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3 years ago

Time period from one noon to the next

1 answer:

3 0

That's the period of time known as one solar "day". We subdivide it into 24 slices which we call "hours". Using this system of time units, the day is about 4 minutes longer than one complete axial rotation of the Earth.

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When it is brought near a neutral object, a charged object can _________ a charge in a neutral object.A) induce, B) cancel,C) ne

Bess [88]

The answer is a induce.

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2 years ago

IF it possible for an object to move for 10 seconds at a high speed and end up with an average velocity of zero? true or false

marishachu [46]

True, if you move something forward at 100 miles an hour but your on something moving backwards 100 miles an hour you up staying in the same location, aka zero velocity.

4 0

3 years ago

Read 2 more answers

How much heat do you need to raise the temperature of 150 g of oxygen from -30c to -15c?

Naddika [18.5K]

The amount of heat needed to raise the temperature of a substance by $\Delta T$ is given by
$Q=m C_s \Delta T$
where
m is the mass of the substance
Cs is its specific heat capacity
$\Delta T$ is the increase in temperature

For oxygen, the specific heat capacity is approximately
$C_s = 0.92 J/(g K)$
The variation of temperature for the sample in our problem is
$\Delta T= -15^{\circ}C-(-30^{\circ} C)=+15^{\circ}C=15 K$
while the mass is m=150 g, so the amount of heat needed is
$Q=m C_s \Delta T=(150 g)(0.92 J/g K)(15 K)=2070 J$

4 0

3 years ago

Read 2 more answers

In a single movable pulley, a load of 500 N is lifted by applying 300 N effort. Calculate MA, VR and efficiency.

galben [10]

Answer:

in pulley there are different kinds.but the most common one are fixed,moveable and compound pulley. in this question we asked about movable pulley.

Explanation:

Given request solutions

L=500N a,M.A=? a,M.A=L/E =5/3

E=300N b,V.R=? b,V.R=2

c,efficiency =? c,£=M.A/V.R=5/6

£=IS NOT THE REAL SYMBOLS OF EFFICIENCY BUT I IS LOOK LIKE THIS

I THINK I HELPED

6 0

2 years ago

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.8×106N , one an angle 11degrees west of north an

sp2606 [1]

Answer: 2,9 ×10¹⁰J.

Explanation:

1. The work is a measure of energy and is calculated as the product of the displacement times the parallel force to such displacement.

That means that the only components of the force that contribute to work are those that result parallel to the displacement.

2. Since it is given that the two tugboats "pull the tanker a distance 0.83km toward the north", that is the displacement, and you have to calculate the net force toward the north.

3. Tugboat #1.

a) Force magnitude: F₁ = 1.8×10⁶N

b) Angle: α = 11° West of North

c) North component of the force F₁: Fy₁ = F₁cos(α) = 1.8×10⁶N × cos(11°) = 1.77×10⁶N

4. Tugboat #2:

a) Force magnitude: F₂ = 1.8×10⁶N

b) Angle: = 11° East of North

c) North component of the force F₂: Fy₂ = F₂cos(β) = 1.8×10⁶N × cos(11°) = 1.77×10⁶N =

5. Total net force, Fn:

Fn = Fy₁ + Fy₂ = 1.77×10⁶N + 1.77×10⁶N = 3.54×10⁶N

6. Work, W:

Displacement, d = 0.83 km = 8,300 m

W = Fn×d = 3.54×10⁶N×8,300m = 29,000 ×10⁶J = 2,9 ×10¹⁰J

The answer is rounded to two significant figures because both data, Force and displacement, have two significant figures.

7 0

2 years ago