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3 years ago

For work to be accomplished we must have

2 answers:

8 0

The answer they are looking for is the last one. However the last two are technically correct but the third one would result in negative work.

Sunny_sXe [5.5K]3 years ago

8 0

Answer:

a force and a movement in the same direction as the force.

Explanation:

As we know that the formula to find the work done is given by

$W = \vec F. \vec d$

here if the force is applied at some angle with displacement then in that case

$W = F d cos\theta$

now we know that if the force and displacement is in opposite direction then the work done is negative while if the force and displacement is in same direction then the work done is positive

So work to be accomplished only when the sign of the work done is positive so it must be in the same direction as that of displacement of the object

so correct answer would be

a force and a movement in the same direction as the force.

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What is the electrical force between q2 and q3? Recall that k = 8. 99 × 109 N•meters squared over Coulombs squared. 1. 0 × 1011

nlexa [21]

Force on the particle is defined as the application of the force field of one particle on another particle. the electrical force between q₁ and q₃ will be –1. 1 × 10¹¹ N.

<h3>What is electric force?</h3>

Force on the particle is defined as the application of the force field of one particle on another particle. It is a type of virtual force.

The electric force in the second case will be the same as in the first case. Therefore the force on the particle will be the same.

$\rm F= K\frac{q_2q_3}{r^2}$

$\rm F= 9\times 10^9 \times \frac{1.6 \times 10^{-13}\times 1.6\times10^{-13}}{(0.5)^2}$

$\rm F= - 1. 1 \times 10^{11 }N$

Hence the electrical force between q₁ and q₃ will be –1. 1 × 10¹¹ N.

To learn more about the electric force refer to the link;

brainly.com/question/1076352

4 0

3 years ago

The cockroach Periplaneta americana can detect a static electric field of magnitude 8.50 kN/C using their long antennae. If the

erica [24]

Answer:

0.235 nC

Explanation:

Given:

$E$ = the magnitude of electric field = $8.50\ kN/C =8.50\times 10^{3}\ N/C$

$F$ = the magnitude of electric force on each antenna = $2.00\ \mu N =2.00\times 10^{-6}\ N$

$q$ = The magnitude of charge on each antenna

Since the electric field is the electric force applied on a charged body of unit charge.

$\therefore E = \dfrac{F}{q}\\\Rightarrow q =\dfrac{F}{E}\\\Rightarrow q =\dfrac{2.00\times 10^{-6}\ N}{8.50\times 10^{3}\ N/C}\\\Rightarrow q =0.235\times 10^{-9}\ C\\\Rightarrow q =0.235\ nC$

Hence, the value of q is 0.235 nC.

4 0

4 years ago

A comet of mass 1.20 × 10¹⁰kg moves in an elliptical orbit around the Sun. Its distance from the Sun ranges between 0.500 AU and

irina [24]

The eccentricity of its orbit is $U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$

Mass is a physical body's total amount of matter. It also serves as a gauge for the body's inertia or resistance to acceleration (change in velocity) in the presence of a net force. The strength of an object's gravitational pull to other bodies is also influenced by its mass.
The kilogram is the SI unit of mass (kg). In science and technology, a body's weight in a given reference frame is the force that causes it to accelerate at a rate equal to the local acceleration of free fall in that frame.
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An object's mass is a crucial indicator of how much stuff it contains. Weight is a measurement of an object's gravitational pull. It is influenced by the object's location in addition to its mass. As a result, weight is a measurement of force.

The length of the semi-major axis is calculated as follows:

where $, $G=6.67 \times 10^{-1} \mathrm{~m}^3 / \mathrm{kgs}$$

$M=1.99 \times 10^{30} \mathrm{~kg}=$ mass of sur

$m=1.20 \times 10^{10} \mathrm{~kg}$ - a mass of the comet

$\begin{aligned}\therefore \quad \text { At aphelion, } r &=50 \times U \\&=50 \times 1.496 \times 10^{11} \mathrm{~m} . \\U=-\frac{6.67 \times 10^{-11} \times \mathrm{m} 1.99 \times 10^{30} \times 1.20 \times 10^{10}}{50 \times 1.496 \times 10^{11}} \\U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$

$U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$

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2 years ago

Which part of the wave has the highest frequency?

trapecia [35]

Answer:

The last part on the right side of the diagram

Explanation:

Im on plato and just got it right :)

6 0

3 years ago

Read 2 more answers

Mass of water= 357g density water= 1.0g/cm3

aliya0001 [1]

$m=357g\\\\\rho=1.0\ \dfrac{g}{cm^3}\\\\\rho=\dfrac{m}{V}\to V=\dfrac{m}{\rho}\\\\\text{substitute}\\\\V=\dfrac{357g}{1.0\frac{g}{cm^3}}=375g\cdot1.0\dfrac{cm^3}{g}=375\ cm^3$

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3 years ago