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Elden [556K]
3 years ago
8

A solid uniform ball of mass 1.0 kg and radius 1.0 cm starts from rest and rolls down a 1.0-m high ramp. there is enough frictio

n on the ramp to prevent the ball from slipping as it rolls down. A). What is the forward speed of the ball when it reaches the bottom of the ramp?
B). What would be the forward speed of the ball if there were no friction on the ramp?
C). Since the ball starts from the Same Height in both cases, why is the speed different?
Physics
1 answer:
Luba_88 [7]3 years ago
4 0
The correct answer is A
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Match the bones with their common name.
mrs_skeptik [129]

1. clavicle = collarbone

2. vertebrae = backbone

3. scapula = shoulder blade

4. femur = thigh

5. humerus = upper arm

6. patella = kneecap

7. cranium = skull

8. tibia = lower leg

9. radius/ulna = forearm

10. phalanges = fingers/toes

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3 years ago
When the voltage is
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Explanation:

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2 years ago
A 1-kg collar (located at point (2,2) from the origin) is pulled along a vertical, frictionless bar with a force of 10 N applied
faltersainse [42]

Answer:

The acceleration of the collar is 10 m/s²

Explanation:

Given;

mass of the collar, m = 1 kg

applied force on the bar, F = 10 N

The acceleration of the collar can be calculated by applying Newton's second law of motion;

F = ma

where;

F is the applied force

m is mass of the object

a is the acceleration

a = F / m

a = 10 / 1

a = 10 m/s²

Therefore, the acceleration of the collar is 10 m/s²

3 0
2 years ago
During the first 6 years of its operation, the Hubble Space Telescope circled the Earth 37,000 times, for a total of 1,280,000,0
madreJ [45]

Answer:

Kilometers\ in\ 1\ Orbit=\frac{1.28*10^9 Km}{3.7*10^4orbits}

Kilometers\ in\ 1\ Orbit=3.46*10^4 Km/Orbit

Explanation:

Given Data:

Numbers of times Telescope cycled around the earth in 6 years=37,000 times

Total Distance traveled in 6 years by the Hubble Space Telescope=1,280,000,000 Km

Find:

Kilometers in one Orbit=?

Solution:

Kilometers in 37,000 Orbits=1,280,000,000 Km

Kilometers in 1 Orbit=1,280,000,000/37,000

In Scientific Notation:

Kilometers\ in\ 3.7*10^4\ Orbits=1.28*10^9 Km

Kilometers\ in\ 1\ Orbit=\frac{1.28*10^9 Km}{3.7*10^4 orbits}

Kilometers in 1 Orbit=34594.594 Km

Kilometers in 1 Orbit in Scientific notation:

Kilometers\ in\ 1\ Orbit=3.46*10^4 Km

8 0
3 years ago
A recent home energy bill indicates that a household used 475475 kWh (kilowatt‑hour) of electrical energy and 135135 therms for
faltersainse [42]

Answer:

Explanation:

Given that,.

A house hold power consumption is

475 KWh

Gas used is

135 thermal gas for month

Given that, 1 thermal = 29.3 KWh

Then,

135 thermal = 135 × 29.3 = 3955.5 KWh

So, total power used is

P = 475 + 3955.5

P =4430.5 KWh

Since 1 hr = 3600 seconds

So, the energy consumed for 1hr is

1KW = 1000W

P = energy / time

Energy = Power × time

E = 4430.5 KWhr × 1000W / KW × 3600s / hr

E = 1.595 × 10^10 J

So, using Albert Einstein relativity equation

E = mc²

m = E / c²

c is speed of light = 3 × 10^8 m/s

m = 1.595 × 10^10 / (3 × 10^8)²

m = 1.77 × 10^-7 kg

Then,

1 kg = 10^6 mg

m = 1.77 × 10^-7 kg × 10^6 mg / kg

m = 0.177mg

m ≈ 0.18 mg

5 0
3 years ago
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