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3 years ago

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A 10-N force is applied horizontally on a box to move it 10 m across a frictionless surface. How much work was done to move the

box?

1 answer:

german3 years ago

4 0

You just multiply these two numbers, it's 100J

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An aluminum wire is held between two clamps under zero tension at room temperature. Reducing the temperature, which results in a

Blababa [14]

Answer:

$\frac{\Delta L}{L} =5.37\times 10^{-4}$

Explanation:

Given:

cross sectional area of the wire, $A=5.75\times 10^{-6}\ m^2$

density of aluminium wire, $\rho=2.7\times 10^3\ kg.m^{-3}$

young's modulus of the material, $E=7\times 10^{10}\ N.m^{-2}$

wave speed, $v=118\ m.s^{-1}$

<u>We have mathematical expression for strain as:</u>

$\frac{\Delta L}{L} =\frac{\sigma}{E}$ ...............................(1)

and since, $\sigma =\frac{T}{A}$

where, T = tension force in the wire

equation (1) becomes:

$\frac{\Delta L}{L} =\frac{T}{A.E}$ ............................(2)

<u>Also velocity ofwave in tensed wire:</u>

$v=\sqrt{\frac{T}{\mu} }$ ...................................(3)

where: $\mu=$ linear mass density of the wire

$\therefore \mu=\rho \times A$

Now, equation (3) becomes

$v=\sqrt{\frac{T}{\rho \times A} }$

$T=v^2.\rho \times A$ ............................(4)

Using eq. (2) & (4) for tension T

$v^2.\rho \times A=A.E\times \frac{\Delta L}{L}$

$\frac{\Delta L}{L} =\frac{v^2.\rho}{E}$

putting the respective values

$\frac{\Delta L}{L} =\frac{118^2\times 2.7\times 10^3}{7\times 10^{10}}$

$\frac{\Delta L}{L} =5.37\times 10^{-4}$

6 0

3 years ago

A man has a mass of 70kg. Calculate his weight on earth where the gravitational strength is 10 N/kg

maw [93]

Answer:

So the weight of the body is 700N

Explanation:

mass of the object=70kg

force of gravity=10N/kg

weight of the object=w-?

as we know that

weight=mass × force of gravity

weight=70kg × 10N/kg

WEIGHT OF THE BODY=700N

I HOPE IT WILL HELP YOU

GOOD LUCK FOR THE ASSIGNMENT

4 0

4 years ago

A 0.80-μm-diameter oil droplet is observed between two parallel electrodes spaced 11 mm apart. The droplet hangs motionless if t

Arisa [49]

A) $2.4\cdot 10^{-16}kg$

The radius of the oil droplet is half of its diameter:

$r=\frac{d}{2}=\frac{0.80 \mu m}{2}=0.40 \mu m = 0.4\cdot 10^{-6}m$

Assuming the droplet is spherical, its volume is given by

$V=\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.4\cdot 10^{-6} m)^3=2.68\cdot 10^{-19} m^3$

The density of the droplet is

$\rho=885 kg/m^3$

Therefore, the mass of the droplet is equal to the product between volume and density:

$m=\rho V=(885 kg/m^3)(2.68\cdot 10^{-19} m^3)=2.4\cdot 10^{-16}kg$

B) $1.5\cdot 10^{-18}C$

The potential difference across the electrodes is

$V=17.8 V$

and the distance between the plates is

$d=11 mm=0.011 m$

So the electric field between the electrodes is

$E=\frac{V}{d}=\frac{17.8 V}{0.011 m}=1618.2 V/m$

The droplet hangs motionless between the electrodes if the electric force on it is equal to the weight of the droplet:

$qE=mg$

So, from this equation, we can find the charge of the droplet:

$q=\frac{mg}{E}=\frac{(2.4\cdot 10^{-16}kg)(9.81 m/s^2)}{1618.2 V/m}=1.5\cdot 10^{-18}C$

C) Surplus of 9 electrons

The droplet is hanging near the upper electrode, which is positive: since unlike charges attract each other, the droplet must be negatively charged. So the real charge on the droplet is

$q=-1.5\cdot 10^{-18}C$

we can think this charge has made of N excess electrons, so the net charge is given by

$q=Ne$

where

$e=-1.6\cdot 10^{-19}C$ is the charge of each electron

Re-arranging the equation for N, we find:

$N=\frac{q}{e}=\frac{-1.5\cdot 10^{-18}C}{-1.6\cdot 10^{-19}C}=9.4 \sim 9$

so, a surplus of 9 electrons.

3 0

4 years ago

A circut contains a 2 microfarads and a 20 microfarads capacitor connected in parallel. What is the total capacitance of the cir

garik1379 [7]

The answer to this question would be:C. 22mf

In this question, there are two capacitors that connected in parallel. The formula to count the capacitor is the opposite of the formula used for the resistor. If constructed in series, the capacitor will have reduced capacitance. For parallel condition the equation would be:

Ctotal= C1+C2
C total= 2 mf+ 20mf= 22mf

5 0

4 years ago

Three uniform spheres are located at the corners of an equilateral triangle. each side of the triangle has a length of 1.20 m. t

Andrei [34K]

Refer to the diagram shown below.

Because of symmetry, equal forces, F, exist between the sphere of mass m and each of the other two spheres.
The acceleration of the sphere with mass m will be vertical as shown.

The gravitational constant is G = 6.67408 x 10⁻¹¹ m³/(kg-s²)

Calculate F.
F = [ (6.67408 x 10⁻¹¹ m³/(kg-s²))*(m kg)*(2.8 kg)]/(1.2 m)²
= 1.2977 x 10⁻¹⁰ m N

The resultant force acting on mass m is
2Fcos(30°) = 2*(1.2977 x 10⁻¹⁰m N)*cos(30°) = 2.2477 x 10⁻¹⁰m N

If the initial acceleration of mass m is a m/s², then
(m kg)(a m/s²) = (2.2477 x 10⁻¹⁰m N)
a = 2.2477 x 10⁻¹⁰ m/s²

Answer:
The magnitude of the acceleration on mass m is 2.25 x 10⁻¹⁰ m/s².
The direction of the acceleration is on a line that joins mass m to the midpoint of the line joining the known masses.

7 0

3 years ago