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3 years ago

10

The speedometer of your automobile shows that you are proceedig at a steady speed of 80 although your automobile is in accelerat

ed motion?

1 answer:

nika2105 [10]3 years ago

7 0

Answer:

<em>The engine must provide power to compensate friction</em>

Explanation:

<u>Accelerated Motion </u>

Newton's first law states that a body will keep its speed or state of rest while no external net force is applied. Our automobile is running through a road which surface exerts a frictional force on the wheels. If we didn't use fuel, the automobile will eventually stop, because the unbalanced friction force causes deceleration. To compensate that force, we must provide power to the engine. When our speedometer shows a constant speed, the net force is zero, but the car needs to accelerate some, just to compensate the friction force.

The total acceleration keeps being zero, but the engine is actually doing work against friction.

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you push a 51 kg box with a force of 485 N. the friction force on the box is 232 N. calculate the acceleration of the crate.

vichka [17]

Answer:

a = 4.96 m/s²

Explanation:

Given,

The mass of the box, m = 51 Kg

The magnitude of the applied force, Fₐ = 485 N

The friction force on the box, Fₓ = 232 N

The net force acting on the box is,

F = Fₐ - Fₓ

Substituting the given values in the above equation

F = 485 - 232

= 253 N

The acceleration of the crate is given by

a = F/m

= 253 / 51

= 4.96 m/s²

Hence, the acceleration of the crate is, a = 4.96 m/s²

3 0

3 years ago

" A bowl of soup at 200Á F. is placed in a room of constant temperature of 60Á F. The

Dahasolnce [82]

<span>T(t)=60+140<span>e<span>−0.075t</span></span></span> <span>T(12)=60+140<span>e<span>−0.075∗12</span></span></span> <span>T(12)=60+140<span>e<span>−0.9</span></span></span> <span><span>T(12)=60+140(0.4065696597)
=116.84
So the temperature will be approximately 117 degrees</span></span>

7 0

3 years ago

What is the energy per photon absorbed during the transition from n = 2 to n = 3 in the hydrogen atom?

adelina 88 [10]

Answer : The energy of one photon of hydrogen atom is, $3.03\times 10^{-19}J$

Explanation :

First we have to calculate the wavelength of hydrogen atom.

Using Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = 10973731.6 m⁻¹

$n_f$ = Higher energy level = 3

$n_i$ = Lower energy level = 2

Putting the values, in above equation, we get:

$\frac{1}{\lambda}=(10973731.6)\left(\frac{1}{2^2}-\frac{1}{3^2} \right )$

$\lambda=6.56\times 10^{-7}m$

Now we have to calculate the energy.

$E=\frac{hc}{\lambda}$

where,

h = Planck's constant = $6.626\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength = $6.56\times 10^{-7}m$

Putting the values, in this formula, we get:

$E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{6.56\times 10^{-7}m}$

$E=3.03\times 10^{-19}J$

Therefore, the energy of one photon of hydrogen atom is, $3.03\times 10^{-19}J$

3 0

3 years ago

A sprinter accelerates from rest to 10.0 m/s in 1.35 s l. What is her acceleration?

ale4655 [162]

Answer:

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Explanation:

sgdfggsfdsgfgsgsmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

8 0

3 years ago

Two masses —m1 and m2— are connected by light cables to the perimeters of two cylinders of radii r1 and r2 respectively, as show

Aleksandr [31]

Answer:

Part a)

Mass of m2 is given as

$m_2 = \frac{20}{3} kg$

Part b)

Angular acceleration is given as

$\alpha = 1.96 rad/s^2$

Part c)

Tension in the rope is given as

$T = 176.6 N$

Explanation:

Part a)

When m1 and m2 both connected to the cylinder then the system is at rest

so we can use torque balance here

$m_1g r_1 = m_2 g r_2$

$20 g(0.5) = m_2 g(1.5)$

$10 = 1.5 m_2$

$m_2 = \frac{20}{3} kg$

Part b)

When block m_2 is removed then system becomes unstable

so force equation of mass m1

$m_1g - T = m_1 a$

also we have

$T r_1 = I\alpha$

now we have

$m_1g = \frac{I a}{r_1^2} + m_1 a$

$a = \frac{m_1g}{\frac{I}{r_1^2} + m_1}$

$a = \frac{20 (9.81)}{\frac{45}{0.5^2} + 20}$

$a = 0.981 m/s^2$

so angular acceleration is given as

$\alpha = \frac{a}{r_1}$

$\alpha = \frac{0.981}{0.5}$

$\alpha = 1.96 rad/s^2$

Part c)

Tension in the rope is given as

$T = \frac{I\alpha}{r_1}$

$T = \frac{45 (1.96)}{0.5}$

$T = 176.6 N$

7 0

3 years ago