Ask question

Ask question

All categories

valentinak56 [21]

2 years ago

10

Suppose that you lift four boxes individually, each at a constant velocity. The boxes have weights of 3.0 N, 4.0 N, 6.0 N, and 2

.0 N, and you do 12 J of work on each. Match each box to the vertical distance through which it is lifted.

1 answer:

Alona [7]2 years ago

8 0

Answer:

The vertical distance of weight 3.0 N = 4 m, vertical distance of weight 4.0 N = 3 m, vertical distance of weight 6.0 N = 2 m, vertical distance of weight 2.0 N = 6 m

Explanation:

Worked : work can be defined as the product of force and distance.

The S.I unit of work is Joules (J).

Mathematically it can be represented as,

W = F×d.................. Equation 1

d = W/F.............................. Equation 2

where W = work, F = force, d = distance.

<em>Given: W = 12 J</em>

(i) for the 3.0 N weight,

using equation 2

d = 12/3

d= 4 m.

(ii) for the 4.0 N weight,

d = 12/4

d = 3 m.

(iii) for the 6.0 N weight,

d = 12/6

d = 2 m.

(iv) for the 2.0 N weight,

d = 12/2

d = 6 m

Therefore vertical distance of weight 3.0 N = 4 m, vertical distance of weight 4.0 N = 3 m, vertical distance of weight 6.0 N = 2 m, vertical distance of weight 2.0 N = 6 m

You might be interested in

An object is located 13.5 cm in front of a convex mirror, the image being 7.05 cm behind the mirror. A second object, twice as t

goldfiish [28.3K]

Answer:

Second object is located at 42.03 cm in front of mirror

Explanation:

In this question we have given,

object distance from convex mirror ,u=-13.5cm

Image distance from convex mirror,v=7.05cm

let focal length of convex mirror be f

we have to find the distance of second object from convex mirror

we know that u, v and f are related by following formula

$\frac{1}{f} =\frac{1}{v}+ \frac{1}{u}$ .............(1)

put values of u and v in equation (1)

we got,

$\frac{1}{f} =\frac{1}{7.05}+ \frac{1}{-13.5}$

$\frac{1}{f}=\frac{13.5-7.05}{13.5\times 7.05}$

$\frac{1}{f}=\frac{6.45}{13.5\times 7.05}\\f=13.5\times 1.09\\f=14.75$

we have given that

second object is twice as tall as the first object

and image height of both objects are same

it means

$o_{2}=2o_{1}\\i_{1}=i_{2}$ .............(2)

we know that

$\frac{v}{u}=\frac{i}{o}\\i=\frac{o\times v}{u}$

therefore,

$i_{1}=\frac{o_{1}\times v}{u}$ .................(3)

put values of v and u in equation 3

$i_{1}=-\frac{o_{1}\times 7.05}{13.5}$

$i_{1}=-0.52o_{1}$

therefore from equation 2

$i_{2}=-0.52o_{1}$

we know that

$i_{2}=\frac{o_{2}\times V}{U}$ .................(4)

put value of $i_{2}$ and $o_{2}$ in equation 4

$-.52o_{1}=\frac{2o_{1}\times V}{U}$

$U=\frac{2o_{1}\times V}{-.52o_{1}} \\U=-3.85V$

we know that U,V and f are related by following formula

$\frac{1}{f} =\frac{1}{V}+ \frac{1}{U}$ .............(5)

put values of f and U in equation 5

we got

$\frac{1}{14.75} =\frac{1}{V}- \frac{1}{3.85V}$

$\frac{1}{14.75} =\frac{2.85}{3.85V}$

$\frac{1}{14.75} =\frac{2.85}{3.85V}\\V=\frac{2.85\times 14.75}{3.85}\\V=10.91 cm$

Therefore,

U=-10.91\times 3.85

U=-42.03 cm

Second object is located at 42.03 cm in front of mirror

4 0

2 years ago

Charge is placed on two conducting spheres that are very far apart and connected by a long thin wire. The radius of the smaller

kobusy [5.1K]

Answer:

σ₁ = $3.167 * 10^{-6}$ C/m²

σ₂ = $7.6 * 10 ^{-6}$ C/m²

Explanation:

The given data :-

i) The radius of smaller sphere ( r ) = 5 cm.

ii) The radius of larger sphere ( R ) = 12 cm.

iii) The electric field at of larger sphere ( E₁ ) = 358 kV/m. = 358 * 1000 v/m

$E_{1} = (\frac{1}{4\pi\epsilon }) (\frac{Q_{1} }{R^{2} } )$

$358000 = 9 * 10^{9 } *\frac{Q_{1} }{0.12^{2} }$

Q₁ = 572.8 $* 10^{-9}$ C

Since the field inside a conductor is zero, therefore electric potential ( V ) is constant.

V = constant

∴ $\frac{Q_{1} }{R} = \frac{Q_{2} }{r}$

$Q_{2} = \frac{r}{R} *Q_{1}$

$Q_{2} = \frac{5}{12} *572.8*10^{-9}$ = $238.666 *10^{-9}$ C

Surface charge density ( σ₁ ) for large sphere.

Area ( A₁ ) = 4 * π * R² = 4 * 3.14 * 0.12 = 0.180864 m².

σ₁ = $\frac{Q_{1} }{A_{1} }$ = $\frac{572.8 *10^{-9} }{0.180864}$ = $3.167 * 10^{-6}$ C/m².

Surface charge density ( σ₂ ) for smaller sphere.

Area ( A₂ ) = 4 * π * r² = 4 * 3.14 * 0.05² =0.0314 m².

σ₂ = $\frac{Q_{2} }{A_{2} }$ = $\frac{238.66 *10^{-9} }{0.0314}$ = $7.6 * 10 ^{-6}$ C/m²

8 0

2 years ago

In ideal flow, a liquid of density 850 kg/m3 moves from a horizontal tube of radius 1.00 cm into a second horizontal tube of rad

Crank

Answer:

a) Q = π r₁ √ 2ΔP / rho [r₁² / r₂² -1] , b) Q = 3.4 10⁻² m³ / s , c) Q = 4.8 10⁻² m³ / s

Explanation:

We can solve this fluid problem with Bernoulli's equation.

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

With the two tubes they are at the same height y₁ = y₂

P₁-P₂ = ½ ρ (v₂² - v₁²)

The flow rate is given by

A₁ v₁ = A₂ v₂

v₂ = v₁ A₁ / A₂

We replace

ΔP = ½ ρ [(v₁ A₁ / A₂)² - v₁²]

ΔP = ½ ρ v₁² [(A₁ / A₂)² -1]

Let's clear the speed

v₁ = √ 2ΔP /ρ[(A₁ / A₂)² -1]

The expression for the flow is

Q = A v

Q = A₁ v₁

Q = A₁ √ 2ΔP / rho [(A₁ / A₂)² -1]

The areas are

A₁ = π r₁

A₂ = π r₂

We replace

Q = π r₁ √ 2ΔP / rho [r₁² / r₂² -1]

Let's calculate for the different pressures

r₁ = d₁ / 2 = 1.00 / 2

r₁ = 0.500 10⁻² m

r₂ = 0.250 10⁻² m

b) ΔP = 6.00 kPa = 6 10³ Pa

Q = π 0.5 10⁻² √(2 6.00 10³ / (850 (0.5² / 0.25² -1))

Q = 1.57 10⁻² √(12 10³/2550)

Q = 3.4 10⁻² m³ / s

c) ΔP = 12 10³ Pa

Q = 1.57 10⁻² √(2 12 10³ / (850 3)

Q = 4.8 10⁻² m³ / s

5 0

2 years ago

Suzette had prepared the graph below to add to her lab

Charra [1.4K]

Answer:

A title

Explanation:

Because this is middle school.

4 0

2 years ago

Read 2 more answers

A 57 kg paratrooper falls through the air. How much force is pulling him down?

just olya [345]

Answer:

Explanation:

Force = Mass * acceleration due to gravity.

Given

Mass of the paratrooper = 57kg

Acceleration due to gravity = 9.81m/s²

Required

Force pulling him down

Substitute unto formula;

F = 57 * 9.81

Force = 559.17 N

Hence the force pulling him down is 559.17N

7 0

2 years ago