The science of life is

2 answers:

8 0

The sciences concerned with the study of living organisms, including physiology, biology, botany, zoology, biochemistry, microbiology, and related subjects.

marin [14]3 years ago

3 0

The answer is Biology.

You might be interested in

A crate pushed along the floor with velocity vâ i slides a distance d after the pushing force is removed. if the mass of the cra

Alenkinab [10]

d The mass is doubled which means that both the momentum and kinetic energy are also doubled. Also the normal force that's acting along with the coefficient of kinetic friction is also doubled. So the friction that's working to slow down the crate is doubled. So the crate will have double the kinetic energy that needs to be dissipated, but the rate of dissipation is also doubled, so the total time required to dissipate the kinetic energy is the same. And since both crates start out with the same velocity and since they'll lose energy (and velocity) at the same proportional rate, they'll take the same distance to slide to a stop.

7 0

3 years ago

Spitting cobras can defend themselves by squeezing muscles around their venom glands to squirt venom at an attacker. Suppose a s

Alenkasestr [34]

Answer: 1.289 m

Explanation:

The path the cobra's venom follows since it is spitted until it hits the ground, is described by a parabola. Hence, the equations for parabolic motion (which has two components) can be applied to solve this problem:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$x$ is the horizontal distance traveled by the venom

$V_{o}=3.10 m/s$ is the venom's initial speed

$\theta=47\°$ is the angle

$t$ is the time since the venom is spitted until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=0.44 m$ is the initial height of the venom

$y=0$ is the final height of the venom (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Let's begin with (2) to find the time it takes the complete path:

$0=0.44 m+3.10 m/s sin\theta(47\°)+\frac{-9.8m/s^{2} t^{2}}{2}$ (3)

Rewritting (3):

$-4.9 m/s^{2} t^{2} + 2.267 m/s t + 0.44 m=0$ (4)

This is a quadratic equation (also called equation of the second degree) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula: