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ICE Princess25 [194]

3 years ago

5

When a quarterback throws a football, the throwing arm is slowed down to rest during the deceleration phase. Of the angular velo

city of the throwing arm is 1800 deg/sec at the instant of ball release, then how much would the arm deceleration change if the deceleration was reduced from 2 sec to 0.5 sec

1 answer:

asambeis [7]3 years ago

3 0

Answer:

The deceleration is 4 times higher for 0.5sec than for 2 sec.

Explanation:

If the deceleration happens from 1800deg/sec to zero in 2 sec we have that the magnitude is:

$a_{1}=w/t_{1}=(1800deg/sec)/2sec=900 deg/sec^{2}$

if now the time to decelerate is 0.5sec:

$a_{2}=w/t_{2}=(1800deg/sec)/0.5sec=3600 deg/sec^{2}$

The deceleration is now 4 times higher than the previous situation.

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A rocket moves upward, starting from rest with an acceleration of +30.0 m/s2 for 5.00 s. It runs out of fuel at the end of this

olchik [2.2K]

Answer:

The distance covered by the rocket after fuel ran out is $3442.04 m$

Explanation:

Given that the rocket moves with an acceleration $a=30m/s^2$

time $t=5 s$

Since the rocket starts from rest initial velocity $u=0 s$

The distance it travelled within this time is given by $s=ut+ \frac{1}{2} at^2$ $=0 \times 5+ \frac{1}{2} (30\times25)=375 m$

Velocity at this point is given by $v=u+at$

$v=0+30\times5=150m/s$

Given that at this height it runs out of fuel but travels further. Here final velocity $v=0$ (maximum height), initial velocity $u=150 m/s$ and time to zero velocity $t=\frac{v}{g} = \frac{150}{9.8} =15.3 s.$

Thus it travels $15.3 seconds$ more after fuel running out. The distance covered during this period is given

$s= ut+\frac{1}{2} gt^2=150 \times 15.3+1/2 \times9.8 \times 15.3^2=3442.04 m$

7 0

3 years ago

A uniform rod of length 50cm and mass 0.2kg is placed on a fulcrum at a distance of 40cm from the left end of the rod. At what d

oksano4ka [1.4K]

Answer:

x = 45 cm

Explanation:

Given that,

The length of a rod, L = 50 cm

Mass, m₁ = 0.2 kg

It is at 40cm from the left end of the rod.

We need to find the distance from the left end of the rod should a 0.6kg mass be hung to balance the rod.

The centre of mass of the rod is at 25 cm.

Taking moments of both masses such that,

$15\times 0.2=x\times 0.6\\\\x=\drac{3}{0.6}\\\\x=5\ cm$

The distance from the left end is 40+5 = 45 cm.

Hence, at a distance of 45 cm from the left end it will balance the rod.

5 0

3 years ago

Dry air is primarily composed of nitrogen. In a classroom demonstration, a physics instructor pours 3.6 L of liquid nitrogen int

neonofarm [45]

Answer:

The value is $V_n = 2.2498 \ m^3$

Explanation:

From the question we are told that

The volume of liquid nitrogen is $V_n = 3.6 \ L= 3.6 *10^{-3} \ m^3$

The density of nitrogen at gaseous form is $\rho_n = 1.2929 \ kg/m^3$ = The dry air at sea level

Generally the density of nitrogen at liquid form is

$\rho _l = 808 \ kg/m^3$

And this is mathematically represented as

$\rho_l = \frac{m}{V_l }$

=> $m = \rho_l * V_l$

Now the density of gaseous nitrogen is

$\rho_n = \frac{m}{V_n }$

=> $m = \rho_n * V_n$

Given that the mass is constant

$\rho_n * V_n = \rho_l * V_l$

$1.2929* V_n = 808 * 3.6*10^{-3}$

=> $V_n = 2.2498 \ m^3$

3 0

3 years ago

Can someone explain please <br> ???

Radda [10]

Answer: A

Explanation:

6 0

3 years ago

Please help with this and explain it,if you can.

vladimir2022 [97]

Answer:

displacement at 45 s = 30

65 s = 50

So the average speed over the interval from 45 s to 65 s is

(50 - 30) cm / 20 s = 1 cm / sec

As a check an average speed of 1 cm / sec for 20 sec will produce a

displacement of 1 cm / sec * 20 sec = 20 cm or from 30 to 50 cm

4 0

3 years ago