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3 years ago

A car accelerates from 25 km/hr in 30 seconds. what Is the acceleration?

2 answers:

7 0

Acceleration is units of distance per time squared, take speed divided by time:

(25 km/hr)/(30 sec) = 0.833 km/hr/sec

these units aren't very helpful, so convert to units more likely to represent what your measuring. probably meters per seconds squared:

(0.833 km/hr/sec)*(1 hr/ 3600sec)*(1000m/km) = 0.23 m/s^2

kaheart [24]3 years ago

6 0

25 km/hr = 6,9 m/s
acceleration = 6,9 / 30 = 0,23

<u>acceleration = 0,23 m / s²</u>

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You drop a 0.375 kg ball from a height of 1.37 m. It hits the ground and bounces up again to a height of 0.67 m. How much energy

Radda [10]

2.57 joule energy lose in the bounce .

<u>Explanation</u>:

when ball is the height of 1.37 m from the ground it has some gravitational potential energy with respect to hits the ground

Formula for gravitational potential energy given by

Potential Energy = mgh

Where ,

m = mass

g = acceleration due to gravity

h = height

Potential energy when ball hits the ground

m= 0.375 kg

h = 1.37 m

g = 9.8 m/s²

$Potential Energy = 0.375\times9.8\times1.37$

Potential Energy = 5.03 joule

Potential energy when ball bounces up again

h= 0.67 m

$Potential Energy = 0.375\times0.67\times9.8$

Potential Energy = 2.46 joule

Energy loss = 5.03 - 2.46 = 2.57 joule

2.57 joule energy lose in the bounce

6 0

2 years ago

A woman living in a third-story apartment is moving out. Rather than carrying everything down the stairs, she decides to pack he

Flura [38]

Answer:

T = 480.2N

Explanation:

In order to find the required force, you take into account that the sum of forces must be equal to zero if the object has a constant speed.

The forces on the boxes are:

$T-Mg=0$ (1)

T: tension of the rope

M: mass of the boxes 0= 49kg

g: gravitational acceleration = 9.8m/s^2

The pulley is frictionless, then, you can assume that the tension of the rope T, is equal to the force that the woman makes.

By using the equation (1) you obtain:

$T=Mg=(49kg)(9.8m/s^2)=480.2N$

The woman needs to pull the rope at 480.2N

8 0

3 years ago

An electron emitted in the beta decay of bismuth-210 has a mean kinetic energy of 390 keV. (a) Find the de Broglie wavelength of

Sauron [17]

Explanation:

Given that,

The mean kinetic energy of the emitted electron, $E=390\ keV=390\times 10^3\ eV$

(a) The relation between the kinetic energy and the De Broglie wavelength is given by :

$\lambda=\dfrac{h}{\sqrt{2meE}}$

$\lambda=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 390\times 10^3}}$

$\lambda=1.96\times 10^{-12}\ m$

(b) According to Bragg's law,

$n\lambda=2d\ sin\theta$

n = 1

For nickel, $d=0.092\times 10^{-9}\ m$

$\theta=sin^{-1}(\dfrac{\lambda}{2d})$

$\theta=sin^{-1}(\dfrac{1.96\times 10^{-12}}{2\times 0.092\times 10^{-9}})$

$\theta=0.010^{\circ}$

As the angle made is very small, so such an electron is not useful in a Davisson-Germer type scattering experiment.

4 0

3 years ago

Enter your answer in the provided box. The mathematical equation for studying the photoelectric effect is hν = W + 1 2 meu2 wher

siniylev [52]

Answer:

$v = 4.44 \times 10^5 m/s$

Explanation:

By Einstein's Equation of photoelectric effect we know that

$h\nu = W + \frac{1}{2}mv^2$

here we know that

$h\nu$ = energy of the photons incident on the metal

$W$ = minimum energy required to remove photons from metal

$\frac{1}{2}mv^2$ = kinetic energy of the electrons ejected out of the plate

now we know that it requires 351 nm wavelength of photons to just eject out the electrons

so we can say

$W = \frac{hc}{351 nm}$

here we know that

$hc = 1242 eV-nm$

now we have

$W = \frac{1242}{351} = 3.54 eV$

now by energy equation above when photon of 303 nm incident on the surface

$\frac{1242 eV-nm}{303 nm} = 3.54 eV + \frac{1}{2}(9.1 \times 10^{-31})v^2$

$4.1 eV = 3.54 eV + (4.55 \times 10^{-31}) v^2$

$(4.1 - 3.54)\times 1.6 \times 10^{-19}) = (4.55 \times 10^{-31}) v^2$

$8.96 \times 10^{-20} = (4.55 \times 10^{-31}) v^2$

$v = 4.44 \times 10^5 m/s$

6 0

3 years ago

During which segments are two states of matter present? the answer is D

Tomtit [17]

Answer:

2,4

Explanation:

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7 0

3 years ago