How can we drop eggs the fewest amount of times without breaking it?

I need help with 23 please.

Ne4ueva [31]

The answer is c.

hope this helps! :)

8 0

3 years ago

A speedboat increases its speed from 14.5 m/s to 29.3 m/s in a distance of 172 m.

dalvyx [7]

Answer:

a. Acceleration, a = 1.88 m/s²

b. Time, t = 7.87 seconds.

Explanation:

Given the following data;

Initial velocity, U = 14.5m/s

Final velocity, V = 29.3m/s

Distance, S = 172m

a. To find the acceleration of the speedboat;

We would use the third equation of motion;

V² = U² + 2aS

Substituting into the formula

29.3² = 14.5² + 2a*172

858.49 = 210.25 + 344a

344a = 858.49 - 210.25

344a = 648.24

a = 648.24/344

Acceleration, a = 1.88 m/s²

b. To find the time;

We would use the first equation of motion;

V = U + at

29.3 = 14.5 + 1.88t

1.88t = 29.3 - 14.5

1.88t = 14.8

Time, t = 14.8/1.88

Time, t = 7.87 seconds.

6 0

3 years ago

A ball is thrown horizontally from the top of a building 0.10 km high. The ball strikes the ground at a point 65 m horizontally

aivan3 [116]

Answer:

option B

Explanation:

given,

height of building = 0.1 km

ball strikes horizontally to ground at = 65 m

speed at which the ball strike = ?

vertical velocity = 0 m/s

time at which the ball strike

$s = \dfrac{1}{2}gt^2$

$t = \sqrt{\dfrac{2s}{g}}$

$t = \sqrt{\dfrac{2\times 100}{9.8}}$

t = 4.53 s

vertical velocity at the time 4.53 s = g × t = 9.8 × 4.53 = 44.39 m/s

horizontal velocity = $\dfrac{65}{4.53}$ =14.35 m/s

speed of the ball = $\sqrt{44.39^2+14.35^2}$

= 46.65 m/s

hence, the speed of the ball just before it strike the ground = 47 m/s

The correct answer is option B

5 0

2 years ago

a baseball is hit 3 feet above ground level at 100 feet per second and at an angle of 45 with respect to the ground. (g=32 feet/

LiRa [457]

Answer:

hmax=81ft

Explanation:

Maximum height of the object is the highest vertical position along its trajectory.

The vertical velocity is equal to 0 (Vy = 0)

$0=V_{y}-g*t=v_{0}*sin(\alpha)-g*th\\$

we isolate th (needed to reach the maximum height hmax)

$th = \frac{v_{0}*sin(\alpha)}{g}$

The formula describing vertical distance is:

$y = Vy * t-g* t^{2} / 2$

So, given y = hmax and t = th, we can join those two equations together:

$hmax = Vy* th-g*th^{2}/2$

$hmax =Vo^{2}*sin(\alpha )^{2}/(2*g)$

if we launch a projectile from some initial height h all you need to do is add this initial elevation

$hmax =h+Vo^{2}*sin(\alpha)^{2}/(2*g)$

$hmax =3+100^{2}*sin(45)^{2}/(2 * 32)=81 ft$

6 0

3 years ago

Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is

Nataly [62]

Answer:

Explanation:

Calculating the exit temperature for K = 1.4

The value of $c_p$ is determined via the expression:

$c_p = \frac{KR}{K_1}$

where ;

R = universal gas constant = $\frac{8.314 \ J}{28.97 \ kg.K}$

k = constant = 1.4

$c_p = \frac{1.4(\frac{8.314}{28.97} )}{1.4 -1}$

$c_p= 1.004 \ kJ/kg.K$

The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :

$0=(h_1-h_2)+\frac{(v_1^2-v_2^2)}{2}$ ------ equation(1)

we can rewrite the above equation as :

$0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}$

$T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}$

where:

$T_1 = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K$

$T_2= 280+\frac{((510)^2-(120)^2)}{2(1.004)} *\frac{1}{10^3}$

$T_2 = 402.36 \ K$

Thus, the exit temperature = 402.36 K

The exit pressure is determined by using the relation: $\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}$

$P_2=P_1(\frac{T_2}{T_1})^\frac{k}{k-1}$

$P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}$

$P_2 = 17.79 \ bar$

Therefore, the exit pressure is 17.79 bar

7 0

3 years ago

How can we drop eggs the fewest amount of times without breaking it?​

How can we drop eggs the fewest amount of times without breaking it?