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tekilochka [14]
2 years ago
7

Nate throws a ball straight up to Kayla, who is standing on a balcony 3.8 m above Nate. When she catches it, the ball is still m

oving upward at a speed of 2.8 m/s. Required:With what initial speed did Nate throw the ball?
Physics
1 answer:
marin [14]2 years ago
8 0

Answer:

9.1m/s  

Explanation:

Nate throws a straight ball to Kayla who is standing at a balcony 3.8m above Nate

When she catches the ball, it is still moving upward with a speed of 2.8m/s

v = 2.8m/s

u = ?

s = 3.8m

a= -9.8(The acceleration has a negative sign because the speed of the ball is declining)    

Therefore the initial speed at which Nate threw the ball can be calculated as follows

v^2= u^2 + 2as

2.8^2= u^2 + 2(-9.8)(3.8)

7.84= u^2 + (-74.48)

7.84= u^2 - 74.48

u^2= 7.84 + 74.48

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u^2= 82.32

u= √82.32

u = 9.1m/s    

Hence the initial speed at which Nate threw the ball is 9.1m/s

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Explanation:

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           Thus a projectile motion is always acted upon by a constant acceleration due to gravity in the down ward direction.

             In the context, Quinn shoots two particle x and y from his sling shot and he observes that both his projectiles travels in a parabola curve in the air. Both the object x and y touches the ground a distance apart from him which is known as the range and it depends upon the velocity of the projectile. Both the projectile reaches a maximum height and then drop on the ground in a parabola shape.

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2 years ago
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A force of 250 N is applied to a hydraulic jack piston that is 0.02 m in diameter. If the piston that supports the load has a di
Vikentia [17]

Answer:

option B

Explanation:

given,

Force exerted by the hydraulic jack piston = F₁ = 250 N

diameter of piston, d₁ = 0.02 m

                                r₁ = 0.01 m

diameter of second piston,  d₂ = 0.15 m

                                r₂ = 0.075 m

mass of the jack to lift = ?

now,

    \dfrac{F_1}{A_1} =\dfrac{F_2}{A_2}

    \dfrac{250}{\pi r_1^2} =\dfrac{F_2}{\pi r_2^2}

    \dfrac{250}{0.01^2} =\dfrac{F_2}{0.075^2}

    F_2= \dfrac{250}{0.01^2}\times {0.075^2}

               F₂ = 14062.5 N

F = m g

m = \dfrac{F}{g}

m = \dfrac{14062.5}{9.8}

m = 1435 Kg

hence, the correct answer is option B

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nataly862011 [7]

Answer:

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Un tubo de acero de 40000 kilómetros forma un anillo que se ajusta bien a la circunferencia de la tierra. Imagine que las person
Darina [25.2K]

Answer:

82.76m

Explanation:

In order to find the distance of the steel ring to the ground, when its temperature has raised by 1°C, you first calculate the radius of the steel tube before its temperature increases.

You use the formula for the circumference of the steel ring:

C=2\pi r    (1)

C: circumference of the ring = 40000 km = 4*10^7m (you assume the circumference is the length of the steel tube)

you solve for r in the equation (1):

r=\frac{C}{2\pi}=\frac{4*10^7m}{2\pi}=6,366,197.724m

Next, you use the following formula to calculate the change in the length of the tube, when its temperature increases by 1°C:

L=Lo[1+\alpha \Delta T]         (2)

L: final length of the tube = ?

Lo: initial length of the tube = 4*10^7m

ΔT = change in the temperature of the steel tube = 1°C

α: thermal coefficient expansion of steel = 13*10^-6 /°C

You replace the values of the parameters in the equation (2):

L=(4*10^7m)(1+(13*10^{-6}/ \°C)(1\°C))=40,000,520m

With the new length of the tube, you can calculate the radius of a ring formed with the tube. You again solve the equation (1) for r:

r'=\frac{C}{2\pi}=\frac{40,000,520m}{2\pi}=6,366,280.484m

Finally, you compare both r and r' radius:

r' - r = 6,366,280.484m - 6,366,197.724m = 82.76m

Hence, the distance to the ring from the ground is 82.76m

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A bicyclist travels the first 800 m of a trip 1.4 minutes, the next 500 m in 1.6 minutes, and finishes up the final 1200 m in 2
soldier1979 [14.2K]

Answer:

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Explanation:

Average speed is defined

         v = (x_{f} -x₀) / Δt

let's look in each section

section 1

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section 2

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section 3

distance interval 1200 m in a time 2 min

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taking the speed of each section we can calculate the average speed

         

the distance traveled

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        Δx = 2500 m

the time spent

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        Δt = 5 min

         v_average = Δx / Δt

         v_average = 2500/5

         v_average = 500 m / min

7 0
2 years ago
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