If the weight of 1kg is 10 N in water what is the density of the stone<br>

What is the wavelength associated with 0.113kg ball traveling with velocity of 43 m/s?

lesya [120]

Answer:

2.73×10¯³⁴ m.

Explanation:

The following data were obtained from the question:

Mass (m) = 0.113 Kg

Velocity (v) = 43 m/s

Wavelength (λ) =?

Next, we shall determine the energy of the ball. This can be obtained as follow:

Mass (m) = 0.113 Kg

Velocity (v) = 43 m/s

Energy (E) =?

E = ½m²

E = ½ × 0.113 × 43²

E = 0.0565 × 1849

E = 104.4685 J

Next, we shall determine the frequency. This can be obtained as follow:

Energy (E) = 104.4685 J

Planck's constant (h) = 6.63×10¯³⁴ Js

Frequency (f) =?

E = hf

104.4685 = 6.63×10¯³⁴ × f

Divide both side by 6.63×10¯³⁴

f = 104.4685 / 6.63×10¯³⁴

f = 15.76×10³⁴ Hz

Finally, we shall determine the wavelength of the ball. This can be obtained as follow:

Velocity (v) = 43 m/s

Frequency (f) = 15.76×10³⁴ Hz

Wavelength (λ) =?

v = λf

43 = λ × 15.76×10³⁴

Divide both side by 15.76×10³⁴

λ = 43 / 15.76×10³⁴

λ = 2.73×10¯³⁴ m

Therefore, the wavelength of the ball is 2.73×10¯³⁴ m.

8 0

3 years ago

A 50.0-g object connected to a spring with a force constant of 35.0 n/m oscillates with an amplitude of 4.00 cm on a frictionles

Dimas [21]

A) The total energy of the system is sum of kinetic energy and elastic potential energy:
$E=K+U= \frac{1}{2}mv^2 + \frac{1}{2}kx^2$
where
m is the mass
v is the speed
k is the spring constant
x is the elongation/compression of the spring

The total energy is conserved, so we can calculate its value at any point of the motion. If we take the point of maximum displacement:
$x=A=4.00 cm = 0.04 m$
then the velocity of the system is zero, so the total energy is just potential energy, and it is equal to
$E=U= \frac{1}{2}kA^2 = \frac{1}{2}(35.0 N/m)(0.04 m)^2=0.028 J$

b) When the position of the object is
$x=1.00 cm = 0.01 m$
the potential energy of the system is
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.01 m)^2 = 1.75 \cdot 10^{-3} J$
and so the kinetic energy is
$K=E-U=0.028 J - 1.75 \cdot 10^{-3}J =0.026 J$
since the mass is $m=50.0 g=0.05 kg$ , and the kinetic energy is given by
$K= \frac{1}{2}mv^2$
we can re-arrange the formula to find the speed of the object:
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.026 J}{0.05 kg} }=1.02 m/s$

c) The potential energy when the object is at
$x=3.00 cm=0.03 m$
is
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J$
Therefore the kinetic energy is
$K=E-U=0.028 J-0.016 J = 0.012 J$

d) We already found the potential energy at point c, and it is given by
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J$

5 0

3 years ago

What is the least count of screw gauge and vernier calliper (9th grade) please help!

nadezda [96]

The least count is the smallest unit of measurement which an instrument can take accurately

7 0

3 years ago

A Jeep Rubicon has a mass of 2,630 kg. With a momentum of 55,200 kg m/s. What was the speed, in m/s, of the vehicle?

Annette [7]

Answer:

The speed of the vehicle was 21 m/s

Explanation:

Momentum can be defined as mass in motion

Momentum depends on the variables mass and velocity

The momentum of an object is equal to the mass of the object times the

velocity of the object

The given is:

→ A Jeep Rubicon has a mass of 2,630 kg

→ With a momentum of 55,200 kg m/s

We need to find the speed of the vehicle

→ The momentum = mass × speed

Substitute the values above in the rule

→ 55,200 = 2,630 × speed

Divide both sides by 2,630

→ speed = 20.99 ≅ 21 m/s

<em>The speed of the vehicle was 21 m/s</em>

4 0

3 years ago

In February 1955, a paratrooper fell 362 m from an airplane without being able to open his chute but happened to land in snow, s

serious [3.7K]

Answer:

s = 0.9689 m

Explanation:

given,

Height of fall of paratroopers = 362 m

speed of impact = 52 m/s

mass of paratrooper = 86 Kg

From from snow on him = 1.2 ✕ 10⁵ N

now using formula

F = m a

a = F/m

$a = \dfrac{1.2 \times 10^5}{86}$

$a =1395.35\ m/s^2$

Using equation of motion

v² = u² + 2 a s

$s =\dfrac{v^2}{2a}$

$s =\dfrac{52^2}{2\times 1395.35}$

s = 0.9689 m

The minimum depth of snow that would have stooped him is s = 0.9689 m

8 0

3 years ago

If the weight of 1kg is 10 N in water what is the density of the stone​

If the weight of 1kg is 10 N in water what is the density of the stone