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3 years ago

Work done by friction?

Physics

2 answers:

Tomtit [17]3 years ago

8 0

increase kinetic energy of body

Alborosie3 years ago

8 0

Answer:

yes it increases kinetic energy of body on which it is applying it also cause heat loss

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What’s the opposite of an electron

Anna007 [38]

As its charge, proton -a positive charged molecule at the center of an atom- is the opposite of the electron -the particle which is orbiting the center of an atom.

7 0

4 years ago

A block is pulled along a horizontal surface at a constant speed by a force (14.1i 0 j 5.1k). The direction k is perpendicular t

Pepsi [2]

Answer:

$W =84.6\ Nm$

Explanation:

given,

F = 14.1 i + 0 j + 5.1 k

displacement = 6 m

Assuming block is moving in x- direction

we know,

dW = F dx

$\int dW = F\int dx$

$W = F\int_0^6 dx$

$W = F[x]_0^6$

$W = 14.1 \times 6$

$W =84.6\ Nm$

hence, work done by the force is equal to $W =84.6\ Nm$

7 0

3 years ago

What is rotation in your own words. and you have to be detailed

attashe74 [19]

Answer:

The action of rotating around something like for ex.. Child A stands in the middle of a room while Child B goes around Child A, Child B is rotating around Child A. Another ex. is; The earth rotates around the sun every 365 days the earth is rotating or going around the sun in a circle. So rotating to me is the act of rotating around something in any way shape or form.

PS: A middle schooler answered this so if you don't wanna believe me or think I'm wrong because I am younger you do you.

7 0

3 years ago

A spider begins to spin a web by first hanging from a ceiling by his fine, silk fiber. He has a mass of 0.025 kg and a charge of

Rasek [7]

Answer:

a) (5.59 × 10³) N/C

b) 0.226 N directed away from the spider.

Explanation:

a) Electric field, E, felt as a result of point charge, Q, at a distance, d away is given by

E = kQ/d²

So, magnitude of the electric field due to the charge on the second spider at the position of the first spider

Q = 4.2 µC = 4.2 × 10⁻⁶ C

k = Coulomb's constant = 8.99 × 10⁹ Nm²/C

d = 2.6 m

E = (8.99 × 10⁹ × 4.2 × 10⁻⁶)/2.6²

E = 5.59 × 10³ N/C

b) Tension in the silk fiber above the spider is the net force due to the weight of spider one and the force of repulsion due the two charges.

Force due to the two charges = Eq

where q now represents the charge of the first spider at the first point, feeling the electric field calculated in (a)

F = 5.59 × 10³ × 3.4 × 10⁻⁶ = 0.01901 N directed upwards. (That is, F = + 0.019 N)

Weight of the spider = mg = 0.025 × 9.8 = 0.245 N directed downwards. (That is, W = -0.245 N)

Net force, T = mg - F = 0.245 - 0.019 = 0.226 N (that is, 0.226 N, directed upwards, away from the spider).

8 0

3 years ago

Calculate the force which will produce an extension of 0.30mm in a steel wire with a length of 4.0m and a cross section area of

Anna [14]

Given data:

* The extension of the steel wire is 0.3 mm.

* The length of the wire is 4 m.

* The area of cross section of wire is,

$A=2\times10^{-6}m^2$

* The young modulus of the steel is,

$Y=2.1\times10^{11}\text{ Pa}$

Solution:

The young modulus of the steel in terms of the force and extension is,

$Y=\frac{F\times l}{A\times dl}$

where F is the force acting on the steel wire,, l is the original length of the wire, dl is the extension of the wire, and A is the area,

Substituting the known values,

$\begin{gathered} 2.1\times10^{11}=\frac{F\times4}{2\times10^{-6}\times0.3\times10^{-3}} \\ F=0.315\times10^2\text{ N} \\ F=31.5\text{ N} \end{gathered}$

Thus, the force which produce the extension of 0.3 mm of the steel wire is 31.5 N.

7 0

1 year ago