Ask question

Ask question

All categories

3 years ago

5

a cyclist coasting down a 5.0 ◦ incline at a constant speed of 6.0 km/h because of air resistance. If the total mass of the bicy

cle + cyclist is 50 kg, how much force must be generated to climb back up the incline at the same speed (and same air resistance)?

1 answer:

Dvinal [7]3 years ago

3 0

Answer:

$F_{net}= 85.41\ N$

Explanation:

mass of the bicycle + cyclist = 50 kg

constant speed = 6 km/h

a cyclist coasting down a 5.0° incline

the downward velocity is constant, so net acceleration must be zero

the air drag must be equal to gravitational force downward along the ramp

$F_a = mg sin \theta$

now for upward motion

$F_{net} = mg sin \theta + air\ drag$

$F_{net} = mg sin \theta + mg sin \theta$

$F_{net} = 2 mg sin \theta$

$F_{net} = 2\times 50 \times 9.8 sin 5^0$

$F_{net}= 85.41\ N$

You might be interested in

Two taut strings of identical mass and length are stretched with their ends fixed, but the tension in one string is 1.10 times g

ollegr [7]

Answer:

The beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

Explanation:

Given;

velocity of wave on the string with lower tension, v₁ = 35.2 m/s

the fundamental frequency of the string, F₁ = 258 Hz

<u>velocity of wave on the string with greater tension;</u>

$v_1 = \sqrt{\frac{T_1}{\mu }$

where;

v₁ is the velocity of wave on the string with lower tension

T₁ is tension on the string

μ is mass per unit length

$v_1 = \sqrt{\frac{T_1}{\mu} } \\\\v_1^2 = \frac{T_1}{\mu} \\\\\mu = \frac{T_1}{v_1^2} \\\\ \frac{T_1}{v_1^2} = \frac{T_2}{v_2^2}\\\\v_2^2 = \frac{T_2v_1^2}{T_1}$

Where;

T₁ lower tension

T₂ greater tension

v₁ velocity of wave in string with lower tension

v₂ velocity of wave in string with greater tension

From the given question;

T₂ = 1.1 T₁

$v_2^2 = \frac{T_2v_1^2}{T_1} \\\\v_2 = \sqrt{\frac{T_2v_1^2}{T_1}} \\\\v_2 = \sqrt{\frac{1.1T_1*(35.2)^2}{T_1}}\\\\v_2 = \sqrt{1.1(35.2)^2} = 36.92 \ m/s$

<u>Fundamental frequency of wave on the string with greater tension;</u>

<u /> $f = \frac{v}{2l} \\\\2l = \frac{v}{f} \\\\thus, \frac{v_1}{f_1} =\frac{v_2}{f_2} \\\\f_2 = \frac{f_1v_2}{v_1} \\\\f_2 =\frac{258*36.92}{35.2} \\\\f_2 = 270.6 \ Hz$ <u />

Beat frequency = F₂ - F₁

= 270.6 - 258

= 12.6 Hz

Therefore, the beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

6 0

3 years ago

You push a disk-shaped platform tangentially on its edge 2.0 m from the axle. The platform starts at rest and has a rotational a

Natali [406]

Answer: 40.84 m

Explanation:

Given

Radius of the disk, r = 2m

Velocity of the disk, v = 7 rad/s

Acceleration of the disk, α = 0.3 rad/s²

Here, we use the formula for kinematics of rotational motion to solve

2α(θ - θ•) = ω² - ω•²

Where,

ω• = 0

ω = v/r = 7/2

ω = 3.5 rad/s

2 * 0.3(θ - θ•) = 3.5² - 0

0.6(θ - θ•) = 12.25

(θ - θ•) = 12.25 / 0.6

(θ - θ•) = 20.42 rad

Since we have both the angle and it's radius, we can calculate the arc length

s = rθ = 2 * 20.42

s = 40.84 m

Thus, the needed distance is 40.84 m

7 0

2 years ago

What is the frequency of a wave that has a wavelength of 20 cm and a speed of 10 m/s?

Inga [223]

Frequency = velocity/wavelength

Frequency = 10/20

Frequency = 0.5 Hz

3 0

3 years ago

A Porsche sports car can accelerate at 8.8 m/s^2. Determine its acceleration in km/h^2.

finlep [7]

Answer:

The acceleration expressed in the new units is $114.048 Km/h^{2}$

Explanation:

To convert from $m/s^{2}$ to $Km/h^{2}$ it is necessary to remember that there are 1000 meters in 1 kilometer and 3600 seconds in 1 hour:

Then by means of a rule of three it is get:

$8.8\frac{m}{s^{2}}.(\frac{1Km}{1000m}).(\frac{3600s}{1h})^{2}$

$8.8\frac{m}{s^{2}}.(\frac{1Km}{1000m}).(\frac{12960000s^{2}}{1h^{2}})$

Hence, the units of meters and seconds will cancel. Notice the importance of square the ratio 3600s/1h, so that way they can match with the other units:

$114.048 Km/h^2$

So the acceleration expressed in the new units is $114.048 Km/h^2$ .

6 0

3 years ago

The energy of the electron in a hydrogen atom can be calculated from the Bohr formula: In this equation stands for the Rydberg e

labwork [276]

Answer:

Wavelength, $\lambda=657\ nm$

Explanation:

The energy of the electron in a hydrogen atom can be calculated from the Bohr formula as :

$E=\dfrac{-R}{n^2}$ .............(1)

Where

R is the Rydberg constant

n is the number of orbit

We need to find the wavelength of the line in the absorption line spectrum of hydrogen caused by the transition of the electron from an orbital with to an orbital with n₁ = 2 to an orbital with n₂ = 3.

Equation (1) can be re framed as :

$\dfrac{1}{\lambda}=R(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})$

$\dfrac{1}{\lambda}=1.096\times 10^7\times (\dfrac{1}{2^2}-\dfrac{1}{3^2})$

$\lambda=6.569\times 10^{-7}\ m$

or

$\lambda=657\ nm$

So, the the wavelength of the line in the absorption line spectrum is 657 nm. Hence, this is the required solution.

3 0

3 years ago