5. What is the elevation along the shoreline (sea level)? O ft 1 ft 10 ft 100 ft

Two uniform solid spheres of the same size, but different mass, are released from rest simultaneously at the same height on a hi

SpyIntel [72]

Answer:

options A and C

Explanation:

Since, the spheres are of same size and rotational speed of the sphere are not dependent on their masses. So, both the sphere will reach the bottom of the at the same time with the same speed. But their kinetic energies are different.

So, options A and C are correct.

4 0

3 years ago

Which statement best describes perigee?

algol13

A. The closet point in the Moon's orbit to Earth . . . . . perigee

B. The farthest point in the Moon's orbit to Earth . . . . . apogee

C. The Sun's orbit that is closest to the Moon . . . . . a meaningless description

D. The closest point in Earth's orbit of the Sun . . . . . perihelion

-- The farthest point in Earth's orbit of the Sun . . . . . aphelion

7 0

4 years ago

A sprinter must average 24.0 mi/h to win a 100-m dash in 9.30 s. What is his wavelength at this speed if his mass is 84.5 kg?

crimeas [40]

Answer:

Wavelength λ = 7.31 × 10^-37 m

Explanation:

From De Broglie's equation;

λ = h/mv

Where;

λ = wavelength in meters

h = plank's constant = 6.626×10^-34 m^2 kg/s

m = mass in kg

v = velocity in m/s

Given;

v = 24 mi/h

Converting to m/s

v = 24mi/h × 0.447 m/s ×1/(mi/h)

v = 10.73m/s

m = 84.5kg

Substituting the values into the equation;

λ = (6.626×10^-34 m^2 kg/s)/(84.5kg × 10.73m/s)

λ = 7.31 × 10^-37 m

7 0

4 years ago

Read 2 more answers

A 2100 g block is pushed by an external force against a spring (with a 22 N/cm spring constant) until the spring is compressed b

Vilka [71]

Answer:

6.5e-4 m

Explanation:

We need to solve this question using law of conservation of energy

Energy at the bottom of the incline= energy at the point where the block will stop

Therefore, Energy at the bottom of the incline consists of the potential energy stored in spring and gravitational potential energy= $\frac{1}{2} kx^{2} +PE1$