The first space craft to drop a weather probe into jupiter's atmosphere was

A small racquet ball launcher is set up 5 meters from a 12 meter tall wall. It launches a racket ball at 30 m/s at an angle of 4

Strike441 [17]

Answer:20.82 m

Explanation:

Given

distance between wall and launcher=5 m

initial velocity=30 m/s

Launching angle $=45^{\circ}$

Height of wall=12 m

maximum height by ball

$h_{max}=\frac{u^2sin^2\theta }{2g}$

$h_{max}=\frac{30^2sin^{2}45}{2\times 9.8}$

h=22.95 m

$y=xtan\theta -\frac{gx^2}{2u^2cos^\theta }$

y=4.72 m

so ball will strike with wall

and for perfect restitution final velocity of ball will be same as the horizontal velocity before its impact, only direction will be opposite and vertical velocity will be zero

thus it seems as if someone throw the ball with horizontal velocity of 30cos45 from a height of 4.72

Time required to cover 4.72 m

$t=\sqrt{\frac{2h}{g}}$

$t=\sqrt{\frac{9.45}{9.8}}$

t=0.981 s

Horizontal distance traveled in this time

$R=ucos45\times t$

$R=30\times cos45\times 0.981$

R=20.82 m

6 0

3 years ago

A car drives at a constant speed of 3m/s around a circle with radius 20m. What is the centripetal acceleration of the car

Kaylis [27]

Centripetal acceleration is the motion inwards towards the center of a circle. It is given by the square of velocity, divided by the radius of the circular path.
ac = v²/r, where ac is the centripetal acceleration in m/s²
Therefore,
Centripetal acceleration = 3²/20m
= 9/20m
= 0.45 m/s²
Thus, the centripetal acceleration is 0.45 m/s²

6 0

3 years ago

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Aluminum rivets used in airplane construction are made slightly larger than the rivet holes and cooled by "dry ice" (solid CO2)

poizon [28]

Answer:

$\rm 4.554\ m m.$

Explanation:

<u>Given:</u>

Initial temperature, $\rm T_i=23.0\ ^\circ C=23 + 273.15 = 296.15\ K.$
Final temperature, $\rm T_f = -78.0\ ^\circ C = -78+273.15=195.15\ K.$
Diameter of the hole, $\rm d = 4.500\ m = 4.500\times 10^{-3}\ m.$
Expansion coefficient of the Aluminum, $\rm \alpha = 2.4\times 10^{-5}\ K^{-1}.$

Let the diameter of the rivet at $\rm T_i=23.0\ ^\circ C$ be $\rm d_o$ , such that,

$\rm d_o=d+\Delta d$

$\rm \Delta d$ is the elongation in the diameter of the rivet.

We know, The change in the diameter of the rivet is related with the change in temperature as:

$\rm \dfrac{\Delta d}{d_o}=\alpha \Delta T.$

where, $\rm \Delta T$ is the change in temperature = $\rm T_f-T_i=-195.15-296.15=-491.30\ K.$

Also, $\rm \Delta d=d_o-d.$

Using these values,

$\rm \dfrac{d_o-d}{d_o}=\alpha \Delta T\\1-\dfrac{d}{d_o}=\alpha \Delta T\\\dfrac{d}{d_o}=1+\alpha \Delta T\\\Rightarrow d_o = \dfrac{d}{1+\alpha \Delta T}\\=\dfrac{4.500\times 10^{-3}}{1+2.4\times10^{-5}\times (-491.30)}\\=4.554\times 10^{-3}\ m.\\=4.554\ m m.$