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10 days ago

For a proton (mass = 1.673 x 10–27 kg) moving with a velocity of 2.83 x 104 m/s, what is the de Broglie wavelength (in pm)?

1 answer:

6 0

The calculated de Broglie wavelength is 14.0 pm. Given the mass of a proton at 1.673 x 10⁻²⁷ kg and its velocity of 2.83 x 10⁴ m/s, the de Broglie wavelength is derived from the formula defined by Planck's constant.

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A proton is accelerated from rest through a potential difference V0 and gains a speed v0. If it were accelerated instead through

inna [2740]

Answer: The resulting speed is $\sqrt{2}v_{0}$ . Option (a) stands as the correct choice. Explanation: Given the context, the potential difference entails calculations linked to speed assessment. If instead accelerated through a different potential difference, the resulting speed will be computed accordingly.

6 0

14 days ago

A 55-kg pilot flies a jet trainer in a half vertical loop of 1200-m radius so that the speed of the trainer decreases at a const

Keith_Richards [2907]

a) $-1.54 m/s^2$

b) 803.4 N

Explanation:

a) At point C (top of the loop), the pilot experiences weightlessness, leading to the normal force from the seat being zero:

N = 0

Consequently, the force balance equation at position C becomes:

$mg=m\frac{v^2}{r}$ where the left term signifies the pilot's weight and the right term represents the centripetal force, with:

= acceleration due to gravity

= jet's velocity at the top $g=9.8 m/s^2$

= loop radius

$v$ By solving for v,

$r=1200 m$

Thus, this is the jet's speed at C.

The speed at position A (bottom) can be derived from $v_C=\sqrt{gr}=\sqrt{(9.8)(1200)}=108.4 m/s$

The distance traveled by the jet corresponds to half the circumference of the circle with radius r, therefore

$v_A=550 km/h =152.8 m/s$

Given the plane's deceleration is consistent, we can obtain it using the following equation:

$s=\pi r=\pi(1200)=3770 m$

b) The pilot experiences a force equal to the normal force from the seat. At point B (half-way through the loop), we find:

$v_C^2-v_A^2=2as\\a=\frac{v_C^2-v_A^2}{2s}=\frac{108.4^2-152.8^2}{2(3770)}=-1.54 m/s^2$ - The normal force from the seat, N, directed towards the center of the loop

- As there are no further forces acting toward the central axis, N must equal the centripetal force:

(1)

where

represents the speed at position B.

To deduce the velocity at B, we note that the distance covered by the jet between positions A and B is a quarter of a circle:

With knowledge of the deceleration, we can implement the equation of motion to find the velocity at the midway point B: $N=m\frac{v_B^2}{r}$

$v_B$

Thus, we then apply eq.(1) to determine the normal force acting on the pilot at B:

$s=\frac{\pi r}{2}=\frac{\pi(1200)}{2}=1885 m$

6 0

16 days ago

The same physics student jumps off the back of her Laser again, but this time the Laser is

Yuliya22 [2968]

a) The student's speed after jumping is 1.07 m/s

b) The final speed of the laser is 10.4 m/s

Explanation:

This issue can be approached through the momentum conservation principle: In the absence of external forces, the combined momentum of the student and the laser must remain unchanged. Hence, we can express:

$p_i = p_f\\0=mv+MV$

where:

The initial momentum is zero

m = 42 kg signifies the mass of the laser

v = 1.5 m/s is the laser's final velocity

M = 59 kg is the mass of the student

V denotes the student's final velocity

Solving this for V, we can determine the student's speed:

$V=-\frac{mv}{M}=-\frac{(42)(1.5)}{59}=-1.07 m/s$

Thus, the student's final speed calculates to 1.07 m/s.

Here, both the laser and the student have a combined speed of 3.1 m/s prior to the student's jump; thus, the initial momentum isn't zero.

<pSo, we formulate the equation of momentum conservation as:

$(m+M)u=mv+MV$

where:

m = 42 kg denotes the mass of the laser

M = 59 kg is the student’s mass

u = 3.1 m/s is their starting velocity

V = -2.1 m/s indicates the student's speed post-jump (she jumps backward)

v signifies the laser's final speed

When we resolve for v, we have:

$v=\frac{(m+M)u-MV}{m}=\frac{(42+59)(3.1)-(59)(-2.1)}{42}=10.4 m/s$

Learn more about momentum:

3 0

1 month ago

Read 2 more answers

A force on a particle depends on position such that F(x) = (3.00 N/m2)x2 + (6.00 N/m)x for a particle constrained to move along

Sav [2826]

Response:

The work performed by the particle traveling from x = 0 to x = 2 m totals 20 J.

Details:

The force impacting a particle, which is restricted to the x-axis, is expressed as follows:

$F(x)=(3\ N/m^2)x^2+(6\ N/m)x$

We need to calculate the work done on a particle moving from x = 0.00 m to x = 2.00 m.

The formula for the work done by the particle is defined as:

$W=\int\limits {F{\cdot} dx}$

$W=\int\limits^2_0 {(3x^2+6x){\cdot} dx} \\\\W={(x^3}+3x^2)_0^2\\\\\W={(2^3}+3(2)^2)\\\\W=20\ J$

Consequently, the work executed by the particle between x = 0 and x = 2 m amounts to 20 J. Thus, this is the solution sought.

3 0

1 month ago

Read 2 more answers

A rock hits the ground at a speed of 15 m/s and leaves a hole 50 cm deep. After it hits the ground, what is the magnitude of the

kicyunya [2911]

Conclusion:

B) 225 m/s^2

Details:

The rock impacts the ground at 15m/s and penetrates 50cm=0.5m into the ground until it comes to a halt.

Assuming the acceleration is constant, the formula we will use is $v^2=v_0^2+2ad$ , which for acceleration is expressed as:

$a=\frac{v^2-v_0^2}{2d}$

Assuming the downward direction is taken as positive (the direction of travel), both the initial velocity and displacement will be considered positive, and we substitute our values aligned with this movement:

$a=\frac{v^2-v_0^2}{2d}=\frac{(0m/s)^2-(15m/s)^2}{2(0.5m)}=-255m/s^2$

The negative sign denotes upward direction.

4 0

28 days ago