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3 years ago

6

An object of mass 30 kg is falling in air and experiences a force due to air resistance of 50 newtons. Determine the net force a

cting on the object:

1 answer:

KengaRu [80]3 years ago

3 0

Explanation:

Draw a free body diagram. There are two forces: weight force pulling down and air resistance pushing up.

Sum the forces in the y direction:

∑F = 50 N − (30 kg) (10 m/s²)

∑F = -250 N

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What are gears and what is a lever ?

blagie [28]

Answer:

agear lever or agear stick is the lever that you use to change gear in acar or other vehicle

5 0

3 years ago

A metal wire breaks when its tension reaches 100 newton. If the radius and length of the wire were both doubled then it would br

serg [7]

Answer:

200 N

Explanation:

Since Young's modulus for the metal, E = σ/ε where σ = stress = F/A where F = force on metal and A = cross-sectional area, and ε = strain = e/L where e = extension of metal = change in length and L = length of metal wire.

So, E = σ/ε = FL/eA

Now, since at break extension = e.

So making e subject of the formula, we have

e = FL/EA = FL/Eπr² where r = radius of metal wire

Now, when the radius and length are doubled, we have our extension as e' = F'L'/Eπr'² where F' = new force on metal wire, L' = new length = 2L and r' = new radius = 2r

So, e' = F'(2L)/Eπ(2r)²

e' = 2F'L/4Eπr²

e' = F'L/2Eπr²

Since at breakage, both extensions are the same, e = e'

So, FL/Eπr² = F'L/2Eπr²

F = F'/2

F' = 2F

Since F = 100 N,

F' = 2 × 100 N = 200 N

So, If the radius and length of the wire were both doubled then it would break when the tension reached 200 Newtons.

7 0

3 years ago

A car accelerates from rest at -3.00m/s^2. What is the velocity at the end of 5.0s? What is the displacement after5.0s?

Andrew [12]

Speed = (acceleration) x (time)
Velocity = (speed) in (direction of the speed)

Speed = (-3 m/s²) x (5 s) = 15 m/s
Velocity =
   (15 m/s) in the direction opposite to the direction you call positive.

Displacement = (distance between start-point and end-point)
   in the direction from start-point to end-point.

Distance = (1/2) (acceleration) (time)²
Distance = (1/2) (3 m/s²) (5 s)²
   = (1/2) (3 m/s²) (25 s²) = 37.5 meters

Displacement =
   37.5 meters in the direction opposite to the direction you call positive.

5 0

3 years ago

Three strings, attached to the sides of a rectangular frame, are tied together by a knot as shown in the figure. The magnitude o

Anna [14]

The magnitude of the tension in the string marked A is 52.5N

Generally, the equation for is mathematically given as

Let's take θ be an angle at A

So, tanθ = 3/8

Let's take α be an angle at B (Below X)

tanα = 5/4

Let's take β be an angle at C (Below x)

tanβ = 1/6

First we take the Horizontal Components

74.9cos(9.46°) = Acos(20.6°) + Bcos(51.3°)

By solving the equation, we get

A = 78.9 - 0.668B … (1)

Now, we take the vertical components

74.9sin(9.46°) + Asin(20.6°) = Bsin(51.3°)

By solving the equation, we get

40.07 = 1.015B

B = 39.5N

By substituting the value of B in equation (1)

A = 78.9 - 0.6668× 39.5

A = 52.5N

Hence, the magnitude of the tension in the string marked A is 52.5N

Learn more about Tension here brainly.com/question/2287912

#SPJ1

8 0

2 years ago

determine the force of gravitational attraction between a 78kg boy sitting 2 meters away from a 65kg girl.

bixtya [17]

Answer:

$F=8.45\times 10^{-8}\ N$

Explanation:

Given that,

Mass of a boy is 78 kg

Mass of a girl is 65 kg

We need to find the force of gravitational attraction between them if they are 2 m away.

The formula for the gravitational force is given by :

$F=\dfrac{Gm_1m_2}{r^2}\\\\F=\dfrac{6.67\times 10^{-11}\times 78\times 65}{(2)^2}\\\\=8.45\times 10^{-8}\ N$

So, the force between them is $8.45\times 10^{-8}\ N$ .

8 0

2 years ago