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Viefleur [7K]
2 years ago
12

Can someone help me?

Physics
1 answer:
Alik [6]2 years ago
6 0

Answer:

Resultant = 3.05N

Explanation:

r =  \sqrt{{5}^{2} + 5^{2} - 2(5)(5) \cos(120) }

r =  \sqrt{25 + 25 - 50(0.8142)}

r =  \sqrt{50 - 40.71}  =  \sqrt{9.29}

r = 3.05

You might be interested in
Scenario
Anvisha [2.4K]

Answer:

1) t = 23.26 s,  x = 8527 m, 2)   t = 97.145 s,  v₀ = 6.4 m / s

Explanation:

1) First Scenario.

After reading your extensive problem, we are going to solve it, for this exercise we must use the parabolic motion relationships. Let's carry out an analysis of the situation, for deliveries the planes fly horizontally and we assume that the wind speed is zero or very small.

Before starting, let's reduce the magnitudes to the SI system

         v₀ = 250 miles/h (5280 ft / 1 mile) (1h / 3600s) = 366.67 ft/s

         y = 2650 m

Let's start by looking for the time it takes for the load to reach the ground.

         y = y₀ + v_{oy} t - ½ g t²

in this case when it reaches the ground its height is zero and as the plane flies horizontally the vertical speed is zero

         0 = y₀ + 0 - ½ g t2

          t = \sqrt{ \frac{2y_o}{g} }

          t = √(2 2650/9.8)

          t = 23.26 s

this is the horizontal scrolling time

          x = v₀ t

          x = 366.67  23.26

          x = 8527 m

the speed at the point of arrival is

         v_y = v_{oy} - g t = 0 - gt

         v_y = - 9.8 23.26

         v_y = -227.95 m / s

Module and angle form

        v = \sqrt{v_x^2 + v_y^2}

         v = √(366.67² + 227.95²)

        v = 431.75 m / s

         θ = tan⁻¹ (v_y / vₓ)

         θ = tan⁻¹ (227.95 / 366.67)

         θ = - 31.97º

measured clockwise from x axis

We see that there must be a mechanism to reduce this speed and the merchandise is not damaged.

2) second scenario. A catapult located at the position x₀ = -400m y₀ = -50m with a launch angle of θ = 50º

we look for the components of speed

           cos θ = v₀ₓ / v₀

           sin θ = v_{oy} / v₀

            v₀ₓ = v₀ cos θ

            v_{oy} = v₀ sin θ

we look for the time for the arrival point that has coordinates x = 0, y = 0

            y = y₀ + v_{oy} t - ½ g t²

            0 = y₀ + vo sin θ t - ½ g t²

            0 = -50 + vo sin 50 t - ½ 9.8 t²

            x = x₀ + v₀ₓ t

            0 = x₀ + vo cos θ t

            0 = -400 + vo cos 50 t

podemos ver que tenemos un sistema de dos ecuación con dos incógnitas

          50 = 0,766 vo t – 4,9 t²

          400 =   0,643 vo t

resolved

          50 = 0,766 ( \frac{400}{0.643 \ t}) t – 4,9 t²

          50 = 476,52 t – 4,9 t²

          t² – 97,25 t + 10,2 = 0

we solve the quadratic equation

         t = [97.25 ± \sqrt{97.25^2 - 4 \ 10.2}] / 2

         t = 97.25 ±97.04] 2

         t₁ = 97.145 s

         t₂ = 0.1 s≈0

the correct time is t1 the other time is the time to the launch point,

         t = 97.145 s

let's find the initial velocity

         x = x₀ + v₀ cos 50 t

         0 = -400 + v₀ cos 50 97.145

         v₀ = 400 / 62.44

         v₀ = 6.4 m / s

5 0
3 years ago
Martin has severe myopia, with a far point on only 17 cm. He wants to get glasses that he'll wear while using his computer whose
marusya05 [52]

Answer:

Explanation:

Far point = 17 cm . That means he can not see beyond this distance .

He wants to see at an object at 65 cm away . That means object placed at 65 has image at 17 cm by concave lens . Using lens formula

1 / v - 1 / u = 1 / f

1 / - 17 - 1 / - 65 = 1 / f

= 1 / 65 - 1 / 17

= -  .0434 = 1 / f

power = - 100 / f

= - 100 x .0434

= - 4.34 D .

6 0
3 years ago
4 what is the difference between an array's size declarator and a subscript?
Ber [7]
What is the difference between<span> a</span>size declarator<span> and a </span>subscript<span>? The </span>size declarator<span> is ... When writing a function that accepts a two-dimensional </span>array<span> as an argument, which </span>size declarator<span> must you provide in the parameter </span>for<span> the</span>array<span>? The second size ...</span>
7 0
3 years ago
What average force is required to stop a 1400 kg car in 6.0 s if the car is traveling at 90 km/h ? Express your answer to two si
devlian [24]

Answer:

F=5833.3 N N

Explanation:

Newton's second law applied to the car

F= m*a Formula (1)

F: Force in Newtons (N)

m : mass in kg

a: acceleration ( m/s²)

kinematics car

vf= v₀ + a*t  Formula (2)

vf : final velocity (m/s)

v₀ : final velocity  (m/s)

a : acceleration  ( m/s²)

t : time t

Equivalences

1 km= 1000m

1 h = 3600 s

Data

m= 1000kg

v₀ = 90 km/h = 90*1000/3600 m/s = 25 m/s

vf= 0

t= 6 s

Problem Development

We calculate the acceleration replacing the data in the formula (2) :

0 =  25 + a*6

a= -25/6 = -4.16 m/s² ( The negative sign indicates that the car is braking)

We calculate the force is required to stop the car replacing the data  in the formula (1)

-F = 1400 kg*(-4.16 m/s²)

F=5833.3 N

8 0
3 years ago
A cube of side 5.0m is in a region where the electric field is directed outward from both of two opposite cube faces, with unifo
guajiro [1.7K]

Answer:

The charge inside the cube is null.

Explanation:

If we apply the gauss theorem with a cubical gaussian surface of the size of the cube:

\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\frac{q_{in}}{\varepsilon_{0}}

If we consider than the direction of the electric field is \vec{E}=E_0\hat{x}, we can solve the problem differentiating  the integral for each face of the cube:

\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\displaystyle\int_{S_1} \vec{E}\,\vec{ds_1}+\displaystyle\int_{S_2} \vec{E}\,\vec{ds_2}+\displaystyle\int_{S_3} \vec{E}\,\vec{ds_3}+\displaystyle\int_{S_4} \vec{E}\,\vec{ds_4}+\displaystyle\int_{S_5} \vec{E}\,\vec{ds_5}+\displaystyle\int_{S_6} \vec{E}\,\vec{ds_6}

\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\displaystyle\int_{S_1} E_0\hat{x}\,\hat{x}ds_1+\displaystyle\int_{S_2} E_0\hat{x}\,\hat{-x}ds_2+\displaystyle\int_{S_3} E_0\hat{x}\,\hat{y}ds_3+\displaystyle\int_{S_4} E_0\hat{x}\,\hat{-y}ds_4+\displaystyle\int_{S_5} E_0\hat{x}\,\hat{z}ds_5+\displaystyle\int_{S_6} E_0\hat{x}\,\hat{-z}ds_6

E₀ is a constant and each surface is equal to each other, so: S_1=S_2=S_i=S

Therefore:

\displaystyle\oint_{S} \vec{E}\,\vec{ds}=E_0\displaystyle\int_{S_1} \,ds_1+E_0\displaystyle\int_{S_2} -1\,ds_2+0+0+0+0=E_0S-E_0S=0

\displaystyle\oint_{S} \vec{E}\,\vec{ds}=0=\frac{q_0}{\varepsilon_0} \longleftrightarrow q_0=0c

3 0
3 years ago
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