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3 years ago

How much work is done if I use 5N of force to move an object 2 meters

Physics

2 answers:

alina1380 [7]3 years ago

6 0

W=Fd. W=5•2=10. It’d be 10 joules.

frez [133]3 years ago

5 0

Answer:

Explanation:the forumla fr work is force times distance so 10 hope this helps god bless

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Which of the following describes the products of a chemical reaction? A. The original materials B. The substances that are chang

lianna [129]

The products of a chemical reaction are the substances that are changed and the chemicals on the right side of a chemical equation. The correct options are B and C.

<h3>What is chemical reaction?</h3>

The chemical reaction is the reaction between two reactants which led to the formation of products.

The products are substances which forms after reaction. The reactants are the substances which are original materials.

The reactants lie on the left side and products lie on the right side of the reaction.

Thus, the correct options are B and C.

Learn more about chemical reaction.

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2 years ago

Thalia is drafting a plan to move a large, perfect sphere concrete sculpture that is in front of her office building. Describe t

Charra [1.4K]

Answer:She would need to first know the weight of the sculpture and what she is going to move it with then she will need to use newton's second law to calculate the amount of force needed to move it

Explanation: I just did the assignment on edgunity

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3 years ago

Read 2 more answers

A car starts from rest and accelerates uniformly at 2.0 m/s2 toward the north. A second car starts from rest 4.0 s later at the

yKpoI14uk [10]

Answer:

the correct solution is 13 s

Explanation:

This is a kinematic problem, let's use accelerated rectilinear motion relationships.

For the first car it has an accelerometer of 2.0 m/s²

x = v₀₁ t + ½ a₁ t²

The second car leaves the same point, but 4.0 seconds later

x = v₀₂ (t-4) + ½ a₂ (t-4)²

With this form we use the same time for both cars.

The initial speeds are zero for both vehicles leave the rest, at the point where they are located has the same position

x = ½ a₁ t²

x = ½ a₂ (t-4)²

Let's solve

a₁ t² = a₂ (t-4)²

a₁/a₂ t² = t² -2 4 t + 16

t² (1- 2.0 / 4.0) - 8 t +16

t² 0.5 - 8 t +16 = 0

t² -16 t + 32 = 0

Let's solve the second degree equation

t = [16 ±√( 16² - 4 32)] / 2

t = ½ (16 ± 11,3)

Solutions

t1 = 13.66 s

t2 = 2.34 s

These are the mathematical solutions for the meeting point, but car 2 leaves after 4 seconds, so the only solution is 13.66 s

the correct solution is 13 s, if you have to select one the nearest 12s

6 0

3 years ago

The volume of the gas at a temperature of 50 Kelvin on this scale is closest to

8_murik_8 [283]

<span>500 cubic centimeters</span>

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3 years ago

Calculate the work against gravity required to build the right circular cone of height 4 m and base of radius 1.2 m out of a lig

Nana76 [90]

Answer:

Work done = 35467.278 J

Explanation:

Given:

Height of the cone = 4m

radius (r) of the cone = 1.2m

Density of the cone = 600kg/m³

Acceleration due to gravity, g = 9.8 m/s²

Now,

The total mass of the cone (m) = Density of the cone × volume of the cone

Volume of the cone = $\frac{1}{3}\pi r^2 h$

thus,

volume of the cone = $\frac{1}{3}\pi 1.2^2\times 4$ = 6.03 m³

therefore, the mass of the cone = 600 Kg/m³ × 6.03 m³ = 3619.11 kg

The center of mass for the cone lies at the $\frac{1}{4}$ times the total height

thus,

center of mass lies at, h' = $\frac{1}{4}\times4=1m$

Now, the work gone (W) against gravity is given as:

W = mgh'

W = 3619.11kg × 9.8 m/s² × 1 = 35467.278 J

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3 years ago