What is nucleic acid

The highest point of a wave is called the

Kipish [7]

The highest point of a wave is called the crest. Among the choices, the correct answer is C. The height of the wave can be determined using the crest and the trough. The trough is the lowest point of a wave. The wavelength is the distance between two crests of a wave.

5 0

3 years ago

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A rocket travels at a speed of 14,000 m/s. What is the distance travelled by the rocket in 150s?

kati45 [8]

1) The distance travelled by the rocket can be found by using the basic relationship between speed (v), time (t) and distance (S):

$v=\frac{S}{t}$

Rearranging the equation, we can write

$S=vt$

In this problem, v=14000 m/s and t=150 s, so the distance travelled by the rocket is

$S=vt=(14000 m/s)(150 s)=2.1 \cdot 10^6 m$

2) We can solve the second part of the problem by using the same formula we used previously. This time, t=300 s, so we have:

$S=vt=(14000 m/s)(300 s)=4.2 \cdot 10^6 m/s$

5 0

4 years ago

If a flea can jump straight up to a height of 0.410 m , what is its initial speed as it leaves the ground?

aivan3 [116]

Initial velocity = $v_0$

acceleration in the downward direction = -9.8 $\frac {m}{s^2}$

Final velocity at the highest point = 0

Maximum height reached = 0.410 m

Now, Using third equation of motion:

$\(v^2 = {v_0}^{2} + 2aH$

$\(0^2 = {v_0}^{2} - 2 \times 9.8 \times 0.410$

${v_0}^{2} = 2 \times 9.8 \times 0.410$

$v_0 = 2.834 \frac {m}{s}$

Speed with which the flea jumps = $2.834 \frac {m}{s}$

4 0

3 years ago

A crate is pushed horizontally by a horizontal force 527.018 N . Sliding friction resists the motion, and the kinetic coefficien

gulaghasi [49]

Answer:

m= 10 kg a = 52 m / s²

Explanation:

For this problem we must use Newton's second law, let's apply it to each axis

X axis

F - fr = ma

The equation for the force of friction is

-fr = miu N

Axis y

N- W = 0

N = mg

Let's replace and calculate laceration

F - miu (mg) = ma

a = F / m - mi g

a = 527.018 / m - 0.17 9.8

We must know the mass of the body suppose m = 10 kg

a = 527.018 / 10 - 1,666

a = 52 m / s²

5 0

4 years ago

The driver of a car going 86.0 km/h suddenly sees the lights of a barrier 44.0 m ahead. It takes the driver 0.75 s to apply the

Lynna [10]

Part a

Answer: NO

We need to calculate the distance traveled once the brakes are applied. Then we would compare the distance traveled and distance of the barrier.

Using the second equation of motion:

$s=ut+0.5at^2$

where s is the distance traveled, u is the initial velocity, t is the time taken and a is the acceleration.

It is given that, u=86.0 km/h=23.9 m/s, t=0.75 s, $a=-10.0 m/s^2$

$\Rightarrow s=23.9 m/s \times 0.75s+0.5\times (-10.0 m/s^2)\times (0.75 s)^2=15.11 m$

Since there is sufficient distance between position where car would stop and the barrier, the car would not hit it.

Part b

Answer: 29.6 m/s

The maximum distance that car can travel is $s=44.0 m$

The acceleration is same, $a=-10.0 m/s^2$

The final velocity, v=0

Using the third equation of motion, we can find the maximum initial velocity for car to not hit the barrier:

$v^2-u^2=2as$

$0-u^2=-2 \time 10.0m/s^2 \times 44.0 m\Rightarrow u=\sqrt{880 m^2/s^2}=29.6 m/s$

Hence, the maximum speed at which car can travel and not hit the barrier is 29.6 m/s.

7 0

3 years ago