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kodGreya [7K]
2 years ago
8

As part of a circus performance, a man is attempting to throw a dart into an apple which is dropped from an overhead platform. U

pon release of the apple, the man has a reflex delay of 235 milliseconds before throwing the dart. If the dart is released with a speed v0 = 13 m/s, at what distance d below the platform should the man aim if the dart is to strike the apple before it hits the ground?
Physics
1 answer:
Tems11 [23]2 years ago
4 0

Answer:

The man has to aim 27.06 cm below the platform.

Explanation:

Given that,

Time = 235 millisecond

Speed = 13 m/s

We need to calculate the distance below the platform

Using equation of motion

s=ut+\dfrac{1}{2}gt^2

Where, u = initial velocity

g = acceleration due to gravity

t = time

Put the value into the formula

s=0+\dfrac{1}{2}\times9.8\times(235\times10^{-3})^2

s=0.2706\ m

s=27.06\ cm

Hence, The man has to aim 27.06 cm below the platform.

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A car is traveling at 20 meters/second and is brought to rest by applying brakes over a period of 4 seconds. What is its average
frez [133]
 (u) = 20 m/s 
(v) = 0 m/s 
<span> (t) = 4 s 
</span>
<span>0 = 20 + a(4) 

</span><span>4 x a = -20 
</span>
so, the answer is <span>-5 m/s^2. or -5 meter per second</span>
8 0
2 years ago
Read 2 more answers
Last year a baseball player made 63 errors. This year he made 42. What percent decrease was there in the number of errors commit
Crank

Answer:

33.33 %

Explanation:

given,

error last year = 63

error this year = 42

percent decrease in the error = ?

to find the percentage difference in the error formula used is

   = \dfrac{difference}{original}\times 100

   = \dfrac{63-42}{63}\times 100

   = \dfrac{21}{63}\times 100

   = 33.33 %

Percentage decrease in the number of error is equal to 33.33%.

7 0
2 years ago
Because the soles of your shoes have cleats, you can exert a forward force of 100 N even on slippery ice. A 10-kg picnic cooler
Brilliant_brown [7]

Answer:

you must throw 3 snowballs

Explanation:

We can solve this exercise using the concepts of conservation of the moment, let's define the system as formed by the refrigerator and all the snowballs. Let's write the moment

Initial. Before bumping that refrigerator

          p₀ = n m v₀

Where n is the snowball number

Final. When the refrigerator moves

         pf = (n m + M) v

The moment is preserved because the forces during the crash are internal

        n m v₀ = (n m + M) v

        n m (v₀ - v) = M v

        n = M/m    v/(vo-v)

Let's look for the initial velocity of the balls, suppose the person throws them with the maximum force if it slides in the snow (F = 100N), let's use the second law and Newton

          F = m a

          a = F / m

The distance the ball travels from zero speed to maximum speed is the extension of the arm (x = 1 m), let's look kinematically for the speed of the balls when leaving the arm

          v₁² = v₀² + 2 a x

          v₁² = 0+ 2 (100/1) 1

          v₁ = 14.14 m / s

This is the initial speed for the crash

         v₀ = v = 14.14 m / s

  Let's calculate

           n = M/m   v/ (v₀-v)

           n = 10/1   3 / (14.14 -3)

          n = 2.7 balls

you must throw 3 snowballs

7 0
2 years ago
calculate the workdone to stretch an elastic string by 40cm if a force of 10N produces an extension of 4cm in it
Lera25 [3.4K]
The force of F=10 N produces an extension of
x=4 cm=0.04 m
on the string, so the spring constant is equal to
k= \frac{F}{x}= \frac{10 N}{0.04 m}=250 N/m

Then the string is stretched by \Delta x=40 cm=0.40 m. The work done to stretch the string by this distance is equal to the variation of elastic potential energy of the string with respect to its equilibrium position:
W= \Delta U= \frac{1}{2}k(\Delta x)^2  = \frac{1}{2}(250 N/m)(0.40 m)^2=20 J
5 0
2 years ago
ASK YOUR TEACHER A 2.0-kg mass swings at the end of a light string with the length of 3.0 m. Its speed at the lowest point on it
Nadya [2.5K]

Answer:

  K_b = 78 J

Explanation:

For this exercise we can use the conservation of energy relations

starting point. Lowest of the trajectory

        Em₀ = K = ½ mv²

final point. When it is at tea = 50º

        Em_f = K + U

        Em_f = ½ m v_b² + m g h

where h is the height from the lowest point

        h = L - L cos 50

        Em_f = ½ m v_b² + mg L (1 - cos50)

energy be conserve

        Em₀ = Em_f

         ½ mv² = ½ m v_b² + mg L (1 - cos50)

         K_b = ½ m v_b² + mg L (1 - cos50)

let's calculate

          K_b = ½ 2.0 6.0² + 2.0 9.8 6.0 (1 - cos50)

          K_b = 36 +42.0

          K_b = 78 J

4 0
2 years ago
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