I NEED SOMEOME TO HELP MEEEE NOWWWW I WILL GIVE YOU. 75 POINTSSSS

I have my exam monday I need some help!

Lana71 [14]

Answer:

Explanation:

If the passage of the waves is one crest every 2.5 seconds, then that is the frequency of the wave, f.

The distance between the 2 crests (or troughs) is the wavelength, λ.

We want the velocity of this wave. The equation that relates these 3 things is

$f=\frac{v}{\lambda}$ and filling in:

$2.5=\frac{v}{2.0}$ so

v = 2.5(2.0) and

v = 5.0 m/s

7 0

3 years ago

A resistance training system that is a variation of circuit training and that alternates upper body and lower body exercises thr

dlinn [17]

Answer:

Peripheral heart action.

Explanation:

The Peripheral heart action purpose is to keep the blood circulation throughout the body at time of workout.

This is specially proceed through the small central muscles which goes around your heart.

After that it is followed by the peripheral muscles which lies in the arms, legs and in the abs part.

6 0

4 years ago

The acceleration of a particle is directly proportional to the square of the time t. When t =0, the particle is at x =24 m. Know

AlexFokin [52]

Answer:

x(t) = ⅟₁₀₈t⁴ + 10t + 24

v(t) = ⅟₂₇t³ + 10

Explanation:

a(t) = C₁t²

velocity is the integral of acceleration

v(t) = ⅓C₁t³ + C₂

position is the integral of velocity

x(t) = (⅟₁₂C₁)t⁴ + C₂t + C₃

x(0) = 24 = (⅟₁₂C₁)0⁴ + C₂0 + C₃

C₃ = 24

x(6) = 96 = (⅟₁₂C₁)6⁴ + C₂6 + 24

72 = 108C₁ + 6C₂

C₂ = 12 - 18C₁

v(6) = 18 = ⅓C₁6³ + C₂

18 = 72C₁ + C₂

18 = 72C₁ + (12 - 18C₁)

6 = 54C₁

C₁ = 1/9

C₂ = 12 - 18(1/9)

C₂ = 10

4 0

3 years ago

In the Bohr theory of the hydrogen atom, an electron moves in a circular orbit about a proton, where the radius of the orbit is

Alchen [17]

Answer:

(a) Electric force will be $77.285\times 10^{-9}N$

(B) Velocity $v=2.1522\times 10^6m/sec$

Explanation:

We have given that radius of the circular orbit $r=0.546\times 10^{-10}m$

Charge on electron $e=1.6\times 10^{-19}C$

(A) According to coulomb's law electric force between two charges is given by

$F=\frac{kq}{r^2}$

So electric force $F=\frac{9\times 10^9\times (1.6\times 10^{-19})^2}{(0.546\times 10^{-10})^2}=77.285\times 10^{-9}N$

We know that centripetal force is given by $F=\frac{mv^2}{r}$

So $\frac{9.11\times 10^{-31}\times v^2}{0.546\times 10^{-10}}=77.285\times 10^{-9}$

$v=2.1522\times 10^6m/sec$

8 0

3 years ago

This is the last question that I need help with, and I'll try and do the rest on my own. Thank you to all of the tutors that hel

ivann1987 [24]

Question 7.

Given:

Mass, m = 2200 kg

Time, t = 8 s

Velocity = 12 m/s

Let's find the average net force required to stop the car.

To find the average net force required to stop the car, let's first find the acceleration of the car.

$v=v_o+at$

Where:

Initial velocity, vo = 12 m/s

Final velocity, v = 0 m/s (velocity of car when it stops)

t is the time = 8 seconds

a is the acceleration.

Rewrite the formula for a:

$a=\frac{v-v_o}{t}$

Hence, we have:

$\begin{gathered} a=\frac{0-12}{8} \\ \\ a=\frac{-12}{8} \\ \\ a=-1.5m/s^2 \end{gathered}$

The acceleration is -1.5 m/s² (negative because the car is deccelerating).

Now, to find the net force, we have:

$\sum ^{}_{}F=ma$

Where:

m is the mass = 2200 kg

a is the acceleration = -1.5 m/s²

Hence, we have:

$\sum ^{}_{}F=2200\times(-1.5)=-3300\text{ N}$

The negative sign indicates that the force acts in the opposite direction the car is travelling.

Therefore, average net force required to stop the car is 3300 Newtons in the opposite direction.

ANSWER:

3300 N in the opposite direction

8 0

1 year ago