Use a(t) =−32 feet per second squared as the acceleration due to gravity. a ball is thrown vertically upward from the ground with an initial velocity of 56 feet per second. for how many seconds will the ball be going upward?

Answers

Answer 1
Answer: Since the ball is moving by uniformly accelerated motion, its vertical velocity at time t is given by
v(t)= v_0 - a t
where we took upward as positive direction, and where v_0 is the initial velocity, a the acceleration and t the time.

The instant at which v(t)=0 is the instant when the ball reverses its velocity (from upward to downward). This means that the difference between the time t at which v(t)=0 and the instant t=0 is the total time during which the ball was going upward:
0=v_0 - at
By plugging numbers into the equation, we find
t= (v_0)/(a)= (56 ft/s)/(32 ft/s^2)=1.75 s

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A water-skier is being pulled by a tow rope attached to a boat. As the driver pushes the throttle forward, the skier accelerates. A 76.2-kg water-skier has an initial speed of 5.0 m/s. Later, the speed increases to 10.4 m/s. Determine the work done by the net external force acting on the skier.

Answers

Answer:

Work done will be equal to 3186.396 J

Explanation:    

We have mass m = 76.2 kg

Initial velocity u = 5 m/sec

Final velocity v = 10.4 m/sec

We have to find the work done

From work energy theorem work done is equal to change in kinetic energy

w=(1)/(2)mv^2-(1)/(2)mu^2

w=(1)/(2)* 76.2* 10.4^2-(1)/(2)* 76.2* 5^2

w = 3168.396 J

So work done will be equal to 3186.396 J

What best describes the bromide ion that forms

Answers

Answer:

it A

Explanation:

Its a negative ion that hss one less valence electron than a netural bromine atom

A^^\-> points in the -x direction with a magnitude of 21. What is the y component of A^^\->

Answers

Answer:

A_y=-36.37^(\circ)

Explanation:

Given that,

Vector A points in the -x direction with a magnitude of 21.

Let the x component is making an angle of 60 degrees with negative x axis. The x component of a vector is given by :

A_x=A\ cos\theta

-21=A\ cos(60)

A=(-21)/(cos(60))

A = -42 units

The y component of a vector is given by :

A_y=A\ sin\theta

A_y=-42\ sin(60)

A_y=-36.37^(\circ)

So, the y component of vector A is (-36.37) degrees. Hence, this is the required solution.

Air is a good conductor of heat. Please select the best answer from the choices provided T F

Answers

If air were a good conductor of heat" then soup will not stay hot for longer because this time convection+conduction will both help to transfer heat away from soup. Because conduction is the transfer of heat through a substances as a result of neighbouring vibrating particles, The particles in air are far apart.

the answer is false. Hope this helps

Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400g over a distance of 0.52 cm as it rapidly bends its thorax, making the "click" that gives it its name. part a assuming the beetle jumps straight up, at what speed does it leave the ground? part b how much time is required for the beetle to reach this speed? part c ignoring air resistance, how high would it go?

Answers

(a) The launching velocity of the beetle is 6.4 m/s

(b) The time taken to achieve the speed for launch is 1.63 ms

(c) The beetle reaches a height of 2.1 m.

(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.

Use the equation of motion,

v^2=u^2+2as

Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.

v^2=u^2+2as\n = (0 m/s)^2+2 (400)(9.8 m/s^2)(0.52*10^-^2 m)\n =40.768 (m/s)^2\n v=6.385 m/s

The launching speed of the beetle is 6.4 m/s.

(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,

v=u+at

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.

v=u+at\n 6.385 m/s = (0 m/s) +400(9.8 m/s^2)t\n t = (6.385 m/s)/(3920 m/s^2) = 1.63*10^-^3s=1.63 ms

The time taken by the beetle to launch itself upwards is 1.62 ms.

(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.

Use the equation of motion,

v^2=u^2+2as

Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.

v^2=u^2+2as\n (0m/s)^2=(6.385 m/s)^2+2(-9.8m/s^2)s\n s=((6.385 m/s)^2)/(2(9.8m/s^2)) =2.08 m

The beetle can jump to a height of 2.1 m



We have that for the Question the Speed,Time and Height are

  • u=6.38m/s
  • T=13sec
  • h=2m

From the question we are told

  • Certain insects can achieve seemingly impossible accelerations while jumping.
  • the click beetle accelerates at an astonishing 400g over a distance of 0.52 cm as it rapidly bends its thorax, making the "click" that gives it its name.

Speed,Time and Height

a)

Generally the equation for the average velocity  is mathematically given as

v^2-u^2=-2ah\n\nTherefore\n\nu=√(2*400*9.8*0.0052)\n\n

  • u=6.38m/s

b)

Generally the equation for the Time of flight  is mathematically given as

T=(2u)/(g)\n\nTherefore\n\nT=(2(6.38))/(9.8)

  • T=13sec

c)

Generally the equation for the air resistance is mathematically given as

v^2-u^2=2gh\n\nTherefore\n\nh=(6.38^2)/(2*9.8)\n\

  • h=2m

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Two resistors are to be combined in parallelto form an equivalent resistance of 400Ω. The resistors are takenfrom available stock on hand as acquired over the years. Readily available are two common resistorsrated at 500±50 Ωand two common resistors rated at 2000 Ω±5%. What isthe uncertainty in an equivalent 400 Ωresistance?(Hint: the equivalent resistance connected in parallel can be obtained by 1212TRRRRR=+)

Answers

Answer:

ΔR_(e) = 84   Ω,     R_(e) = (40 ± 8) 10¹   Ω

Explanation:

The formula for parallel equivalent resistance is

          1 / R_(e) = ∑ 1 / Ri

In our case we use a resistance of each

           R₁ = 500 ± 50  Ω

          R₂ = 2000 ± 5%

This percentage equals

        0.05 = ΔR₂ / R₂

        ΔR₂ = 0.05 R₂

        ΔR₂ = 0.05 2000 = 100   Ω

We write the resistance

        R₂ = 2000 ± 100    Ω

We apply the initial formula

        1 / R_(e) = 1 / R₁ + 1 / R₂

        1 / R_(e) = 1/500 + 1/2000 = 0.0025

        R_(e)  = 400    Ω

Let's look for the error  (uncertainly) of Re

      R_(e) = R₁R₂ / (R₁ + R₂)

       R’= R₁ + R₂

       R_(e) = R₁R₂ / R’

Let's look for the uncertainty of this equation

      ΔR_(e) / R_(e) = ΔR₁ / R₁ + ΔR₂ / R₂ + ΔR’/ R’

The uncertainty of a sum is

      ΔR’= ΔR₁ + ΔR₂

We substitute the values

     ΔR_(e) / 400 = 50/500 + 100/2000 + (50 +100) / (500 + 2000)

     ΔR_(e) / 400 = 0.1 + 0.05 + 0.06

     ΔR_(e) = 0.21 400

     ΔR_(e) = 84   Ω

Let's write the resistance value with the correct significant figures

    R_(e) = (40 ± 8) 10¹   Ω