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Korolek [52]
2 years ago
9

Camping equipment weighing 6000 n is pulled across a frozen lake by means of a horizontal rope. the coefficient of kinetic frict

ion is 0.05. how much work is done by the campers in pulling the equipment 1000 m if its speed is increasing at the constant rate of 0.20 m/s2?

Physics
1 answer:
Triss [41]2 years ago
8 0
To solve the problem you must make a free body diagram and see what forces act on the body. In vertical direction you have the normal force w and the weight mg. In horizontal direction you have the strength due to friction and strength due to the increase in speed. Attached solution.

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Alicia can row 6 miles downstream in the same time it takes her to row 4 miles upstream. She rows downstream 3 miles/hour faster
m_a_m_a [10]
Let us assume the upstream rowing rate of Alicia = x
Let us assume the downstream rowing rate of Alicia = y
We already know that
Travelling time = Distance traveled/rowing rate
Then
6/(x + 3) = 4/x
6x = 4x + 12
6x - 4x = 12
2x = 12
x = 6
Then
Rowing rate of Alicia going upstream = 6 miles per hour
Rowing rate of Alicia going downstream = 9 miles per hour.
4 0
2 years ago
Read 2 more answers
A large container, 120 cm deep is filled with water. If a small hole is punched in its side 77.0 cm from the top, at what initia
prisoha [69]

Answer:

The water will flow at a speed of 3,884 m/s

Explanation:

Torricelli's equation

v = \sqrt{2gh}

*v = liquid velocity at the exit of the hole

g = gravity acceleration

h = distance from the surface of the liquid to the center of the hole.

v = \sqrt{2*9,8m/s^2*0,77m} = 3,884 m/s

6 0
2 years ago
A 5.0-n projectile leaves the ground with a kinetic energy of 220 j. at the highest point in its trajectory, its kinetic energy
NikAS [45]
First, we get the difference between the kinetic energies such that,
             difference = (220J - 120J)
             difference = 100 J
The difference in kinetic energy is the equivalent of the potential energy which is calculated through the equation,
              PE = mgh
To calculate for the height, we derive the equation in a form,
           h = PE/mg
The product of the mass and acceleration due to gravity is the weight. 
                   h = (100 J) / (5 N)
                   h = 20 m

<em>Hence, the answer is 20 m. </em>
3 0
2 years ago
(YOU WILL GET BRAINLIEST)Matter may be classified as a pure substance or a mixture. Where on the Venn diagram would you insert t
inessss [21]
I think it would be B because it is matter, since it has atoms, and it contains subatomic particles, which are smaller than atoms
3 0
2 years ago
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Cesium-137 undergoes beta decay and has a half-life of 30.0 years. How many beta particles are emitted by a 14.0-g sample of ces
Mandarinka [93]

Answer: 0.81\times 10^{16} beta particles

Explanation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass = 14.0 g

Molar mass = 137 g/mol

\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

1 mole of cesium contains atoms =  6.023\times 10^{23}

0.102 moles of cesium contains atoms =  \frac{6.023\times 10^{23}}{1}\times 0.102=0.614\times 10^{23}

The relation of atoms with time for radioactivbe decay is:

N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}

Where N_t =atoms left undecayed

N_0 = initial atoms

t = time taken for decay = 3 minutes

{t_{\frac{1}{2}}} = half life = 30.0 years = 1.577\times 10^7 minutes

The fraction that decays  :  1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}

Amount of particles that decay is  = 0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}

Thus 0.81\times 10^{16} beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.

7 0
2 years ago
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