Ask question

Ask question

All categories

1 year ago

11

A longitudinal wave is observed to be moving along a slinky. Adjacent crests are 2.4 m apart. Exactly 6 crests are observed to m

ove past a given point in 9.1 s. Determine the wavelength, frequency, and speed of this wave.

1 answer:

Umnica [9.8K]1 year ago

8 0

Answer:

5

Explanation:

You might be interested in

What is the magnitude of the relative angle φ

melomori [17]

Complete question is;

A ski jumper travels down a slope and leaves the ski track moving in the horizontal direction with a speed of 24 m/s. The landing incline below her falls off with a slope of θ = 59◦ . The acceleration of gravity is 9.8 m/s².

What is the magnitude of the relative angle φ with which the ski jumper hits the slope? Answer in units of ◦

Answer:

14.08°

Explanation:

The time covered will be given by the formula;

t = (2V_x•tan θ)/g

t = (2 × 24 × tan 59)/9.8

t = 8.152 s

Now, the slope of the flight path at the point of impact will be given by the formula;

tan α = V_y/V_x

We are given V_x = 24 m/s

V_y will be gotten from the formula;

v = gt

Thus;

V_y = gt

V_y = 9.8 × (8.152) = 78.89 m/s

Thus;

tan α = 78.89/24

tan α = 3.2871

α = tan^(-1) 3.2871

α = 73.08°

Thus ;

Relative angle φ = α - θ = 73.08 - 59 = 14.08°

6 0

2 years ago

Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above i

SCORPION-xisa [38]

Answer:

The value is $r = 5.077 \ m$

Explanation:

From the question we are told that

The Coulomb constant is $k = 9.0 *10^{9} \ N\cdot m^2 /C^2$

The charge on the electron/proton is $e = 1.6*10^{-19} \ C$

The mass of proton $m_{proton} = 1.67*10^{-27} \ kg$

The mass of electron is $m_{electron } = 9.11 *10^{-31} \ kg$

Generally for the electron to be held up by the force gravity

Then

Electric force on the electron = The gravitational Force

i.e

$m_{electron} * g = \frac{ k * e^2 }{r^2 }$

$\frac{9*10^9 * (1.60 *10^{-19})^2 }{r^2 } = 9.11 *10^{-31 } * 9.81$

$r = \sqrt{25.78}$

$r = 5.077 \ m$

7 0

2 years ago

Why are we unable to work long without food

nataly862011 [7]

Answer:

Without food your not able to produce energy thats why you cannot work so long

7 0

2 years ago

The length of a wire 2.00 m is measured as 2.02m. What is the percentage error in the measurement?

n200080 [17]

Answer:

1%

Explanation:

Percent error can be found by dividing the absolute error (difference between measure and actual value) by the actual value, then multiplying by 100.

$Percent Error=\frac{V_{measured}- V_{true} } {V_{true}} *100$

The measured value is 2.02 meters and the actual value is 2.00 meters.

$V_{measured}=2.02\\\\V_{true}=2.00$

$Percent Error=\frac{2.02-2.00}{2.00} *100$

First, evaluate the fraction. Subtract 2.00 from 2.02

$Percent Error=\frac{0.02}{2.00}*100$

Next, divide 0.02 by 2.00

$PercentError=0.01 *100$

Finally, multiply 0.01 and 100.

$Percent Error=1\\Percent Error= 1 \%$

The percent error is 1%.

6 0

2 years ago

A block of mass 3m is placed on a frictionless horizontal surface, and a second block of mass m is placed on top of the first bl

tatuchka [14]

By Newton's second law, assuming <em>F</em> is horizontal,

• the net <u>horizontal</u> force on the <u>larger</u> block is

<em>F</em> - <em>µmg</em> = 3<em>mA</em>

where <em>µmg</em> is the magnitude of friction felt by the larger block due to rubbing with the smaller one, <em>µ</em> is the coefficient of static friction between the two blocks, and <em>A</em> is the block's acceleration;

• the net <u>vertical</u> force on the <u>larger</u> block is

4<em>mg</em> - 3<em>mg</em> - <em>mg</em> = 0

where 4<em>mg</em> is the mag. of the normal force of the surface pushing up on the combined mass of the two blocks, 3<em>mg</em> is the weight of the larger block, and <em>mg</em> is the weight of the smaller block;

• the net <u>horizontal</u> force on the <u>smaller</u> block is

<em>µmg</em> = <em>ma</em>

where <em>µmg</em> is again the friction between the two blocks, but notice that this points in the same direction as <em>F</em>. It is the only force acting on the smaller block in the horizontal direction, so (b) static friction is causing the smaller block to accelerate;

• the net <u>vertical</u> force on the <u>smaller</u> block is

<em>mg</em> - <em>mg</em> = 0

where <em>mg</em> is the magnitude of both the normal force of the larger block pushing up on the smaller one, and the weight of the smaller block.

(You should be able to draw your own FBD's based on the forces mentioned above.)

(c) Solve the equations above for <em>A</em> and <em>a</em> :

<em>A</em> = (<em>F</em> - <em>µmg</em>) / (3<em>m</em>)

<em>a</em> = <em>µg</em>

5 0

1 year ago