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A high jumper jumps 2.04 m. If the jumper has a mass of 67 kg, what is his gravitational potential energy at the highest point i

Mariulka [41]

Answer: 1339.5 joules

Explanation:

Gravitational potential energy, GPE is the energy possessed by the jumper as he moves against gravity.

Thus, GPE = Mass m x Acceleration due to gravity g x Height h

Since Mass = 67kg

g = 9.8m/s^2

h = 2.04 metres

Thus, GPE = 67kg x 9.8m/s^2 x 2.04m

GPE = 1339.5 joules

Thus, the gravitational potential energy at the highest point is 1339.5 joules

3 0

3 years ago

A particle has a de Broglie wavelength of 2.1 × 10-10m. Then its kinetic energy increases by a factor of 3. What is the particle

svlad2 [7]

Answer:

$\lambda' =1.21\times 10^{-10}\ m$

Explanation:

given,

de broglie wavelength of the particle = 2.1 × 10⁻¹⁰m

Kinetic energy increase factor = 3

de broglie wavelength of new particle = ?

we know,

$P = \dfrac{h}{\lambda}$

Kinetic energy

$K =\dfrac{p^2}{2m}$

$K =\dfrac{h^2}{2m\lambda^2}$ ...........(1)

Kinetic energy of other particle

$3K =\dfrac{h^2}{2m\lambda'^2}$ ............(2)

dividing equation (1)/(2)

$\dfrac{1}{3} =\dfrac{\lambda'^2}{\lambda^2}$

$\lambda' = \dfrac{\lambda'}{\sqrt{3}}$

$\lambda' = \dfrac{2.1\times 10^{-10}}{\sqrt{3}}$

$\lambda' =1.21\times 10^{-10}\ m$

6 0

3 years ago

What describes how to compare the energy of two different waves

Mazyrski [523]

Answer:

the answer is C

Explanation:

If you need an explanation let me know and I'll be sure to provide one

3 0

3 years ago

Read 2 more answers

A proton moving to the right at 6.2 × 105 ms-1 enters a region where there is an electric field of 62 kNC-1 directed to the left

aniked [119]

Answer:

2.06 x 10^-7 s

Explanation:

For going opposite to the applied electric field

u = 6.2 x 10^5 m/s, v = 0 , q = 1.6 x 19^-19 C, E = 62000 N/C,

m = 1.67 x 106-27 Kg, t1 = ?

Let a be the acceleration.

a = q E / m = (1.6 x 10^-19 x 62000) / (1.67 x 10^-27) = 6 x 10^12 m/s^2

Use first equation of motion

v = u + a t1

0 = 6.2 x 10^5 - 6 x 10^12 x t1

t1 = 1.03 x 10^-7 s

Let the distance travelled is s.

Use third equation of motion

V^2 = u^2 - 2 a s

0 = (6.2 x 10^5)^2 - 2 x 6 x 10^12 x s

s = 0.032 m

Now when the proton moves in the same direction of electric field.

Let time taken be t2

use second equation of motion

S = u t + 1/2 a t^2

0.032 = 0 + 1/2 x 6 x 10^12 x t2^2

t2 = 1.03 x 10^-7 s

total time taken = t = t1 + t2

t = 1.03 x 10^-7 + 1.03 x 10^-7 = 2.06 x 10^-7 s

7 0

3 years ago

A 1200 kg car moving at 5.8 m/s is initially traveling north in the positive y direction. After completing a 90° right-hand turn

Keith_Richards [23]

Answer

given,

mass of car = 1200 Kg

initial velocity in north = 5.8 m/s

90° right-hand turn in time = 3.3 s

car stop in, time = 270 ms

a) impulse = change in momentum

Initial momentum = ( m x v) j

= ( 5.8 x 1200 ) j

= ( 6960) j

final momentum after turn is 90°

= 6960 i

impulse = 6960 i - 6960 j

impulse = 6960 ( i - j )

b) due to collision

initial momentum after turning = 6960 i

final momentum is equal to zero

impulse = 0 - 6960 i

impulse = - 6960 i

c) during turn

I = F Δ t

$F = \dfrac{I}{t}$

$F = \dfrac{6960 ( \hat{i} -\hat{j})}{3.3}$

$F =2109.1 ( \hat{i} -\hat{j})$

magnitude of force

$F =2109.1\sqrt{2}$

$F =2982.72\ N$

d) during collision

$F = \dfrac{I}{t}$

$F = \dfrac{-6960 \hat{i} }{0.27}$

$F =-25777.78\hat{i}\ N$

magnitude of average force is equal to

$F =25777.78\ N$

e) angle between average force

$F =2109.1 ( \hat{i} -\hat{j})$

$\theta =tan^{-1}(\dfrac{-1}{1})$

$\theta =-45^0$

4 0

4 years ago