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2 years ago

3. How much work is done when you pull a 6 N wagon for 5 meters?

Physics

1 answer:

8090 [49]2 years ago

3 0

Answer:

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question

force = 6 N

distance = 5 m

We have

workdone = 6 × 5 = 30

We have the final answer as

Hope this helps you

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You have two resistors with the same cross sectional area and resistivity. Resistor A has length L1 and resistor B has length L2

oee [108]

Answer:

Explanation:

Given

Resistor A has length $L_1$

and Resistor B has Length $L_2$

and Resistance is given by

$R=\frac{\rho L}{A}$

Considering $\rho$ and A to be constant thus

$R_2>R_1$ because $L_2>L_1$

(a)When they are connected in series

As the current in series is same and power is $i^2R$

therefore $P_2>P_1$ as R is greater for second resistor

(b)if they are connected in Parallel

In Parallel connection Voltage is same

$P=\frac{V^2}{R}$

resistance of 2 is greater than 1 thus Power delivered by 1 is greater than 2

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3 years ago

A car travels 150 meters East of its original starting position in 15s. What is the cars velocity?

Andrew [12]

Answer:

it is 347

Explanation:

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2 years ago

Read 2 more answers

A long, straight, horizontal wire carries a left-to-right current of 40 A. If the wire is placed in a uniform magnetic field of

sladkih [1.3K]

Answer:

The magnitude of the resultant of the magnetic field is $4.11\times10^{-5}\ T$

Explanation:

Given that,

Current = 40 A

Magnetic field $B=3.7\times10^{-5}\ T$

Distance = 22 cm

We need to calculate the magnetic field

Using formula of magnetic field

$B'=\dfrac{\mu_{0}I}{2\pi r}$

Where, r = distance

I = current

Put the value into the formula

$B'=\dfrac{4\pi\times10^{-7}\times20}{2\pi\times0.22}$

$B'=1.8\times10^{-5}\ T$

We need to calculate the magnitude of the resultant of the magnetic field

Using formula of resultant

$B''=\sqrt{B^2+B'^2}$

Put the value into the formula

$B''=\sqrt{(3.7\times10^{-5})^2+(1.8\times10^{-5})^2}$

$B''=4.11\times10^{-5}\ T$

Hence, The magnitude of the resultant of the magnetic field is $4.11\times10^{-5}\ T$

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3 years ago

The speed of a certain electron is 995 km s−1 . If the uncertainty in its momentum is to be reduced to 0.0010 per cent, what unc

Tpy6a [65]

Answer:

The uncertainty in the location that must be tolerated is $1.163 * 10^{-5} m$

Explanation:

From the uncertainty Principle,

Δ $_{y}$ Δ $_{p}$ $= \frac{h}{2\pi }$

The momentum P $_{y}$ = (mass of electron)(speed of electron)

= $(9.109 * 10^{-31}kg)(995 * 10^{3} m/s)$

= $9.0638 * 10^{-25}kgm/s$

If the uncertainty is reduced to a 0.0010%, then momentum

= $9.068 * 10^{-30}kgm/s$

Thus the uncertainty in the position would be:

Δ $_{y} = \frac{h}{2\pi } * \frac{1}{9.068 * 10^{-30} }$

Δ $_{y} \geq 1.163 * 10^{-5}m$

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