Sultan throws a ball horizontally from his window, 12 m above the garden. It reaches the ground after

An airplane accelerates from rest down a runway at 3.20 m/s2 for 32.8 seconds until it lifts off the ground. Determine the dista

vagabundo [1.1K]

Answer:

s=1721.344m ,v=104.96m/s.

Explanation:

using thr equation of motion;

$s=ut+\frac{1}{2}at^{2}$

u=0, plane starts from rest,

$s=\frac{1}{2}at^{2}$

$a=3.2m/s^{2}, t=32.8s \\ s=\frac{1}{2}*3.2*32.8^{2}$

s=1721.344m

v=u+at

v=0 +3.2*32.8

v=104.96m/s

8 0

3 years ago

A bucket of water with total mass 23 kg is attached to a rope, which in turn is wound around a 0.050-m radius pulley at the top

adoni [48]

Answer:

$I = 0.083 kg m^2$

Explanation:

Mass of the bucket, m = 23 kg

Radius of the pulley, r = 0.050 m

The bucket is released from rest, u = 0 m/s

The time taken to fall, t = 2 s

Speed, v = 8.0 m/s

Moment of Inertia of the pulley, I = ?

Using the equation of motion:

v = u + at

8 = 0 + 2a

a = 8/2

a = 4 m/s²

The relationship between the linear and angular accelerations is given by the equation:

$a = \alpha r$

Angular acceleration, $\alpha = a/r$

$\alpha = 4/0.050\\\alpha = 80 rad/s^2$

Since the bucket is falling, it can be modeled by the equation:

mg - T = ma

T = mg - ma = m(g-a)

T = 23(9.8 - 4)

The tension, T = 133.4 N

The equation for the pulley can be modeled by:

$T* r = I * \alpha\\133.4 * 0.050 = I * 80\\6.67 = 80 I\\I = 6.67/80\\I = 0.083 kg m^2$

6 0

3 years ago

A certain radioactive nuclide decays with a disintegration constant of 0.0178 h-1.

umka21 [38]

Explanation:

Given that,

The disintegration constant of the nuclide, $\lambda=0.0178\ h^{-1}$

(a) The half life of this nuclide is given by :

$t_{1/2}=\dfrac{ln(2)}{\lambda}$

$t_{1/2}=\dfrac{ln(2)}{0.0178}$

$t_{1/2}=38.94\ h$

(b) The decay equation of any radioactive nuclide is given by :

$N=N_oe^{-\lambda t}$

$\dfrac{N}{N_o}=e^{-\lambda t}$

Number of remaining sample in 4.44 half lives is :

$t_{1/2}=4.44\times 38.94$

$t_{1/2}=172.89\ h^{-1}$

So, $\dfrac{N}{N_o}=e^{-0.0178\times 172.89}$

$\dfrac{N}{N_o}=0.046$

(c) Number of remaining sample in 14.6 days is :

$t_{1/2}=14.6\times 24$

$t_{1/2}=350.4\ h^{-1}$

So, $\dfrac{N}{N_o}=e^{-0.0178\times 350.4}$

$\dfrac{N}{N_o}=0.0019$

Hence, this is the required solution.

4 0

4 years ago

Can anyone help me with questions a and c

Ganezh [65]

Answer:

a)

n=sin i/sin r

n= -0.305/-0.428

n=0.713

b)

sin c=1/n

sin c=1/0.713

sin c= 1.403

c=sin⁻¹(1.403)

c= 40.842°

Explanation:

i hope it will be helpful

plzzz mark as brainliest

5 0

3 years ago

Jupiter has approximately 100 times the volume of earth?<br><br> True<br><br> False

seropon [69]

False this is incorrect because Jupiter is 1,300 the volume of earth.

6 0

3 years ago

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