) 1.597-------------------(4)

Step 2:  Finding V_{o}:

From equation(4)

67.1 =V_{o}(1) 1.597

V_0 = \frac{6.71}{1.597}

V_{o}=42.01 m/s (8)  

Substituting V_{o} in (3):

V=42.01 -(9.8)(1.597)

v =42 .01 - 15.3566  

V=26.359 m/s

5 0
2 years ago
The central star of a planetary nebula emits ultraviolet light with wavelength 104nm. This light passes through a diffraction gr
Gala2k [10]

Answer: 31.33 degrees

Explanation:

The diffraction angles \theta_{n} when we have a slit divided into n parts are obtained by  the following equation:

dsin\theta_{n}=n\lambda   (1)

Where:

d is the width of the slit

\lambda  is the wavelength of the light

n is an integer different from zero.

Now, the first-order diffraction angle is given when n=1, hence equation (1) becomes:

dsin\theta_{1}=\lambda   (2)

Now we have to find the value of \theta_{1}:

sin\theta_{1}=\frac{\lambda}{d}  

\theta_{1}=arcsin(\frac{\lambda}{d})   (3)

We know:

\lambda=104nm=104(10)^{-9}m

In addition we are told the diffraction grating has 5000 slits per mm, this means:

d=\frac{1mm}{5000}=\frac{1(10)^{-3}m}{5000}

Substituting the known values in (3):

\theta_{1}=arcsin(\frac{104(10)^{-9}m}{\frac{1(10)^{-3}m}{5000}})

\theta_{1}=arcsin(0.52)

<u>Finally:</u>

\theta_{1}=31.33\º >>>This is the first-order diffraction angle

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Answer:

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