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3 years ago

HURRY ! THE BEST I WILL GIVE THE BRAINLIEST

Physics

1 answer:

erik [133]3 years ago

6 0

The speed of both cars is the same ... 80 km per hour.

But their velocities are different, because DIRECTION is part of velocity, and their directions are different.

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A sample of helium (He) occupies 8.0 liters at 1 atm and 20.0◦C. What pressure is necessary to change the volume to 1.0 liters a

nevsk [136]

Apply the combined gas law

PV/T = const.

P = pressure, V = volume, T = temperature, PV/T must stay constant.

Initial PVT values:

P = 1atm, V = 8.0L, T = 20.0°C = 293.15K

Final PVT values:

P = ?, V = 1.0L, T = 10.0°C = 283.15K

Set the PV/T expression for the initial and final PVT values equal to each other and solve for the final P:

1(8.0)/293.15 = P(1.0)/283.15

P = 7.7atm

7 0

3 years ago

A 0.500-nm x-ray photon is deected through 134 in a Compton scattering event. At what angle (with respect to the incident beam)

natita [175]

Answer:

The angle of recoil electron with respect to incident beam of photon is 22.90°.

Explanation:

Compton Scattering is the process of scattering of X-rays by a charge particle like electron.

The angle of the recoiling electron with respect to the incident beam is determine by the relation :

$\cot\phi = (1+\frac{hf}{m_{e}c^{2} })\tan\frac{\theta }{2}$ ....(1)

Here ∅ is angle of recoil electron, θ is the scattered angle, h is Planck's constant, $m_{e}$ is mass of electron, c is speed of light and f is the frequency of the x-ray photon.

We know that, f = c/λ ......(2)

Here λ is wavelength of x-ray photon.

Rearrange equation (1) with the help of equation (1) in terms of λ .

$\cot\phi = (1+\frac{h}{m_{e}c\lambda })\tan\frac{\theta }{2}$

Substitute 6.6 x 10⁻³⁴ m² kg s⁻¹ for h, 9.1 x 10⁻³¹ kg for $m_{e}$ , 3 x 10⁸ m/s for c, 0.500 x 10⁻⁹ m for λ and 134° for θ in the above equation.

$\cot\phi = (1+\frac{6.6\times10^{-34} }{9.1\times10^{-31}\times3\times10^{8}\times0.5\times10^{-9} })\tan\frac{134 }{2}$

$\cot\phi=2.37$

$\phi$ = 22.90°

8 0

3 years ago

An Apple falls from a tree and one-half second later hits the ground

docker41 [41]

There isnt enough information to answer the question, the missing variable is "distance from said falling spot and ground"

8 0

3 years ago

A car is 200 m from a stop sign and traveling toward the sign at 40.0 m/s. At this time, the driver suddenly realizes that she m

victus00 [196]

Answer:

The acceleration of the car will be $a=9600m/sec^$

Explanation:

We have given that distance from stop sign s = 200 m

Time t = 0.2 sec

We have to find the constant acceleration

Now from second equation of motion $s=ut+\frac{1}{2}at^2$

$200=40\times 0.2+\frac{1}{2}\times a\times 0.2^2$

$a=9600m/sec^$

So the acceleration of the car will be $a=9600m/sec^$

6 0

3 years ago

A mass m attached to a horizontal massless spring with spring constant k, is set into simple harmonic motion. its maximum displa

Lesechka [4]

At the point of maximum displacement (a), the elastic potential energy of the spring is maximum:
$U_i= \frac{1}{2} ka^2$
while the kinetic energy is zero, because at the maximum displacement the mass is stationary, so its velocity is zero:
$K_i =0$
And the total energy of the system is
$E_i = U_i+K= \frac{1}{2}ka^2$

Viceversa, when the mass reaches the equilibrium position, the elastic potential energy is zero because the displacement x is zero:
$U_f = 0$
while the mass is moving at speed v, and therefore the kinetic energy is
$K_f = \frac{1}{2} mv^2$
And the total energy is
$E_f = U_f + K_f = \frac{1}{2} mv^2$

For the law of conservation of energy, the total energy must be conserved, therefore $E_i = E_f$ . So we can write
$\frac{1}{2} ka^2 = \frac{1}{2}mv^2$
that we can solve to find an expression for v:
$v= \sqrt{ \frac{ka^2}{m} }$

6 0

3 years ago

HURRY ! THE BEST I WILL GIVE THE BRAINLIEST​

HURRY ! THE BEST I WILL GIVE THE BRAINLIEST