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2 years ago

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Physics

1 answer:

Zigmanuir [339]2 years ago

6 0

Answer:

it’s 2.5 m/s

Explanation:

i’m too lazy but trust

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A person kicks a ball, giving it an initial velocity of 20.0 m/s up a wooden ramp. When the ball reaches the top, it becomes air

Alex Ar [27]

Answer:

(a) Height is 4.47 m

(b) Height is 4.37 m

Solution:

As per the question:

Initial velocity of teh ball, $v_{o} = 20.0 m/s$

Angle made by the ramp, $\theta = 22.0^{\circ}$

Distance traveled by the ball on the ramp, d = 5.00 m

Now,

(a) At any point on the projectile before attaining maximum height, the velocity can be given by the eqn-3 of motion:

$v^{2} = v_{o}^{2} - 2gH$

where

H = $dsin22^{\circ} = 5sin22^{\circ}$

g = $9.8 m/s^{2}$

$v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ}$

$v = \sqrt{400 - 19.6\times 5sin22^{\circ}}$ = 19.06 m/s

Now, maximum height attained is given by:

$h = \frac{(vsin\theta)^{2}}{2g}$

$h = \frac{(19sin(22^{\circ}))^{2}}{2\times 9.8} = 2.60 m$

Height from the ground = $5sin22^{circ} + 2.86 = 1.87 + 2.60 = 4.47m$

(b) now, considering the coefficient of friction bhetween ramp and the ball, $\mu = 0.150$ :

velocity can be given by the eqn-3 of motion:

$v^{2} = v_{o}^{2} - 2gH - \mu gd$

$v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ} - 0.150\times 9.8\times 5$

$v = \sqrt{400 - 19.6\times 5sin22^{\circ} - 0.150\times 9.8\times 5}$ = 18.7 m/s

Now, maximum height attained is given by:

$h = \frac{(vsin\theta)^{2}}{2g}$

$h = \frac{(18.7sin(22^{\circ}))^{2}}{2\times 9.8} = 2.50 m$

Height from the ground = $5sin22^{circ} + 2.86 = 1.87 + 2.50 = 4.37 m$

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2 years ago

Why scientists collect data about the moon?

Strike441 [17]

Explanation:

Scientists collect data for decades after Apollo 11's return to earth. Finding include that the moon is moving farther away from earth and that the universal force of gravity is stable.

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3 years ago

1. Each of four boys were given

GalinKa [24]

Answer:

the answer is A because

from tate 4 dozen is 48 and from joe the sixth multiple of eight is 48

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2 years ago

A leaky 10-kg bucket is lifted from the ground to a height of 10 m at a constant speed with a rope that weighs 0.9 kg/m. Initial

sladkih [1.3K]

Answer:

The work done shall be 14715 Joules

Explanation:

The work done by a force 'F' in a displacement 'dy' is given by

$W=m(y)g\times dy$

At any position 'y' the weight shall be sum of weft of water and weight of string

$\therefore m(y)=m_{water}(y)+m_{string}(y)\\\\m(y)=30(1-\frac{y}{10})+0.9y$

Thus applying values we get

$W=\int m(y)g\times dy\\\\W=\int_{0}^{10}(30(1-\frac{y}{10}+0.9y)\times g)dy\\\\Solving\\W=294.3\int_{0}^{10}(1-\frac{y}{10}+0.9y)dy\\\\W=14715J$

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2 years ago

In the first law of Thermodynamics ΔE = Q - W, what does ΔE stand for???

Alexxx [7]

<span>Δ</span>E = q + w

q = heat (quantity of)

q and w can be positive or negative depending on if work/heat is being absorbed/done on the system or released/done by the system

7 0

2 years ago

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