All categories

2 years ago

A car’s tire rotates 5.25 times in 3 seconds. What is the tangential velocity of the tire?

1 answer:

4 0

The tangential velocity of the car's tire is the product of the angular velocity and radius of the car's tire which is 11(r) m/s.

<h3>Angular velocity of the tire</h3>

The angular velocity of the tire is the rate of change of angular displacement of the tire with time.

The magnitude of the angular velocity of the tire is calculated as follows;

ω = 2πN

where;

N is the number of revolutions per second

ω = 2π x (5.25 / 3)

ω = 11 rad/s

<h3>Tangential velocity of the tire</h3>

The tangential velocity of the car's tire is the product of the angular velocity and radius of the car's tire.

The magnitude of the tangential velocity is caculated as follows;

v = ωr

where;

r is the radius of the car's tire

v = 11r m/s

Learn more about tangential velocity here: brainly.com/question/25780931

You might be interested in

A 12.0N force with a fixed orientation does work on a

kvasek [131]

Answer:

(a) $\theta=62.31^{\circ}$

(b) $\theta=117.68^{\circ}$

Explanation:

It is given that,

Force acting on the particle, F = 12 N

Displacement of the particle, $d=(2.00i -4.00j+3.00k)\ m$

Magnitude of displacement, $d=\sqrt{2^2+4^2+3^2}= 5.38\ m$

(a) If the change in the kinetic energy of the particle is +30 J. The work done by the particle is given by :

$W=Fd\ cos\theta$

$\theta$ is the angle between force and the displacement

According to work energy theorem, the charge in kinetic energy of the particle is equal to the work done.

So,

$cos\theta=\dfrac{W}{Fd}$

$cos\theta=\dfrac{+30\ J}{12\times 5.38}$

$\theta=62.31^{\circ}$

(b) If the change in the kinetic energy of the particle is (-30) J. The work done by the particle is given by :

$cos\theta=\dfrac{W}{Fd}$

$cos\theta=\dfrac{-30\ J}{12\times 5.38}$

$\theta=117.68^{\circ}$

Hence, this is the required solution.

8 0

2 years ago

Seesaw unbalanced force explain

wolverine [178]

Like a seesaw, it shows that the forces aren’t equal because if it was the seesaw would stay put

3 0

2 years ago

John is carrying a shovelful of snow. The center of mass of the 3.00 kg of snow he is holding is 15.0 cm from the end of the sho

Andru [333]

Answer:

James is correct here as the force of hand pushing upwards is always more than the force of hand pushing down

Explanation:

Here we know that one hand is pushing up at some distance midway while other hand is balancing the weight by applying a force downwards

so here we can say

Upwards force = downwards Force + weight of snow

while if we find the other force which is acting downwards

then for that force we can say that net torque must be balanced

so here we have

$F_{down} L_1 = W_{snow} L_2$

so here we have

$F_{down} = \frac{L_2}{L_1} (W_{snow})$

so here we can say that upward force by which we push up is always more than the downwards force

8 0

2 years ago

What is the value of n in the balmer series for which the wavelength is 410.2 nm.?

m_a_m_a [10]

The answer is n= 6.

What is Balmer series?

The Balmer series is the portion of the emission spectrum of hydrogen that represents electron transitions from energy levels n > 2 to n = 2. These are four lines in the visible spectrum. They are also known as the Balmer lines. The four visible Balmer lines of hydrogen appear at 410 nm, 434 nm, 486 nm and 656 nm.

For the Balmer series, the final energy level is always n=2. So, the wavelengths 653.6, 486.1, 434.0, and 410.2 nm correspond to n=3, n=4, n=5, and n=6 respectively. Since the last wavelength, 410.2 nm, corresponds to n=6, the next wavelength should logically correspond to n=7.

To solve for the wavelength, calculate the individual energies, E2 and E7, using E=-hR/(n^2). Then, calculate the energy difference between E2 (which is the final) and E7 (which is the initial). Finally, use lamba=hc/E to get the wavelength.

To learn more about emission spectrum click on the link below:

brainly.com/question/24213957

#SPJ4

3 0

1 year ago

The graph in the accompanying figure (Figure 1) shows the magnitude of the force exerted by a given spring as a function of the

serious [3.7K]

The elastic potential energy stored in the stretched spring is 1 J.

<h3>What is Hooke's law?</h3>

Hooke's law states that; provided the elastic limit is not exceeded, the extension of the spring is directly proportional to the force on the spring.

Given that;

Force on the spring = 350 Newton

Distance stretched = 7 centimeters or 0.07 m

Hence;

F = ke

k = F/e = 350 Newton/0.07 m = 5000 N/m

Work done in stretching a spring = 1/2ke^2

= 0.5 × 5000 × (2 × 10^-2)^2 =1 J

Learn more about elastic potential energy: brainly.com/question/156316

4 0

1 year ago