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VladimirAG [237]
1 year ago
15

A train travels 8.81 m/s in a -51.0° direction. The train accelerated for 2.23 s, changing its velocity to 9.66 m/s in a 37.0° d

irection. What is (delta)x? ​

Physics
1 answer:
Evgen [1.6K]1 year ago
7 0

The value of the \rm \triangle x will be 29.14 m. Delta represent the change.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The velocity in the part 1;

V₁ = 8.81

The velocity in the part 2;

V₂ = 9.66 m/sec

The resultant of the two vectors of velocity gives the velocity for the connecting path.

\rm V = \sqrt{8.81 ^2 + 9.66^2 } \\\\ V= 13.07 \ m/sec

The  time period for the train accelerates;

\rm  \triangle x = v \times t \\\\ \triangle x = 13.07 \ m/sec  \times 2.23 \ sec \\\\  \triangle x =  29.14 \ m

Hence the value of the \rm \triangle x will be 29.14 m.

To learn more about the velocity, refer to the link: brainly.com/question/862972.

#SPJ1

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Degger [83]

The magnitude of the force acting on the object lying on a flat surface without moving is 10 N.

The given parameters;

  • magnitude of force on the object, F = 10 N
  • angle between the object and the horizontal flat surface = 0⁰

Apply Newton's second law of motion to determine the magnitude of the force on the object.

Due to the position of the object, the magnitude of the force acting on it is calculated as;

\Sigma F_{net} = F\sin(\theta ) + F cos(\theta)\\\\\Sigma F_{net} = 10 sin(0) + 10cos(0)\\\\\Sigma F_{net} = 10 \ N

Therefore, the magnitude of the force acting on the object is 10 N.

Learn more here: brainly.com/question/19887955

7 0
2 years ago
An airplane is flying 340 km/hr at 12o east of north. the wind is blowing 40 km/hr at 34o south of east. what is the plane's act
jok3333 [9.3K]
Define an x-y coordinate system such that
The positive x-axis = the eastern direction, with unit vector  \hat{i}.
The positive y-axis = the northern direction, with unit vector \hat{j}.

The airplane flies at 340 km/h at 12° east of north. Its velocity vector is
\vec{v}_{1} = 340(sin(15^{o})\hat{i} + cos(15^{o})\hat{j} ) = 88\hat{i} + 328.4\hat{j}

The wind blows at 40 km/h in the direction 34° south of east. Its velocity vector is
\vec{v}_{2} =40(cos(34^{o})\hat{i} - sin(24^{o})]\hat{j}) = 33.1615\hat{i} -22.3677\hat{j})

The plane's actual velocity is the vector sum of the two velocities. It is
\vec{v}=\vec{v}_{1}+\vec{v}_{2} = 121.1615\hat{i}+306.0473\hat{j}

The magnitude of the actual velocity is
v = √(121.1615² + 306.0473²) = 329.158 km/h

The angle that the velocity makes north of east is
tan⁻¹ (306.04733/121.1615) = 21.6°

Answer:
The actual velocity is 329.2 km/h at 21.6° north of east.
5 0
3 years ago
Sam is observing the velocity of a car at different times. After two hours, the velocity of the car is 54 km/h. After four hours
MariettaO [177]
With that information you can only suppose a uniformly accelerated motion.  This is, acceleration is constant.

Then, acceleration = change in velocity / change in time = (58 -54)km/h / 2 h = 4km/h / 2 h = 2 km/h^2

Then the equation for velocity, V is

V = Vo + a*t = Vo + 2 (km/h^2)  * t = Vo +  2t

Vo is the initial velocity, which you can find using V = 54km/h and t = -2

Vo = V after 2 hours - a*(2hours) = 54km/h - 2(km/h^2)*2h = 54km/k - 4km/h = 50km/h

Then, the equation is: V = 50 km/h + 2t

Valid for constant acceleration. 
5 0
3 years ago
Two conducting spheres of different sizes are at the same potential. The radius of the larger sphere is four times larger than t
Sergeeva-Olga [200]

Answer:

0.8

Explanation:

The two spheres have the same potential, V.

Let the radius of the larger sphere be R and the radius of the smaller sphere be r,

=> R = 4r

Let the charge on the smaller sphere be q. Hence, the larger sphere will have charge Q - q.

The potential of the smaller sphere will be:

V_S = \frac{kq}{r}

The potential of the larger sphere will be:

V_L = \frac{k(Q - q)}{R}

Inputting R = 4r,

V_L = \frac{k(Q - q)}{4r}

Since V_S = V_L = V,

\frac{k(Q - q)}{4r} = \frac{kq}{r}

=> Q - q = 4q

=> 5q = Q

q = 0.2Q

The fraction of the charge Q that rests on the smaller sphere is 0.2

The charge of the larger sphere is:

Q - q = Q - 0.2Q = 0.8Q

∴ The fraction of the total charge Q that rests on the larger sphere is 0.8

7 0
3 years ago
Calculate the rate of heat conduction through a layer of still air that is 1 mm thick, with an area of 1 m, for a temperature of
max2010maxim [7]

Answer:

The rate of heat conduction through the layer of still air is 517.4 W

Explanation:

Given:

Thickness of the still air layer (L) = 1 mm

Area of the still air = 1 m

Temperature of the still air ( T) = 20°C

Thermal conductivity of still air (K) at 20°C = 25.87mW/mK

Rate of heat conduction (Q) = ?

To determine the rate of heat conduction through the still air, we apply the formula below.

Q =\frac{KA(\delta T)}{L}

Q =\frac{25.87*1*20}{1}

Q = 517.4 W

Therefore, the rate of heat conduction through the layer of still air is 517.4 W

6 0
3 years ago
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