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3 years ago

A car slows down from 27.7 m/s to 10.9 m/s in 2.37 s. What is its acceleration?

Physics

2 answers:

frosja888 [35]3 years ago

7 0

Answer:

$\boxed {\boxed {\sf -7.09 \ m/s^2}}$

Explanation:

We are asked to find the acceleration of a car. Acceleration is the change in velocity with respect to time. We will use the following formula:

$a= \frac {v_f-v_i}{t}$

In this formula, $v_f$ is the final velocity, $v_i$ is the initial velocity, and $t$ is the time. The car slows down from 27.7 meters per second to 10.9 meters per second in 2.37 seconds. Therefore:

$v_f$ = 10.9 m/s
$v_i$ = 27.7 m/s
$t$ = 2.37 s

Substitute the values into the formula.

$a= \frac{ 10.9 m/s-27.7 m/s}{2.37 \ s}$

Solve the numerator.

10.9 m/s - 27.7 m/s= -16.8 m/s

$a= \frac{-16.8 \ m/s}{2.37 \ s}$

Divide.

$a= -7.088607595 \ m/s/s$

$a= -7.088607595 \ m/s^2$

The original measurements of velocity and time have 3 significant figures, so our answer must have the same. For the number we found, that is the hundredth place. The 8 in the thousandth place tells us to round the 8 in the hundredth place up to a 9.

$a \approx - 7.09 \ m/s^2$

The acceleration of the car is approximately -7.09 meters per second squared. The acceleration is negative because the car is slowing down.

Sunny_sXe [5.5K]3 years ago

4 0

Answer:

- 7.088 m/s²

Explanation:

As we know that,

★ Acceleration = Change in velocity/Time

→ a = (v - u)/t

Here,

Initial velocity (u) = 27.7 m/s
Final velocity (v) = 10.9 m/s

→ a = (10.9 m/s - 27.7 m/s)/2.37 s

→ a = -16.8/2.37 m/s²

→ a = -7.088 m/s² [Answer]

Negative sign denotes that the velocity is decreasing.

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Alex787 [66]

Answer:

Explanation:

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3 years ago

An object weighs 60.0 kg on the surface of the earth. How much does it weigh 4R from the surface? (5R from the center)

Alecsey [184]

"60 kg" is not a weight. It's a mass, and it's always the same
no matter where the object goes.

The weight of the object is

   (mass) x (gravity in the place where the object is) .

On the surface of the Earth,

   Weight = (60 kg) x (9.8 m/s²)

      =    588 Newtons.

Now, the force of gravity varies as the inverse of the square of the distance from the center of the Earth.
On the surface, the distance from the center of the Earth is 1R.
So if you move out to 5R from the center, the gravity out there is

(1R/5R)² = (1/5)² = 1/25 = 0.04 of its value on the surface.

The object's weight would also be 0.04 of its weight on the surface.

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3 years ago

Calculate the radiative and collisional energy losses (in keV/micron) for a 1.9 MeV electron in lead and determine the rad./coll

Andrew [12]

Answer:

Explanation:

During an energy transfer, the collision loss for an electron can be determined by using the formula:

$Q = \dfrac{4mME }{(m+M)^2}$

However; from the total stopping power & power loss of the electron;

$\dfrac{radiational \ energy \ loss}{colisional \ energy \ loss } = \dfrac{ZE}{800}$

where;

Z = atomic no. for lead = 82

E = 1.9 MeV

∴

radiational energy loss = collisional energy loss $=\dfrac{82 \times 1.9}{800}$

= 0.19475

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Initially, the conventional lead protection however reduces the mild clarity at the same moment as plexiglass is useful for light transmittance and readability.

Moreover, the traditional lead with its high density and thickness reduces observation features, in the meantime, the plexiglass is a whole lot higher than the stated.

Finally, plexiglass contains a high dimensional balance with an excessive dielectric constant.

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3 years ago

An elevator provides 21 000 w of power during a 12 s ride. how much work does the elevator do?

Mice21 [21]

Answer:

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3 years ago

Ultraviolet light of wavelength 270 nm strikes a metal whose work function is 2.3 eV.What is the shortest de Broglie wavelength

soldier1979 [14.2K]

Answer:

The shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm

Explanation:

Given;

wavelength of ultraviolet light, λ = 270 nm

work function of the metal, φ = 2.3 eV = 2.3 x 1.602 x 10⁻¹⁹ J = 3.685 x 10⁻¹⁹ J

The energy of the ultraviolet light is given by;

$E = \frac{hc}{\lambda}\\\\E = \frac{(6.626*10^{-34} )(3*10^{8}) }{270*10^{-9} }\\\\E = 7.362 * 10^{-19} \ J$

The energy of the incident light is related to kinetic energy of the electron and work function of the metal by the following equation;

E = φ + K.E

K.E = E - φ

K.E = (7.362 x 10⁻¹⁹ J) - (3.685 x 10⁻¹⁹ J )

K.E = 3.677 x 10⁻¹⁹ J

K.E = ¹/₂mv²

mv² = 2K.E

velocity of the electron is given by;

$V = \sqrt{\frac{2K.E}{m} }\\\\V = \sqrt{\frac{2(3.677*10^{-19}) }{9.1*10^{-31} } }\\\\V = 8.99*10^{5} \ m/s$

the shortest de Broglie wavelength for the electrons is given by;

$\lambda = \frac{h}{mv}\\ \\\lambda = \frac{6.626*10^{-34} }{(9.1*10^{-31})( 8.99*10^{5} )}\\\\\lambda = 8.10*10^{-10} \ m\\\\\lambda = 0.81 \ nm$

Therefore, the shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm

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4 years ago