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Greeley [361]
2 years ago
11

A stone weighing 1.5 kilograms is resting on a rock at a height of 20 meters above the ground. The stone rolls down 10 meters an

d comes to rest on a patch of moss. The gravitational potential energy of the stone on the moss is joules. (Use PE = m × g × h, where g = 9.8 N/kg.)
Physics
2 answers:
Andrews [41]2 years ago
6 0
The potential energy of the moss at its position is 147 Joules. 
34kurt2 years ago
4 0
Given that the equation for the potential energy of an object is: PE = mgh, where, m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object above the ground, simply substitute the given values to obtain the potential energy. In this case, m = 1.5 kg, g = 9.8 N/kg, and h = 10 m. The height is set to 10 m since this is the height of the moss above the ground after it fell. Thus, the potential energy of the moss at its position is 147 Joules. 
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NeX [460]

Answer:

1.056 x 10⁷ lb-ft

Explanation:

v = Speed of the bike = 20 mph

t = time of travel = 2 h

d = distance traveled by cyclist

Distance traveled by cyclist is given as

d = v t

d = (20) (2)

d = 40 miles

We know that, 1 mile = 5280 ft

d = 40 (5280) ft

d = 211200 ft

F = force applied by cyclist = 50 lb

W = work done by cyclist

Work done by cyclist is given as

W = F d

W = (50) (211200)

W = 1.056 x 10⁷ lb-ft

5 0
2 years ago
A charge q = 3 × 10-6 C of mass m = 2 × 10-6 kg, and speed v = 5 × 106 m/s enters a uniform magnetic field. The mass experiences
NeX [460]

Answer:

Magnetic field, B = 0.004 mT

Explanation:

It is given that,

Charge, q=3\times 10^{-6}\ C

Mass of charge particle, m=2\times 10^{-6}\ C

Speed, v=5\times 10^{6}\ m/s

Acceleration, a=3\times 10^{4}\ m/s^2

We need to find the minimum magnetic field that would produce such an acceleration. So,

ma=qvB\ sin\theta

For minimum magnetic field,

ma=qvB

B=\dfrac{ma}{qv}

B=\dfrac{2\times 10^{-6}\ C\times 3\times 10^{4}\ m/s^2}{3\times 10^{-6}\ C\times 5\times 10^{6}\ m/s}

B = 0.004 T

or

B = 4 mT

So, the magnetic field produce such an acceleration at 4 mT. Hence, this is the required solution.

4 0
2 years ago
A machine produces photo detectors in pairs. Tests show that the first photo detector is acceptable with probability 3/5. When t
klasskru [66]

Answer:

a.a. \ \frac{7}{25}

b.\ \ \ P(D_1D_2)=\frac{6}{25}

Explanation:

a. Find the probability that exactly one photo detector of a pair is acceptable:

Let A_i=i^{th} photo is accepted and the probability D_i=i^{th} is defected.

Therefore:

P(A_i)=3/5,\ P(A_2|A_1)=4/5,\ \ P(A_2|D_1)=2/5\\\\\\=P(A_1D_2)+P(D_1A_2)\\\\=\frac{3}{5}\times\frac{1}{5}+\frac{2}{5}\times\frac{2}{5}\\\\=\frac{7}{25}

#The probability of exactly one photo detector of a pair is accepted is 7/25

b.Find the probability that both photo detectors in a pair are defective,P(D1D2):

P(D_1D_2)=\frac{2}{5}\times \frac{3}{5}\\\\=\frac{6}{25}

Hence, from out tree diagram,the probability that both photo detectors in a pair are defective is 6/25

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nataly862011 [7]

Answer:

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Answer:

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Explanation:

8 0
2 years ago
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