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4 years ago

Give an example of a balanced force and explain what makes it a balanced force

1 answer:

5 0

When two forces are the same strength but act in opposite direction, they are called balanced forces. Again, tug-of-war is a perfect example. If the people on each side of the rope are pulling with the same strength, but in the opposite direction, the forces are balanced. The result is no motion.

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Maria is trying to meet her friends at the pool but construction is blocking her path from the parking lot directly to the pool.

pychu [463]

Answer:

Her displacement is 35 feet. Her distance is 75 feet.

If she walks 20 ft N, 35 ft E, and 20 ft S, it makes a rectangle with one missing side, which is 35. Her displacement is therefore 35, and 20 plus 35 plus 20 equals 75, which is the distance she walked.

Hope this helps!

6 0

3 years ago

Which statement describes how a machine can help make work easier? It can put out more force than the input force by decreasing

Neko [114]

Answer:

it can be

Explanation:

7 0

3 years ago

The drawing shows a tire of radius R on a moving car

masha68 [24]

Answer:

H / R = 2/3

Explanation:

Let's work this problem with the concepts of energy conservation. Let's start with point P, which we work as a particle.

Initial. Lowest point

Em₀ = K = 1/2 m v²

Final. In the sought height

$Em_{f}$ = U = mg h

Energy is conserved

Em₀ = $Em_{f}$

½ m v² = m g h

v² = 2 gh

Now let's work with the tire that is a cylinder with the axis of rotation in its center of mass

Initial. Lower

Em₀ = K = ½ I w²

Final. Heights sought

Emf = U = m g R

Em₀ = $Em_{f}$

½ I w² = m g R

The moment of inertial of a cylinder is

I = $I_{cm}$ + ½ m R²

I= ½ $I_{cm}$ + ½ m R²

Linear and rotational speed are related

v = w / R

w = v / R

We replace

½ $I_{cm}$ w² + ½ m R² w² = m g R

moment of inertia of the center of mass

$I_{cm}$ = ½ m R²

½ ½ m R² (v²/R²) + ½ m v² = m gR

m v² ( ¼ + ½ ) = m g R

v² = 4/3 g R

As they indicate that the linear velocity of the two points is equal, we equate the two equations

2 g H = 4/3 g R

H / R = 2/3

7 0

3 years ago

A ball is thrown into the air with an upward velocity of 32 feet per second. Its height, h, in feet after t

LekaFEV [45]

Differentiate the expression, to obtain expression for velocity. Set velocity to 0, this when max height is reached. Obtain the tmax from that expression.

h(t) = –16t² + 32t + 6
h'(t) = –32t² + 32
0 = –32t² + 32
t max= 1

hmax = –16(1)² + 32(1) + 6
hmax = 22

Therefore, first option is the correct answer.

8 0

3 years ago

Read 2 more answers

Calculate the heat energy released when 13.3 g of liquid mercury at 25.00 C is converted to solid mercury at its melting point.C

dmitriy555 [2]

Answer:

$-270.321012\ J$