A 1.5m wire carries a 4 A current when a potential difference of 63 V is applied. What is the resistance of the wire?

If a lawn mower is pushed with a distance of 30 meters and 12N-m of work is exerted, calculate the force.

Nimfa-mama [501]

Answer:

Explanation:

W = FΔx so filling in:

12 = F(30) so

F = .4N

7 0

2 years ago

An object of mass 2 kg starts from rest and is allowed to slide down a

WARRIOR [948]

Answer: B

Explanation:

Given that an object of mass 2 kg starts from rest and is allowed to slide down a frictionless incline so that its height changes by 20 m.

The parameters given from the question are:

Mass M = 2kg

Height h = 20m

Let g = 9.8m/s^2

At the bottom of the incline plane, the object will experience maximum kinetic energy.

From conservative of energy, maximum K.K.E = maximum P.E

Maximum P.E = mgh

Maximum P.E = 2 × 9.8 × 20 = 392 J

But

K.E = 1/2mv^2

Substitute the values of energy and mass into the formula

392 = 1/2 × 2 × V^2

V^2 = 392

V = sqrt( 392 )

V = 19.8 m/s

V = 20 m/s approximately

7 0

3 years ago

Hello people ~

lozanna [386]

Answer:

It’s called a conservative field.

Explanation:

I think it’s going to be the conservative field because in the question it talks about how it is able to become possible to define potential at a point in an electric field because electric field.

3 0

2 years ago

Read 2 more answers

An electron in a mercury atom changes from energy level b to a higher energy level when the

emmainna [20.7K]

The energy is 3.06 electronvolts, E = 3.06eV

1eV = 1.6 * 10^-19 J

3.06 eV = 3.06* 1.6 * 10^-19 J = 4.896 * 10^-19 J

4 0

2 years ago

A river flows due east at 1.60 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant v

Citrus2011 [14]

Answer:

part (a) $v\ =\ 10.42\ at\ 81.17^o$ towards north east direction.

part (b) s = 46.60 m

Explanation:

Given,

velocity of the river due to east = $v_r\ =\ 1.60\ m/s.$
velocity of the boat due to the north = $v_b\ =\ 10.3\ m/s.$

part (a)

River is flowing due to east and the boat is moving in the north, therefore both the velocities are perpendicular to each other and,

Hence the resultant velocity i,e, the velocity of the boat relative to the shore is in the North east direction. velocities are the vector quantities, Hence the resultant velocity is the vector addition of these two velocities and the angle between both the velocities are $90^o$

Let 'v' be the velocity of the boat relative to the shore.

$\therefore v\ =\ \sqrt{v_r^2\ +\ v_b^2}\\\Rightarrow v\ =\ \sqrt{1.60^2\ +\ 10.3^2}\\\Rightarrow v\ =\ 10.42\ m/s.$

Let $\theta$ be the angle of the velocity of the boat relative to the shore with the horizontal axis.

Direction of the velocity of the boat relative to the shore. $\therefore Tan\theta\ =\ \dfrac{v_b}{v_r}\\\Rightarrow Tan\theta\ =\ \dfrac{10.3}{1.60}\\\Rightarrow \theta\ =\ Tan^{-1}\left (\dfrac{10.3}{1.60}\ \right )\\\Rightarrow \theta\ =\ 81.17^o$

part (b)

Width of the shore = w = 300m

total distance traveled in the north direction by the boat is equal to the product of the velocity of the boat in north direction and total time taken

Let 't' be the total time taken by the boat to cross the width of the river. $\therefore w\ =\ v_bt\\\Rightarrow t\ =\ \dfrac{w}{v_b}\\\Rightarrow t\ =\ \dfrac{300}{10.3}\\\Rightarrow t\ =\ 29.12 s$

Therefore the total distance traveled in the direction of downstream by the boat is equal to the product of the total time taken and the velocity of the river $\therefore s\ =\ u_rt\\\Rightarrow s\ =\ 1.60\times 29.12\\\Rightarrow s\ =\ 46.60\ m$

7 0

3 years ago