Organ system that removes waste from your body

A ball is dropped from a height of 80m. The time, in seconds, it takes to reach the ground is:__________.

Aleonysh [2.5K]

Answer:

4.04 s

Explanation:

h = vi + 1/2 a t ^2

HERE h = 80 m , vi = 0 , a =9.81 m/s^2

80 = 0 + 1/2 × 9.81 × t ^2

80 = 4.905 t^2

t^2 = 80/4.905

t ^2 = 16.30988

t = square root of 16.30988

t = 4.0385 s

t = 4.04 s

4 0

3 years ago

Can someone tell me the answer to this?

larisa86 [58]

Answer:

<h2>yes </h2><h3>bt pic nyi clear is Lia pata nyi lag taha</h3>

3 0

3 years ago

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A catapult with a radial arm 3.81 m long accelerates a ball of mass 18.2 kg through a quarter circle. The ball leaves the appara

saw5 [17]

Answer:

(a) $\alpha = 53.73 m/s^2$

(b) I =428 $kgm^2$

(c) $\tau = 428 \times 53.73 = 22996 .44Nm$

Explanation:

GIVEN

mass = 18.2 kg

radial arm length = 3.81 m

velocity = 49.8 m/s

mass of arm = 22.6 kg

we know using relation between linear velocity and angular velocity

$\omega = \frac{v}{l}$

$\omega = \frac{49.8}{3.81} \\\omega = 12.99 rad/s$

for angular acceleration, use the following equation.

$\omega _{f}^2 = \omega_{i}^2+2\alpha\theta$

since $\omega _{i} = 0$

here for one circle is 2 π radians. therefore for one quarter of a circle is π/2 radians

so for one quarter $\theta = \frac{\pi }{2}$

$(12.99)^2 = 2\alpha(\frac{\pi }{2})$

on solving

$\alpha = \frac{168.74}{\pi }\\\alpha = 53.73 m/s^2$

(b)

For the catapult,

moment of inertia

$I = \frac{1}{2}MR^2$

$I = \frac{1}{2} \times 22.6\times 3.81 \times 3.81\I = 164kg m^2$

For the ball,

$I = MR^2$

$I = 18.2 \times 14.51$

$I = 264 kgm^2$

so total moment of inertia = 428 $kgm^2$

(c)

$\tau = I\alpha$

$\tau = 428 \times 53.73 = 22996 .44Nm$

3 0

4 years ago

A car travels from Boston to Hartford in 4 hours. The two cities are 240 kilometers apart. What was the average speed of the car

Y_Kistochka [10]

Answer:

Average speed is 60 km/hour

Explanation:

When we need to calculate average speed, we use this equation:

$V = \frac{x_{f} - x_{o}}{t_{f} - t_{o}}$

Where: $x_{o} = 0 km$ position at the beginning

$x_{f} = 240 km$ at the end

$t_{o} = 0 hours$

$t_{f} = 4 hours$

Then: $V = \frac{240 km - 0 km}{4 hours - 0 hours}$

$V = \frac{240 km}{4 hours}$

Finally V = 60 km/hour

6 0

3 years ago

Suppose you are holding a basketball while standing still on a skateboard. You and the skateboard have a mass of 50kg. You throw

Nutka1998 [239]

Your acceleration before you throw the ball is zero. After the throw, you use the equation

F = m * a 

To solve for acceleration 

a = F/m 

a = (10 N) / (50 kg) 

a = 0.2 m/s^2

4 0

4 years ago

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