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Step2247 [10]
3 years ago
12

A man climbs a ladder. Which two quantities can be used to calculate the energy stored of the man at the top of the ladder.

Physics
1 answer:
Dvinal [7]3 years ago
8 0

Answer:The answer must be The weight of the man and the vertical distance moved.

Explanation: you calculate it by the force you applied times the distance you moved

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Hi,please i need help with this one in physics.hurry and correct i need it.<br>Topic:Equilibrium.​
Rudik [331]

The half-meter rule (easy math) is 0.5 meters or 50 centimeters since a meter is 1 meters long, which is equivalent to 100 centimeters. Therefore, we shall apply the 50 cm rule.

A 50 cm rule's center of mass is now 25 cm away.

Additionally, according to the data, the object is pivoted at 15 cm, while the 40 g object is hung at 2 cm from the rule's beginning. Using a straightforward formula, we can compare the two situations: the distance from the pivot to the center of the mass times the mass of the 40 g object divided by 2 cm must equal the distance from the pivot to the center of the mass times mass of the 10 x g object

The result of the straightforward computation must be 52g.

Most simplified version:

the center of mass of the rule is at the 25 cm mark

⇒ 40 g * (15 cm - 2 cm)

⇒ = M * (25 cm - 15 cm)

#SPJ2

5 0
2 years ago
The normal force of a parked car is 15,000 Newtons. The coefficient of static friction between the rubber of the tires and the a
Shtirlitz [24]

Answer:

11250 N

Explanation:

From the question given above, the following data were obtained:

Normal force (R) = 15000 N

Coefficient of static friction (μ) = 0.75

Frictional force (F) =?

Friction and normal force are related by the following equation:

F = μR

Where:

F is the frictional force.

μ is the coefficient of static friction.

R is the normal force.

With the above formula, we can calculate the frictional force acting on the car as follow:

Normal force (R) = 15000 N

Coefficient of static friction (μ) = 0.75

Frictional force (F) =?

F = μR

F = 0.75 × 15000

F = 11250 N

Therefore, the frictional force acting on the car is 11250 N

3 0
3 years ago
Air (14.5 lb) undergoes a polytropic process in a closed system from p1 = 80 lbf/in2, υ1 = 4 ft3/lb to a final state where p2 =
Yanka [14]
The energy transfer in terms of work has the equation:

W = mΔ(PV)

To be consistent with units, let's convert them first as follows:

P₁ = 80 lbf/in² * (1 ft/12 in)² = 5/9 lbf/ft²
P₂ = 20 lbf/in² * (1 ft/12 in)² = 5/36 lbf/ft²
V₁ = 4 ft³/lbm
V₂ = 11 ft³/lbm

W = m(P₂V₂ - P₁V₁)
W = (14.5 lbm)[(5/36 lbf/ft²)(4 ft³/lbm) - (5/9 lbf/ft²)(11 lbm/ft³)]
W = -80.556 ft·lbf

In 1 Btu, there is 779 ft·lbf. Thus, work in Btu is:
W = -80.556 ft·lbf(1 Btu/779 ft·lbf)
<em>W = -0.1034 BTU</em>


4 0
3 years ago
If all else stays the same, which would cause an increase in the gravitational force on a space shuttle?
maks197457 [2]
"decreasing the distance of the space shuttle from Earth" 

F = Gm(1)m(2)/R²
where R is the distance between the 2 objects, as it decreases, the force increases. 

5 0
4 years ago
Read 2 more answers
A person walks in the following pattern: 3.1 km north, then 2.4 km west, and finally 5.2 km south. a) Sketch the vector diagram
zysi [14]

Answer:

d = 3.19 km

direction is given as

\theta = 41.2 degree South of West

Explanation:

Part b)

displacement is given as

d_1 = 3.1 \hat j

d_2 = 2.4 \hat i

d_3 = 5.2(-\hat j)

now we will have

d = d_1 + d_2 + d_3

d = 2.4 \hat i + (3.1 - 5.2)\hat j

d = 2.4 \hat i - 2.1 \hat j

total displacement is given as

d = \sqrt{2.4^2 + 2.1^2}

d = 3.19 km

direction is given as

tan\theta = \frac{-2.1}{2.4}

\theta = 41.2 degree South of West

3 0
4 years ago
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