All categories

3 years ago

A projectile is fired at 0=433.0 m/s

Physics

1 answer:

White raven [17]3 years ago

5 0

Answer:

10,600 m

Explanation:

Use range equation:

R = v₀² sin(2θ) / g

R = (433.0 m/s)² sin(2 × 64.8°) / (9.8 m/s²)

R = 14,700 m

The range is shortened by 28.1%.

0.719R = 10,600 m

You might be interested in

It is impossible to destroy __________.

fredd [130]

C.energy
The law of conservation of energy says that energy can neither be created nor destroyed.

6 0

3 years ago

Read 2 more answers

A 67.5-kg person throws a 0.0410-kg snowball forward with a ground speed of 34.0 m/s. A second person, with a mass of 57.5 kg, c

lora16 [44]

Answer:

Final velocity of the first person is 3.43m/s and that of the second person is 0.0242m/s

Explanation:

Let the momentum of the first person, the ball second person be Ma, Mb and Mc.

From the principle of the conservation of momentum, sum of the momentum before collision is equal to the sum of the momentum after collision.

Ma1 + Mb1 = Ma2 + Mb2.

The ball and the first person are both moving together with a common velocity 3.45m/s.

Let the velocity of the first person be v1

Therefore

67.5×3.45+ 0.041×3.45= 67.5v1 + 0.041×34

233.02 = 1.39+ 67.5v1

67.5v1 = 233.02 - 1.39 = 231.61

v1 = 231.61 / 67.5

v1 = 3.43m/s

The second person and the ball move together with a common velocity after catching the ball.

For the second person and the ball let their final common velocity be v

Mb2 + Mc2 = Mb3 + Mc3

0.041 × 34 + 57.5 ×0 = (57.5 + 0.041)×v

57.541v = 1.39

v = 1.39 /57.541

v = 0.0242m/s

5 0

4 years ago

If the coefficient of static friction between your coffee cup and the horizontal dashboard of your car is µs = 0.800, how fast c

Cloud [144]

Answer:

Before start of slide velocity will be 14.81 m/sec

Explanation:

We have given coefficient of static friction $\mu =0.8$

Angle of inclination is equal to $\Theta =tan^{-1}\mu$

$\Theta =tan^{-1}0.8=38.65^{\circ}$

$tan{38.65^{\circ}}=0.8$

Radius is given r = 28 m

Acceleration due to gravity $g=9.8m/sec^2$

We know that $tan\Theta =\frac{v^2}{rg}$

$0.8=\frac{v^2}{28\times 9.8}$

$v^2=219.52$

$v=14.816m/sec$

So before start of slide velocity will be 14.81 m/sec

3 0

3 years ago

Water drops from the nozzle of a shower onto the floor 81 inches below. The drops fall at regular intervals of time, the first d

Alexxx [7]

Answer:

0.91437 m

0.22859 m

Explanation:

g = Acceleration due to gravity = 9.81 m/s² = a

$s=81\ inches=81\times 0.0254=2.0574\ m$

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2.0574=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.0574\times 2}{9.81}}\\\Rightarrow t=0.64764\ s$

When the time intervals are equal, if four drops are falling then we have 3 time intervals.

So, the time interval is

$t'=\dfrac{t}{3}\\\Rightarrow t'=\dfrac{0.64764}{3}\\\Rightarrow t'=0.21588\ s$

For second drop time is given by

$t''=2t'\\\Rightarrow t''=2\times 0.21588\\\Rightarrow t''=0.43176\ s$

Distance from second drop

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut''+\dfrac{1}{2}at''^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.43176^2\\\Rightarrow s=0.91437\ m$

Distance from second drop is 0.91437 m

Distance from third drop

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut'+\dfrac{1}{2}at'^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.21588^2\\\Rightarrow s=0.22859\ m$

Distance from third drop is 0.22859 m

6 0

3 years ago

Completeaza urmatoarul enunt.<br> La ... unui corp moleculele lui se misca mai incet.

SashulF [63]

Translation.............

5 0

3 years ago