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Ray Of Light [21]
3 years ago
6

If the mercury barometer is 75.58 cm at the base of the mountain and 66.37 cm at the summit. What is the height of the mountain?

Physics
1 answer:
Licemer1 [7]3 years ago
3 0

Answer:

5016.2446

Explanation:

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Wich usage of water uses the least amount in a year in the united states
Lapatulllka [165]

Live Stock because in 2010 live stock used 10,000 millions gallons of water per day but everything else was higher and irrigation is the highest with 115,000 million gallons per day.

6 0
3 years ago
Read 2 more answers
A ball that is attached to a string travels in a horizontal, circular path, as shown in Figure 1. At time t0 , the ball has a sp
Levart [38]

Answer:

F₁ = 4 F₀

Explanation:

The force applied on the string by the ball attached to it, while in circular motion will be equal to the centripetal force. Therefore, at time t₀, the force on ball F₀ is given as:

F₀ = mv₀²/r   --------------- equation (1)

where,

F₀ = Force on string at t₀

m = mass of ball

v₀ = speed of ball at t₀

r = radius of circular path

Now, at time t₁:

v₁ = 2v₀

F₁ = mv₁²/r

F₁ = m(2v₀)²/r

F₁ = 4 mv₀²/r

using equation (1):

<u>F₁ = 4 F₀</u>

5 0
3 years ago
A student draws a picture of the products and reactants of a chemical reaction. What, if anything, is wrong with the drawing?
ASHA 777 [7]

The complete question is: A student draws a picture of the products and reactants of a chemical reaction. What, if anything, is wrong with the drawing?

A) The drawing is wrong because there are more chemicals on the products side.

B) The drawing is correct because there are 12 compounds on each side of the arrow.

C) The drawing is wrong because there are different compounds on each side of the arrow.

D) The drawing is correct because there are 12 atoms of each type on each side of the arrow.

Answer:

Option D is correct

Explanation:

In the diagram attached below, it can be seen that there are 12 atoms of element which combine with 12 atoms of another element forming a compound. For the drawing to be correct, there should be 12 atoms of each type of element on both the reactants as well as product side, which is the case. There cannot be imbalance in the number of atoms of different elements on the two sides for a chemical reaction to occur.

Hence, option D is correct.

5 0
3 years ago
Three blocks are shown: Three blocks are shown. Block A has mass 3 kilograms, length 6 centimeters, height 4 centimeters, and wi
Yakvenalex [24]

Answer:

Block A has the greatest density.

Explaination:

Block A density:0.0625 kg/cm3

Block B density:0.020833 kg/cm3

Block C density:0.041667 kg/cm3

5 0
3 years ago
A 36.0 kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130
konstantin123 [22]

Answer:

(a) W = 650J

(b) Wf = 529.2J

(c) W = 0J

(d) W = 0J

(e) ΔK = 120.8J

(f) v2 = 2.58 m/s

Explanation:

(a) In order to find the work done by the applied force you use the following formula:

W=Fd      (1)

F: applied force = 130N

d: distance = 5.0m

W=(130N)(5.0m)=650J

The work done by the applied force is 650J

(b) The increase in the internal energy of the box-floor system is given by the work done of the friction force, which is calculated as follow:

W_f=F_fd=\mu Mgd       (2)

μ: coefficient of friction = 0.300

M: mass of the box = 36.0kg

g: gravitational constant = 9.8 m/s^2

W_f=(0.300)(36.0kg)(9.8m/s^2)(5.0m)=529.2J

The increase in the internal energy is 529.2J

(c) The normal force does not make work on the box because the normal force is perpendicular to the motion of the box.

W = 0J

(d) The same for the work done by the normal force. The work done by the gravitational force is zero because the motion of the box is perpendicular o the direction of the gravitational force.

(e) The change in the kinetic energy is given by the net work on the box. You use the following formula:

\Delta K=W_T         (3)

You calculate the total work:

W_T=Fd-F_fd=(F-F_f)d     (4)

F: applied force = 130N

Ff: friction force

d: distance = 5.00m

The friction force is:

F_f=(0.300)(36.0kg)(9.8m/s^2)=105.84N

Next, you replace the values of all parameters in the equation (4):

W_T=(130N-105.84N)(5.00m)=120.80J

\Delta K=120.80J

The change in the kinetic energy of the box is 120.8J

(e) The final speed of the box is calculated by using the equation (3):

W_T=\frac{1}{2}M(v_2^2-v_1^2)       (5)

v2: final speed of the box

v1: initial speed of the box = 0 m/s

You solve the equation (5) for v2:

v_2 = \sqrt{\frac{2W_T}{M}}=\sqrt{\frac{2(120.8J)}{36.0kg}}=2.58\frac{m}{s}

The final speed of the box is 2.58m/s

5 0
3 years ago
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