How do unicellular organisms reproduce?

South African frogs are capable of jumping as far as 10.0m in one hop. Suppose one of these frogs makes exactly 15 of these jump

kenny6666 [7]

Average speed = (total distance) / (time to cover the distance)

Total distance = (15 x 10m) = 150 m
Time to cover the distance = 60 s

Average speed = (150m) / (60s) = (150/60) m/s = 2.5 m/s

We have no way to describe the frog's velocity, because we have
no information about the direction of any this activity.

3 0

3 years ago

Convection cell that occurs at the lowest latitudes is called what

Gre4nikov [31]

Convection cell that occurs at the lowest latitudes is called the Hadley cells. This is an atmospheric convection cell which is large-scale. The air rises at the equator and sinks at the medium latitudes which are 30 degrees north or south.

7 0

4 years ago

2. If you exert a force of 10.0 N to lift a box a distance of 0.9 m, how much work have you done?

KonstantinChe [14]

<span> force of 10.0 N
</span>
<span>distance of 0.9 m
w=f*d
w=10*0.9
=9.0 j</span>

6 0

3 years ago

Read 2 more answers

An object is released from rest and falls a distance h during the first second of time. How far will it fall during the next sec

Viefleur [7K]

Answer:

E. 3h

Explanation:

We know that

u = 0 m/s.

velocity after t = 1s

v = u+gt = 0+9.81 x 1s= 9.81 m/s

distance covered in 1st sec

= =>> ut+0.5 x g x t²

=>>0 + 0.5x 9.81 x 1 = 4.90m

Let 4.90 be h

distance travelled in 2nd second will now be used

So velocity after t = 1s

=>>1 x t+ 0.5 x g x t²

=>9.81x 1 + 0.5 x 9.81 x 1 = 3 x 4.90

So since h= 4.90

Then the ans is 3x h = 3h

3 0

3 years ago

An ultrasound unit is being used to measure a patient's heartbeat by combining the emitted 2.0 MHz signal with the sound waves r

alexandr402 [8]

Hi there,
for this question we have:
Signal 2.0 MHz = Emitted so we can call it $f_e$
and we need the Reflected = $f_{r}$
In this question, we have a source which goes to the heart and a reflected which comes back from the heart and we need the speed of the reflected.
So you should know that the speed of reflected is lower than the source(Emitted).
we also know: ΔBeat frequency(max) = 560 Hz = $f_{b}$
so we have:
$f_{e}$ - $f_{r}$ = $f_{b}$
so frequency of Reflected is:
2.0 × 10^6 Hz - 560 Hz = 1.99 × 10^6 Hz = $f_{r}$
now you know that Lambda = v/f
so if we find the lambda with our Emitted then we can find v with the Reflected:
Lambda = 1540(m/s) / 2.0 × 10^6 Hz = 7.7 × 10^-4 m
=> $v_{max}$ = (lambda)( $f_{r}$
=> 7.7 × 10^-4m (1.99 × 10^6Hz) = 1532 m/s
so the $v_{max}$ is equal to 1532 m/s :)))
This question is solved by two top teachers as fast as they could :))
I hope this is helpful
have a nice day

8 0

4 years ago