Ammonia and _________ are waste products of animal cells. A. oxygen B. urine C. hydrogen D. carbon dioxide

A projectile is launched into the air with the initial speed of vi = 40 m/s at a launch angle of 20 degrees above the horizontal

Sphinxa [80]

The range of the projectile is 188 m

Explanation:

The motion of the arrow in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

The path of a projectile is the combination of these two motions: see figure in attachment.

In order to find the horizontal range of the projectile, we just need to calculate the horizontal distance travelled.

We have:

t = 5.0 s (time of fligth of the projectile)

and the horizontal velocity is constant, and it is given by

$v_x = v_i cos \theta$

where

$v_i = 40 m/s$ is the initial velocity

$\theta=20^{\circ}$ is the angle of projection

Substituting,

$v_x = (40)(cos 20^{\circ})=37.6 m/s$

And therefore, the range of the projectile is:

$d=v_x t = (37.6)(5.0)=188 m$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

5 0

3 years ago

What is the relationship between the strength of a bond (single vs double vs triple) and its wave-number on an IR spectrum? As t

Elanso [62]

<u>Explanation</u>

The relationship between the strength of a bond (single vs double vs triple) and its wave-number on an IR spectrum as the bond strength increases the wave number increases.

STRENGTH OF BONDS TRIPLE>DOUBLE>SINGLE

WAVE NUMBER SINGLE>DOUBLE>TRIPLE

wave number for single bond is greatest because it has greatest bond frequency among the three( more the frequency greater is the wave number).

8 0

3 years ago

A far-sighted person has a near-point of 80 cm. To correct their vision so that they can see objects that are as close as 10 cm

bekas [8.4K]

Answer:

$f = 8.89 cm$

Explanation:

As we know that Far sighted person has near point shifted to 80 cm distance

so he is able to see the object 80 cm

now the distance of lens from eye is 2 cm

and the person want to see the objects at distance 10 cm

so here the image distance from lens is 80 cm and the object distance from lens is 8 cm

now from lens formula we have

$\frac{1}{d_i} + \frac{1}{d_0} = \frac{1}{f}$

$-\frac{1}{80} + \frac{1}{8} = \frac{1}{f}$

$f = 8.89 cm$

3 0

3 years ago

You pull a rope to the left with 300 N and a friend pulls the rope to the right with 425 N

eimsori [14]

I’m assuming you’re supposed to calculate the resultant force?

425N (right) -300N (left)
=125 N to the right

6 0

3 years ago

Suppose a grower sprays (8.2x10^1) kg of water at 0 °C onto a fruit tree of mass 180 kg. How much heat is released by the water

____ [38]

Since there is no temperature change which drives heat flow, thus no heat will be released by the water.

<h3>Heat released by the water when it freezes</h3>

The heat released by the water when it freezes is calculated as follows;

Q = mcΔФ

where;

m is mass of water
c is specific heat capacity of water
ΔФ is change in temperature = Фf - Фi

when water freezes, the temperature, Фf = 0 °C

Q = 82 x 4200 x (0 - 0)

Q = 0

Since there is no temperature change which drives heat flow, thus no heat will be released by the water.

Learn more about heat flow here: brainly.com/question/14437874

6 0

2 years ago