Which is the product of cellular respiration? A. ATTP B. light C. oxygen D.sugar

When learning a new exercise what is the first thing you should focus on?

Brums [2.3K]

Answer:

If your form is correct (b)

3 0

2 years ago

Read 2 more answers

A ceiling fan with 90-cm-diameter blades is turning at 64 rpm . Suppose the fan coasts to a stop 28 s after being turned off. Wh

mixas84 [53]

Answer:

the speed of the tip of a blade 10 s after the fan is turned off is 16.889 m/s.

Explanation:

Given;

diameter of the ceiling fan, d = 90 cm = 0.9 m

angular speed of the fan, ω = 64 rpm

time taken for the fan to stop, t = 28 s

The distance traveled by the ceiling fan when it comes to a stop is calculated as;

$d = vt = \omega r\times t= ( \frac{64 \ rev}{\min} \times \frac{2 \pi \ rad}{rev} \times \frac{1 \min}{60 \ s} \times 0.9 \ m) \times 28 \ s\\\\d = 168.89 \ m$

The speed of the tip of a blade 10 s after the fan is turned off is calculated as;

$v = \frac{d}{t} \\\\v = \frac{168.89}{10} \\\\v = 16.889 \ m/s$

Therefore, the speed of the tip of a blade 10 s after the fan is turned off is 16.889 m/s.

4 0

3 years ago

A food packet is dropped from a helicopter during a flood-relief operation from a height of 750 meters. Assuming no drag (air fr

Leviafan [203]

1. Velocity at which the packet reaches the ground: 121.2 m/s

The motion of the packet is a uniformly accelerated motion, with constant acceleration $a=g=9.8 m/s^2$ directed downward, initial vertical position $d=750 m$ , and initial vertical velocity $v_0 = 0$ . We can use the following SUVAT equation to find the final velocity of the packet after travelling for d=750 m:

$v_f^2 -v_i^2 =2ad$

substituting, we find

$v_f^2 = 2ad\\v_f = \sqrt{2ad}=\sqrt{2(9.8 m/s^2)(750 m)}=121.2 m/s$

2. height at which the packet has half this velocity: 562.6 m

We need to find the heigth at which the packet has a velocity of

$v_f=\frac{121.2 m/s}{2}=60.6 m/s$

In order to do that, we use again the same SUVAT equation substituting $v_f$ with this value, so that we find the new distance d that the packet travelled from the helicopter to reach this velocity:

$v_f^2-v_i^2=2ad\\d=\frac{v_f^2}{2a}=\frac{(60.6 m/s)^2}{2(9.8 m/s^2)}=187.4 m$

Which means that the heigth of the packet was

$h=750 m-187.4 m=562.6 m$

3 0

3 years ago

Why is Energy not created nor destroyed?<br> (Need the answer ASAP!!)

eduard

Since both heat and work can be measured and quantified, this is the same as saying that any change in the energy of a system must result in a corresponding change in the energy of the surroundings outside the system. In other words, energy cannot be created or destroyed.

4 0

3 years ago

What distance is covered by an airplane traveling at a velocity of 660 miles per hour in 3.5 hours?

N76 [4]

As per the question, the velocity of the airplane [v] = 660 miles per hour.

The total time taken by airplane [t] = 3.5 hours.

We are asked to determine the total distance travelled by the airplane during that period.

The distance covered [ S] by a body is the product of velocity with the time.

Mathematically distance covered = velocity × total time

S = v × t

= 660 miles/hour ×3.5 hours

= 2310 miles.

Hence, the total distance travelled by the airplane in 3.5 hour is 2310 miles.

4 0

3 years ago