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1 year ago

8

A wave has wavelength of 8m and a speed of 360 m/s. What is the frequency of the wave?

1 answer:

Naily [24]1 year ago

8 0

Answer:

$45~ \text{Hz}.$

Explanation:

$\text{Frequency,}~ f = \dfrac{v}{\lambda}\\\\\\~~~~~~~~~~~~~~~~~~=\dfrac{360~ \text{ms}^{-1}}{8~ \text m}\\\\\\~~~~~~~~~~~~~~~~~~=45~ \text s^{-1}\\\\\\~~~~~~~~~~~~~~~~~~~=45~ \text{Hz}$

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A woman pushes a 3.2kg textbook on a table as she walks forward 3.5 meters. How much work does the vertical component of the for

Effectus [21]

Answer:

The done by the vertical component of force is zero.

Explanation:

Given data,

The mass of the textbook the women pushes horizontally, m = 3.2 kg

The displacement of the textbook, S = 3.5 m

Let the force of the women acts on the book horizontally,

Therefore, the horizontal component of force is maximum and the vertical component is zero.

If F is the force applied by the women, then the horizontal and vertical component of the force is,

$F_{x}=F Cos\theta$

$F_{y}=F Sin\theta$

Since the force is acting along with the horizontal x component, the vertical component of the force is zero.

Hence, the done by the vertical component of force is zero.

4 0

2 years ago

A man kicks the 150-g ball such that it leaves the ground at an angle of 60° and strikes the ground at the same elevation a dist

egoroff_w [7]

Answer:

$313.92\ \text{m/s}$

$47.088\ \text{kg m/s}$

Explanation:

m = Mass of ball = 150 g

$\theta$ = Angle of kick = $60^{\circ}$

$x$ = Displacement of ball in x direction = 12 m

Range of projectile is given by

$x=\dfrac{u\sin^2\theta}{2g}\\\Rightarrow u=\dfrac{2xg}{\sin^2\theta}\\\Rightarrow u=\dfrac{2\times 12\times 9.81}{\sin^260^{\circ}}\\\Rightarrow u=313.92\ \text{m/s}$

The velocity of the ball instantly after the man kicks the ball is $313.92\ \text{m/s}$

Impulse is given by

$J=mu\\\Rightarrow J=0.15\times 313.92\\\Rightarrow J=47.088\ \text{kg m/s}$

The impulse of his foot on the ball is $47.088\ \text{kg m/s}$ .

3 0

2 years ago

Consider a car engine running at constantspeed. That is, the crankshaft of the en-gine rotates at constant angular velocity whil

Irina-Kira [14]

Answer:

The value is $|v| = 7.39 \ m/s$

Explanation:

From the question we are told that

The angular speed is $w = 2030 \ rpm = \frac{2030 * 2 * \pi }{ 60} = 212.61 \ rad/s$

The distance between the minimum and maximum external position is $d = 6.95 \ cm = 0.0695 \ m$

Generally the amplitude of the crank shaft is mathematically represented as

$A = \frac{d}{2}$

=> $A = \frac{0.0695}{2}$

=> $A = 0.03475 \ m$

Generally the maximum speed of the piston is mathematically represented as

$|v| = A * w$

=> $|v| = 0.03475 * 212.61$

=> $|v| = 7.39 \ m/s$

7 0

2 years ago

A rock with a mass of 80grams dropped into a graduated cylinder with 30 ml of water in it the readingof water became 70 ml what

muminat

Explanation:

$b = m \div v \\ 80 \div 30 \times 10 { -}^{2}$

5 0

2 years ago

Two spherical asteroids have the same radius R. Asteroid 1 has mass M and asteroid 2 has mass 1.97·M. The two asteroids are rele

nekit [7.7K]

Answer:

$0.536\sqrt{\frac{GM}{R}}$

Explanation:

We are given that

Mass of one asteroid 1, $m_1=M$

Mass of asteroid 2, $m_2=1.97 M$

Initial distance between their centers,d=13.63 R

Radius of each asteroid=R

d'=R+R=2R

Initial velocity of both asteroids

$u=0$

We have to find the speed of second asteroid just before they collide.

According to law of conservation of momentum

$(m_1+m_2)u=m_1v_1+m_2v_2$

$(M+1.97 M)\times 0=Mv_1+1.97Mv_2$

$Mv_1=-1.97 Mv_2$

$v_1=-1.97v_2$

According to law of conservation of energy

$Gm_1m_2(\frac{1}{d'}-\frac{1}{d})=\frac{1}{2}m_1v^2_1+\frac{1}{2}m_2v^2_2$

$GM(1.97M)(\frac{1}{2R}-\frac{1}{13.63R})=\frac{1}{2}M(-1.97v_2)^2+\frac{1}{2}(1.97M)v^2_2$

$1.97M^2G(\frac{13.63-2}{27.26R})=\frac{1}{2}Mv^2_2(3.8809+1.97)$

$1.97MG(\frac{11.63}{27.26 R})=\frac{1}{2}(5.8509)v^2_2$

$v^2_2=\frac{1.97GM\times11.63\times 2}{27.26R\times 5.8509}$

$v_2=\sqrt{\frac{1.97GM\times11.63\times 2}{27.26R\times 5.8509}}$

$v_2=0.536\sqrt{\frac{GM}{R}}$

Hence, the speed of second asteroid = $0.536\sqrt{\frac{GM}{R}}$

8 0

2 years ago