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enot [183]
3 years ago
13

How do you determine the speed or velocity of an object using a distance vs. time graph

Physics
1 answer:
Taya2010 [7]3 years ago
4 0
The speed of an object can be determined from the distance vs time graph.

You know that speed = distance/time

in the graph, distance/time = slope of the curve.

So SPEED IS GIVEN BY THE SLOPE of the curve in the graph.

● If the distance vs time curve is a straight line, parallel to time axis(x-axis), slope is 0. That means speed is 0. So the object is at rest.

● If the distance vs time curve is a straight line, with some non-zero slope; That means speed is nonzero and constant. So the object is in uniform motion.

● If the distance vs time curve is a curved, the slope is changing. That means speed is changing. So the object is in an accelerated motion.
You might be interested in
Determine the weight in newtons of a woman whose weight in pounds is 157. Also, find her mass in slugs and in kilograms. Determi
DerKrebs [107]

Answer:

Weight of the woman in Newton

   x =  698.6 \  N

 Mass of the woman in slug

  Mass =  4.86 \  slug

 Mass of the woman in kg

   Mass =  16 \  kg

  My weight in Newton

  W =  784 \  N

Explanation:

From the question we are told that

   The weight of the woman in pounds is  W = 157 \ lb

Converting to Newton

    1 N  =  0.22472 lb

    x N  =  157

=>   x =  \frac{157 *  1}{0.22472}

=>   x =  698.6 \  N

Obtaining the mass in slug

   Mass =  \frac{W}{g}

Here  g =  32.2 ft/s^2

So

     Mass =  \frac{157 }{32.2}

       Mass =  4.86 \  slug

Obtaining the mass in kilogram

     Mass =  \frac{W}{g}

Here  g =  9.8 \  m/s

So

   Mass =  \frac{157 }{9.8}

   Mass =  16 \  kg

Generally weight is mathematically represented as

     W =  m *  g

Given that my mass is   80 kg   then my weight is  

     W =   80 *9.8

      W =  784 \  N

6 0
3 years ago
A vertical spring has a length of 0.25 m when a 0.175 kg mass hangs from it, and a length of 0.775 m when a 2.075 kg mass hangs
Contact [7]

Answer:

A) 35.5N/m b) 20.1cm

Explanation:

Using Hooke's law;

F = Ke where F is the weight of the object = mass of the object in kg * acceleration due to gravity in m/s^2 and k if the force constant of the spring in N/m and e is the extension of the spring which original length of the spring - new length after extension in meters

For the first body, m*g = K * (0.25- li)

Where li is the initial length of the spring

0.175*9.81 = k(0.25-li)

1.72 = k(0.25-li) as equation 1

For the second body, m *g = K* ( 0.775-li)

2.075*9.81 = k (0.775-li) equation 2

20.36 = k(0.775-li)

Make li subject of the formula;

li = 0.775 - 20.36/k

Substitute for li in equation 1

1.72 = k(0.25- (0.775 - 20.36/k))

1.72 = k ( 0.25 - 0.775 + 20.36/k)

Open the bracket with k

1.72 = 0.25k - 0.775k + 20.36 (since k cancel k)

Collect the like terms:

1.72 - 20.36 = - 0.525k

- 18.64 = -0.525k

Divide both side by -0.525

-18.64/-0.525 = -0.525/-0.525k

K = 35.5N/m

B) substitute for k in using

li = 0.775 - 20.36/k

li = 0.775 - 20.36/35.5

li = 0.775 - 0.574

li = 0.201 in meters

li = 0.201 * 100 centimeters = 20.1cm

4 0
3 years ago
Help me PLEASE! 25 points!!!!! :)
NeTakaya

Answer:

A.

Explanation:

All the arrows align to the point right

7 0
2 years ago
The electric field in a certain region of Earth’s atmo- sphere is directed vertically down. At an altitude of 300 m the field ha
I am Lyosha [343]

Answer:

Net charge contained in the cubeq= 3.536×10^-6C

Explanation:

Formular for total flux in a cube is given as:

Total flux= E300Acos(180) + E200Acos(0)

Where A is crossectional area

Total flux= A(E200-E300)

Total flux= q/Eo

q= Eo×total flux

q=(8.84×10^-12)×(100)^2×(100-60)

q= 3.536×10^-6C

6 0
3 years ago
Read 2 more answers
A 95.0 kg satellite moves on a circular orbit around the Earth at the altitude h=1.20*103 km. Find: a) the gravitational force e
Natali5045456 [20]

Answer:

a) F = 660.576\,N, b) a_{c} = 6.953\,\frac{m}{s^{2}}, c) v \approx 7255.423\,\frac{m}{s}, \omega = 9.583\times 10^{-4}\,\frac{rad}{s}, d) T \approx 1.821\,h

Explanation:

a) The gravitational force exerted by the Earth on the satellite is:

F = G\cdot \frac{m\cdot M}{r^{2}}

F = \left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot \frac{(95\,kg)\cdot (5.972\times 10^{24}\,kg)}{(7.571\times 10^{6}\,m)^{2}}

F = 660.576\,N

b) The centripetal acceleration of the satellite is:

a_{c} = \frac{660.576\,N}{95\,kg}

a_{c} = 6.953\,\frac{m}{s^{2}}

c) The speed of the satellite is:

v = \sqrt{a_{c}\cdot R}

v = \sqrt{\left(6.953\,\frac{m}{s^{2}} \right)\cdot (7.571\times 10^{6}\,m)}

v \approx 7255.423\,\frac{m}{s}

Likewise, the angular speed is:

\omega = \frac{7255.423\,\frac{m}{s} }{7.571\times 10^{6}\,m}

\omega = 9.583\times 10^{-4}\,\frac{rad}{s}

d) The period of the satellite's rotation around the Earth is:

T = \frac{2\pi}{\left(9.583\times 10^{-4}\,\frac{rad}{s} \right)} \cdot \left(\frac{1\,hour}{3600\,s} \right)

T \approx 1.821\,h

6 0
3 years ago
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