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cupoosta [38]
3 years ago
11

What's the concept of mass ?

Physics
1 answer:
Marat540 [252]3 years ago
6 0
Masa is a concept that identifies that magnitude of a physical nature that allows indicating the amount of matter in a body. Within the International System , its unit is the kilogram ( kg . ) 
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A plane is flying east at 115 m/s. The wind accelerates it at 2.88 m/s^2 directly northwest. After 25.0s, what is the magnitude
kondaur [170]

Answer:

81.8 m/s

Explanation:

The initial velocity of the plane is:

v_0=115 m/s (toward east)

So, decomposing along the x- and y- directions:

v_{x0} = 115 m/s\\v_{y0} = 0

(we took east as positive x-direction and north as positive y-direction)

The acceleration is

a=2.88 m/s^2 (northwest, so the angle with the positive x-direction is 135 degrees)

Decomposing it along the two directions:

a_x = a cos 135^{\circ} = (2.88 m/s^2)(cos 135^{\circ})=-2.04 m/s^2\\a_y = a sin 135^{\circ} = (2.88 m/s^2)(sin 135^{\circ})=2.04 m/s^2

So the two components of the velocity after a time t = 25.0 s will be

v_x = v_{x0} + a_x t = 115 m/s + (-2.04 m/s^2)(25.0 s)=64 m/s\\v_y = v_{y0} + a_y t = 0 m/s + (2.04 m/s^2)(25.0 s)=51 m/s

So, the magnitude of the velocity of the plane will be

v=\sqrt[v_x^2+v_y^2}=\sqrt{(64 m/s)^2+(51 m/s)^2}=81.8 m/s

6 0
3 years ago
If the box is a distance 1.81 m from the rear of the truck when the truck starts, how much time elapses before the box falls off
alexira [117]
If the box is a distance 1.81 m from the rear of the truck when the truck starts,<span> ... Force of Friction = mu_s * Normal Force( </span>M<span> * G) ... The </span>box starts<span> moving! ... Now that the </span>box<span> is moving, the bed of the </span>truck<span> pulls at it with 17.4 ... out how </span>long<span> it will take the </span>box<span> to reach the back of the </span>truck<span>. ... T^2 = 2 * </span>1.81<span> / .64</span>
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4 years ago
What is the best course of treatment for people with phobias
balandron [24]
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6 0
3 years ago
Read 2 more answers
you push with an 18-N horizontal force on a 5-kg box of coffee resting on a on a horizontal surface. the force of friction on th
KatRina [158]

The acceleration is found to be 2 m/s², final velocity after 10 seconds is 20 m/s and the final position after 10 seconds is 100 m away from the starting point.

Answer:

Explanation:

As per Newton's second law of motion, acceleration of any object is directly proportional to the net external unbalanced force acting on that object and inversely proportional to the mass of the object.

Since there are two forces acting on the box in opposite direction, the net force will be the difference of horizontal and frictional force acting on the object.

Net force = Horizontal force - Frictional force = 18 N - 8 N = 10 N.

Now, from second law of motion, Acceleration = \frac{Net force}{Mass}

So, acceleration = 10 N /5 kg = 2 m/s².

Since, acceleration exerted by the box is found to be 2 m/s², we can determine the final velocity of the object after 10 seconds using the first equation of motion.

v = u + at, Here v is the final velocity and u is the initial velocity which is zero for the present case. Other parameters like a is found to be 2 m/s² and time is 10 seconds.

So Final velocity v = 0+(2×10)=20 m/s.

And the final position can be determined using the second equation of motion.

s = ut+1/2at²

Final position = (0×10)+(0.5×2×10×10)= 100 m.

So the final position is 100 m.

Thus, the acceleration is found to be 2 m/s², final velocity after 10 seconds is 20 m/s and the final position after 10 seconds is 100 m away from the starting point.

3 0
3 years ago
A string, 0.28 m long and vibrating in its third harmonic, excites an open pipe that is 0.82 m long into its second overtone res
Rama09 [41]

Answer:

117.8 m/s

Explanation:

Given that:

String length, L = 0.28

pipe length, L' = 0.82

Speed of sound in air, v = 345 m/s

n = 3 (3rd harmonic)

Frequency, f of 3rd harmonic ;

f = (v*n) / 2L - - - - (1)

for the pipe: ; 3rd harmonic

f = (v*n) / 2L' - - - (2)

Equating (1) and (2)

(v*n) / 2L = (v*n) / 2L'

2L' * v * n = v* n * 2L

v = vL / L'

v = (345 * 0.28) / 0.82

v = 96.6 / 0.82

v = 117.80487

v = 117.8 m/s

7 0
3 years ago
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