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svlad2 [7]
2 years ago
10

Suppose the initial position of an object is zero, the starting velocity is 3 m/s and the final velocity was 10 m/s. The object

moves with constant acceleration. Which part of a velocity vs. time graph can be used to calculate the displacement of the object? A. the area of the rectangle under the line B. the area of the rectangle above the line C. the area of the rectangle plus the area of the triangle under the line D. the area of the rectangle plus the area of the triangle above the line

Physics
2 answers:
Tatiana [17]2 years ago
6 0

Answer:

C. the area of the rectangle plus the area of the triangle under the line

Explanation:

Based on the information provided, the velocity vs. time graph is a line with a positive slope and a y-intercept of (0, 3).  The displacement is the area under this line.  This area can be divided into a triangle and a rectangle.  So of the options available, C is the correct one.

vesna_86 [32]2 years ago
6 0

Answer:

C

Explanation:

There is no explanation for the answer.

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A 6.0-cm-diameter, 11-cm-long cylinder contains 100 mg of oxygen (O2) at a pressure less than 1 atm. The cap on one end of the c
butalik [34]

Answer:

The temperature of the gas is 1197.02 K

Explanation:

From ideal gas law;

PV = nRT

Where;

P is the pressure of the gas

V is the volume of the gas

R is ideal gas constant = 8.314 L.kPa/mol.K

T is the temperature of the gas

n is the number of moles of gas

Volume of the gas in the cylindrical container = πr²h

Given;

r = 6/2 = 3 cm = 0.03 m

h = 11 cm = 0.11 m

V = π × (0.03)² × 0.11 = 3.11 × 10⁻⁴ m³ = 0.311 L

number of moles of oxygen gas = Reacting mass / molar mass

=\frac{0.1}{32} = 0.003125, moles

T = \frac{PV}{nR} = \frac{100X0.311}{0.003125X8.314} =1197.02K

Therefore, the temperature of the gas is 1197.02 K

6 0
2 years ago
You are flying a hang glider at 14 mph in the northeast direction (45°). The wind is blowing at 4 mph from due north.
Afina-wow [57]

Answer:

<em>a) 17.05 mph</em>

<em>b) 54.7°  northeast direction</em>

<em>c) 10.71 mph</em>

<em>The direction is -22.58° relative to the east.</em>

<em></em>

<em>To head northeast, you must either increase your gliding speed or increase your angle relative to the x-axis greater than 45°.</em>

Explanation:

The question is a little confusing but, I guess the correct question should be;

You are flying a hang glider at 14 mph in the northeast direction (45°). The wind is blowing at 4 mph due north.

a) What is your airspeed?

b) What angle (direction) are you flying?

c) The wind increases to 14 mph from north. Now what is your airspeed and what direction are you flying? If your destination is to the northeast, how would you change your speed or direction so you might make it there?

NB: The difference in the question and my suggestion is highlighted boldly.

Your speed = 14 mph

direction is 45° northeast

Th wind speed = 4 mph

direction is north

We resolve the your speed and the wind speed into the horizontal and vertical components

For vertical the component component

V_{y} = 14(sin 45) + 4 = 9.89 + 4 = 13.89 mph

For the horizontal speed component

V_{x} = 14(cos 45) + 0 = 9.89 + 0 = 9.89 mph

Resultant speed = \sqrt{V^{2} _{y}+V^{2} _{x}  }

==> \sqrt{13.89^{2} +9.89^{2}   } = <em>17.05 mph  This is your airspeed</em>

b) To get your direction, we use

tan ∅ = V_{y} /V_{x}

tan ∅ = 13.89/9.89 = 1.413

∅ = tan^{-1}(1.413) = <em>54.7°  northeast direction</em>

c) If the wind increases to 14 mph from the north, then it means the wind blows due south. As before, only the vertical component is affected .

In this case,

V_{y} = 14(sin 45) - 14 = 9.89 - 14 = -4.11 mph

Resultant speed = \sqrt{V^{2} _{y}+V^{2} _{x}  }

==> \sqrt{4.11^{2} +9.89^{2}   } = <em>10.71 mph  This is your airspeed</em>

Your direction will be,

tan ∅ = V_{y} /V_{x}

tan ∅ = -4.11/9.89 = -0.416

∅ = tan^{-1}(-0.416) =<em> -22.58°  this is the angle you'll travel relative to the east.</em>

<em>To head northeast, you must either increase your gliding speed or increase your angle relative to the x-axis greater than 45°.</em>

5 0
2 years ago
Given equal time periods, which statement is correct?
Mekhanik [1.2K]
The correct option is B.
Nuclear fission and fusion are two different types of nuclear reactions, through which energy may be obtained. Nuclear fission involves the splitting of a molecule into two different part in order to generate energy while nuclear fusion reaction involves the joining together of two elements in other to form one product. Nuclear fission generate much radioactive waste than nuclear fusion. The radioactive waste that is obtainable during nuclear fusion is less than 1% of that produce by nuclear fission.
3 0
2 years ago
Read 2 more answers
A moving sidewalk 95 m in length carries passengers at a speed of 0.53 m/s. One passenger has a normal walking speed of 1.24 m/s
Archy [21]

Answer:

a) t = 1.8 x 10² s

b) t = 54 s

c) t = 49 s

Explanation:

a) The equation for the position of an object moving in a straight line at constan speed is:

x = x0 + v * t

where

x = position at time t

x0 = initial position

v = velocity

t = time

In this case, the origin of our reference system is at the begining of the sidewalk.

a) To calculate the time the passenger travels on the sidewalk without wlaking, we can use the equation for the position, using as speed the speed of the sidewalk:

x = x0 + v * t

95 m = 0m + 0. 53 m/s * t

t = 95 m/ 0.53 m/s

t = 1.8 x 10² s

b) Now, the speed of the passenger will be her walking speed plus the speed of th sidewalk (0.53 m/s + 1.24 m/s = 1.77 m/s)

t = 95 m/ 1.77 m/s = 54 s

c) In this case, the passenger is located 95 m from the begining of the sidewalk, then, x0 = 95 m and the final position will be x = 0. She walks in an opposite direction to the movement of the sidewalk, towards the origin of the system of reference ( the begining of the sidewalk). Then, her speed will be negative ( v = 0.53 m/s - 2*(1.24 m/s) = -1.95 m/s. Then:

0 m = 95 m -1.95 m/s * t

t = -95 m / -1.95 m/s = 49 s

3 0
2 years ago
A simple watermelon launcher is designed as a spring with a light platform for the watermelon. When an 8.00 kg watermelon is put
olga_2 [115]

To solve this problem it is necessary to apply the concepts related to the Force from Hooke's law, the force since Newton's second law and the potential elastic energy.

Since the forces are balanced the Spring force is equal to the force of the weight that is

F_s = F_g

kx = mg

Where,

k = Spring constant

x = Displacement

m = Mass

g = Gravitational Acceleration

Re-arrange to find the spring constant

k = \frac{mg}{x}

k = \frac{8*9.8}{0.1}

k = 784N/m

Just before launch the compression is 40cm, then from Potential Elastic Energy definition

PE = \frac{1}{2} kx^2

PE =\frac{1}{2} 784*0.4^2

PE = 63.72J

Therefore the energy stored in the spring is 63.72J

6 0
2 years ago
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