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3 years ago

The small intestine _____.

Physics

2 answers:

icang [17]3 years ago

6 0

Answer:

digests food

Explanation:

VladimirAG [237]3 years ago

3 0

Answer:

the small intestine digest food

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In a physics lab experiment, a compressed spring launches a 24 g metal ball at a 35o angle above the horizontal. Compressing the

Levart [38]

Answer:

k = 45.95 N/m

Explanation:

First, we will find the launch speed of the ball by using the formula for the horizontal range of the projectile.

$R = \frac{v_{o}^{2}\ Sin\ 2\theta}{g} \\\\v_{o}^{2} = \frac{Rg}{Sin\ 2\theta}\\$

where,

Vo = Launch Speed = ?

R = Horizontal Range = 5.3 m

θ = Launch Angle = 35°

Therefore,

$v_{o}^{2} = \frac{(5.3\ m)(9.81\ m/s^{2})}{Sin\ 2(35^{o})}\\$

v₀² = 55.33 m²/s²

Now, we know that the kinetic energy gain of ball is equal to the potential energy stored by spring:

$Kinetic\ Energy\ Gained\ By\ Ball = Elastic\ Potential\ Energy\ Stored\ in \ Spring\\\frac{1}{2}mv_{o}^{2} = \frac{1}{2}kx^{2}\\\\k = \frac{mv_{o}^{2}}{x^2} \\$

where,

k = spring constant = ?

x = compression = 17 cm = 0.17 m

m = mass of ball = 24 g = 0.024 kg

Therefore,

$k = \frac{(0.024\ kg)(55.33\ m^2/s^2)}{(0.17\ m)^2} \\$

4 0

3 years ago

When Arti kicks a football, two forces interact. Arti's foot exerts a force on the ball. What exerts a force on Arti's foot?

vagabundo [1.1K]

The football and air resistance between contact

7 0

3 years ago

A diver 40 m deep in 10 degrees C fresh water exhales a 1.5 cm diameter bubble.

zzz [600]

Answer:

0.0257259766982 m

Explanation:

$P_2$ = Atmospheric pressure = 101325 Pa

$d_1$ = Initial diameter = 1.5 cm

$d_2$ = Final diameter

$\rho$ = Density of water = 1000 kg/m³

h = Depth = 40 m

The pressure is

$P_1=P_2+\rho gh\\\Rightarrow P_1=101325+1000\times 9.81\times 40\\\Rightarrow P_1=493725\ Pa$

From ideal gas law we have

$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow \dfrac{P_1\dfrac{4}{3\times8}\pi d_1^3}{T_1}=\dfrac{P_2\dfrac{4}{3\times8}\pi d_2^3}{T_2}\\\Rightarrow \dfrac{P_1d_1^3}{T_1}=\dfrac{P_2d_2^3}{T_2}\\\Rightarrow d_2=(\dfrac{P_1d_1^3T_2}{P_2T_1})^{\dfrac{1}{3}}\\\Rightarrow d_2=(\dfrac{493725\times 0.015^3\times (20+273.15)}{101325\times (10+273.15)})^{\dfrac{1}{3}}\\\Rightarrow d_2=0.0257259766982\ m$

The diameter of the bubble is 0.0257259766982 m

8 0

3 years ago

A coin is dropped into a wishing well. It takes 1.1 seconds for a splash to be heard. Calculate the depth of the wishing well

Zigmanuir [339]

Answer:

If the wishing well was in a vacuum, then s=ut + 0.5 a t^2 (s=distance, ... wishing well if you drop a coin into it and hear the splash 10 seconds

Explanation:

8 0

3 years ago

You have probably noticed that carrying a person in a pool of water is much easier than carrying a person through air. To unders

Kruka [31]

Answer:

The answer is "0.91238 and 744.8"

Explanation:

In this scenario it is easier to take a person to the water-pool than to transport the people in the air, as the person's strength is increased by water upwards:

$f_b \to m \to mg =person \\\\F_B \ in\ air = v\ & air\ g \\\\$

$=0.076 \times 1.225 \times 9.8 \\\\ =0.91238 \ N\\\\$

$F_B \ in \ water = v \& water \ g \\\\$

$=0.076 \times 1000 \times 9.8\\\\= 744.8 \ N\\$

6 0

3 years ago