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il63 [147K]
3 years ago
6

7. You start to walk toward the east towards home at a constant speed of 4 km/hr. At the same time, Someone else leaves your hom

e driving west at 28 km/hr. The total distance for your walk is 3.2 km. a. Draw a position-time graph that shows the motion of you and the other person. b. From the graph, determine how long you will travel before you meet the other person. c. You do NOT need a trend line for this graph but you do need to have the other required parts.

Physics
1 answer:
amm18123 years ago
4 0

A) position time graph for both is shown

here one of the graph is of lesser slope which means it is moving with less speed while other have larger slope which shows larger speed

At one point they intersects which is the point where they both will meet

B) Let the two will meet after time "t"

now we can say that

if they both will meet after time "t"

then the total distance moved by you and other person will be same as the distance between you and home

so it is given as

v_1t + v_2t = d

4*t + 28*t = 3.2 km

t = \frac{3.2}{32} = 0.1 hr

so they will meet after t = 6 min

so from position time graph we can see that two will meet after t = 6 min where at this position two graphs will intersect


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A 50.0-kg box rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.300 and the
Dimas [21]

Answer:

98N and 147N

Explanation:

We have the following information:

m=50kg\\\mu_s =0.4\\\mu_k = 0.2\\F=140N

We can find the static fricton force as follow,

F=\mu_s * N

Where N is the normal force (mg)

F=0.3*50*9.8\\F=147N

Static friction force at 147N is greater than the force applied hence body does not move.

F=\mu_k N = 0.2*50*9.8= 98N

3 0
3 years ago
Determine the maximum theoretical speed that may be achieved over a distance of 66 m by a car starting from rest, knowing that t
My name is Ann [436]

Answer:

v=32.49 m/s

Explanation:

Given that

Distance ,d= 66 m

Initial speed of the car ,u = 0 m/s

Coefficient of friction ,μ = 0.8

Lets take the total mass of the car = m

The acceleration of the car is given as

a = μ g                               ( g= 10 m/s² )

Now by putting the values in the above equation we get

a= 0.8 x 10 m/s²

a= 8 m/s²

We know that ,final speed is given as

v²= u ²+ 2 a d

Now putting the value

v²=0² + 2 x 8 x 66

v²= 1056

v=32.49 m/s

3 0
3 years ago
An astronaut has a momentum of 280 kg and travels 10 m/s. what is the mass of the astronaut?
Kamila [148]

Answer:

The answer is

<h2>28 kg</h2>

Explanation:

The mass of an object given it's momentum and velocity / speed can be found by using the formula

m =  \frac{p}{v}  \\

where

m is the mass

p is the momentum

v is the speed or velocity

From the question

p = 280 kg/ms

v = 10 m/s

The mass of the object is

m =  \frac{280}{10}  = 28 \\

We have the final answer as

<h3>28 kg</h3>

Hope this helps you

3 0
3 years ago
Please answer I will give good rating and mark brainlest
lina2011 [118]
The answer is C. You must divide your wavelength and your frequency to get your answer.
8 0
3 years ago
A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 58.8m/s2 . The accel
Zanzabum
Split the operation in two parts. Part A) constant acceleration 58.8m/s^2, Part B) free fall.

Part A)
Height reached, y = a*[t^2] / 2 = 58.8 m/s^2 * [7.00 s]^2 / 2 = 1440.6 m

Now you need the final speed to use it as initial speed of the next part.

Vf = Vo + at = 0 + 58.8m/s^2 * 7.00 s = 411.6 m/s

Part B) Free fall

Maximum height, y max ==> Vf = 0

Vf = Vo - gt ==> t = [Vo - Vf]/g = 411.6 m/s / 9.8 m/s^2 = 42 s

ymax = yo + Vo*t - g[t^2] / 2

ymax = 1440.6 m + 411.6m/s * 42 s - 9.8m/s^2 * [42s]^2 /2
ymax = 1440.6 m + 17287.2m - 8643.6m = 10084.2 m

Answer: ymax = 10084.2m
8 0
3 years ago
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