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Elena-2011 [213]
3 years ago
12

The peak of the trajectory occurs at time t1. This is the point where the ball reaches its maximum height ymax. At the peak the

ball switches from moving up to moving down, even as it continues to travel horizontally at a constant rate.. . Part C. . What are the values of the velocity vector components v1,x and v1,y (both in m/s) as well as the acceleration vector components a1,x and a1,y (both in m/s2)? Here the subscript 1 means that these are all at time t1.. . A) 0, 0, 0, 0. B) 0, 0, 0, -9.80. C) 15.0, 0, 0, 0. D) 15.0, 0, 0, -9.80. E) 0, 26.0, 0, 0. F) 0, 26.0, 0, -9.80. G) 15.0, 26.0, 0, 0. H) 15.0, 26.0, 0, -9.80
Physics
1 answer:
Digiron [165]3 years ago
7 0
<span>Part B
What are the values of the intial velocity vector components v0,x and v0,y (both in m/s) as well as the acceleration vector components a0,x and a0,y (both in m/s2)? Here the subscript 0 means "at time t0."
15.0, 26.0, 0, -9.80

</span><span>Part C
What are the values of the velocity vector components v1,x and v1,y (both in m/s) as well as the acceleration vector components a1,x and a1,y (both in m/s2)? Here the subscript 1 means that these are all at time t1.
 15, 26, 0, -9.81</span><span>

</span>
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Answer: conduction :it transfers heat between objects that are in direct contact with eachother
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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

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Um corpo de massa 50g recebe 300 calorias e sua temperatura sobe de -10°C até 20°C. Determine a capacidade térmica do corpo e o
kykrilka [37]

The specific heat capacity of the substance is 0.963 J/g^{\circ}C

Explanation:

When an object of mass m is supplied with a certain amount of energy Q, its temperature increases according to the equation:

Q=mC_s \Delta T

where

m is the mass of the object

C_s is its specific heat capacity

\Delta T is the increase in temperature of the object

In this problem, we have

Q=300 cal \cdot 4.814 = 1444.2 J

m = 50 g

\Delta T = 20C-(-10C)=30^{\circ}C

Therefore, we can solve for C_s to find its specific heat capacity:

C_s = \frac{Q}{m\Delta T}=\frac{1444.2}{(50)(30)}=0.963 J/gC

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

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A 1.1-kg uniform bar of metal is 0.40 m long and has a diameter of 2.0 cm. When someone bangs one end of this bar, a 1.5 MHz sho
lyudmila [28]

Answer:

\lambda = 2.22\times 10^{-3}\ m

Explanation:

Given,

mass of the bar = 1.1 Kg

length of rod, l = 0.40 m

diameter of the rod, d = 2 cm

frequency, f = 1.5 MHz

time, t = 0.12 ms

wavelength of the shock wave = ?

Speed of the wave =\dfrac{L}{t}=\dfrac{0.40}{0.12\times 10^{-3}}

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wavelength of the wave

\lambda = \dfrac{v}{f}= \dfrac{3333.33}{1.5\times 10^6}

\lambda = 2.22\times 10^{-3}\ m

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