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3 years ago

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Andrea was watching her brother in the ocean and noticed that the waves were coming into the beach at a frequency of 0.777778 Hz

. How many waves hit the beach in 27 s? Answer in units of waves.

1 answer:

Stolb23 [73]3 years ago

3 0

8 wave units I guess I tried it should be the answer though

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the purpose of fuses and circuit breakers is (first answer)

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2 years ago

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A 6 kg box with initial speed 5 m/s slides across the floor and comes to a stop after 1.9 s. What is the coefficient of kinetic

Ilia_Sergeevich [38]

Answer:

$\mu_k=0.27$

Explanation:

According to the free body diagram, in this case, we have:

$\sum F_x:-F_k=ma\\\sum F_y:N=mg$

Recall that the force of friction is given by:

$F_k=\mu_k N$

Replacing and solving for the coefficient of kinetic friction:

$-\mu_kN=ma\\-\mu_k(mg)=ma\\\mu_k=-\frac{a}{g}$

We have an uniformly accelerated motion. Thus, the acceleration is defined as:

$a=\frac{v_f-v_0}{t}\\a=\frac{0-5\frac{m}{s}}{1.9s}\\a=-2.63\frac{m}{s^2}$

Finally, we calculate $\mu_k$ :

$\mu_k=-\frac{-2.63\frac{m}{s^2}}{9.8\frac{m}{s^2}}\\\mu_k=0.27$

4 0

3 years ago

A 53.0-kg skater is traveling due east at a speed of 2.90 m/s. A 72.0-kg skater is moving due south at a speed of 6.20 m/s. They

soldier1979 [14.2K]

Answer:

a. $\thta=71^{\circ}$

b. $v_f=3.78 m/s$

Explanation:

We are given that

$m_1=53 kg$

$v_1=2.9 m/s$

$m_2= 72 kg$

$v_2=6.2 m/s$

a.We have to find the angle

$\theta=tan^{-1}{\frac{m_2v_2}{m_1v_1}=\frac{72(6.2)}{53(2.9)}=71^{\circ}$

$\theta=71^{\circ}$

b. We have to find the speed $v_f$

According to law of conservation of momentum

$m_1v_1=(m_1+m_2)v_fcos\theta$

$53(2.9)=(53+72)v_fcos 71^{\circ}=40.7 v_f$

$v_f=\frac{153.7}{40.7}=3.78 m/s$

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3 years ago

Which the answer It's science

earnstyle [38]

The answer is A. ive done a 5-k race, so its for sure 3 miles.

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3 years ago

A shopping cart given an initial velocity of 2.0 m/s undergoes a constant acceleration to a velocity of 13 m/s. What is the magn

olga55 [171]

Answer:

The acceleration is a = 2.75 [m/s^2]

Explanation:

In order to solve this problem we must use kinematics equations.

$v_{f} = v_{i} + a*t\\$

where:

Vf = final velocity = 13 [m/s]

Vi = initial velocity = 2 [m/s]

a = acceleration [m/s^2]

t = time = 4 [s]

Now replacing:

13 = 2 + (4*a)

(13 - 2) = 4*a

a = 2.75 [m/s^2]

5 0

3 years ago