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iogann1982 [59]

3 years ago

7

A dog pushes a toy horizontally on a frictionless floor with a net force of 2.0\, \text N2.0N for 1.0\,\text m1.0m. How much kin

etic energy does the toy gain?

2 answers:

TiliK225 [7]3 years ago

6 0

Answer: 2 J

Explanation: Kinetic energy is possessed by a body due to virtue of its motion. A body gains kinetic energy when some work is done on it.

W = Δ K.E.

We know that work is said to be done on a body when a force causes displacement of the body.

W = F.s

It is given that net force on the toy, F = 2.0 N which displaces the toy by, s = 1.0 m.

Hence, work done on the toy, W = 2.0 N × 1.0 m = 2 J which is equal to the kinetic energy gained by the toy.

IRINA_888 [86]3 years ago

4 0

Answer:

2.0 J

Explanation:

Since the surface is frictionless, we can solve the problem by using the work-energy theorem, which states that the work done on the toy is equal to the kinetic energy gained by the toy:

$W=\Delta E_k$

where

W is the work done on the toy

$\Delta E_k$ is the kinetic energy gained by the toy

The work done can be calculated as the product between the force applied on the toy and its displacement, so:

$W=Fd=(2.0 N)(1.0 m)=2.0 J$

Therefore, according to the work-energy theorem, the toy has gained 2.0 J of kinetic energy.

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A flat disk of radius 0.50 m is oriented so that the plane of the disk makes an angle of 30 degrees with a uniform electric fiel

nexus9112 [7]

Answer:

The electric flux is $280\ \rm N.m^2/C$

Explanation:

Given:

Radius of the disc R=0.50 m
Angle made by disk with the horizontal $\theta=30^\circ$
Magnitude of the electric Field $E=713.0\ \rm N/C$

The flux of the Electric Field E due to the are dA in space can be found out by using Gauss Law which is as follows

$\phi=\int E.dA$

where

$\phi$ is the total Electric Flux
E is the Electric Field
dA is the Area through which the electric flux is to be calculated.

Now according to question we have

$=EA\cos\theta \\=713\times 3.14\times 0.5^2 \times \cos60^\circ\\=280\ \rm N.m^2/C$

Hence the electric flux is calculated.

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3 years ago

Suppose scientists discover two new moons.The average surface temperature of one of the moons is –180°C, but the temperature can

s344n2d4d5 [400]

The answer is a in the center of active volcanos

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3 years ago

Read 2 more answers

A Hall probe, consisting of a rectangular slab of current-carrying material, is calibrated by placing it in a known magnetic fie

Citrus2011 [14]

Answer:

(a) 0.345 T

(b) 0.389 T

Solution:

As per the question:

Hall emf, $V_{Hall} = 20\ mV = 0.02\ V$

Magnetic Field, B = 0.10 T

Hall emf, $V'_{Hall} = 69\ mV = 0.069\ V$

Now,

Drift velocity, $v_{d} = \frac{V_{Hall}}{B}$

$v_{d} = \frac{0.02}{0.10} = 0.2\ m/s$

Now, the expression for the electric field is given by:

$E_{Hall} = Bv_{d}sin\theta$ (1)

And

$E_{Hall} = V_{Hall}d$

Thus eqn (1) becomes

$V_{Hall}d = dBv_{d}sin\theta$

where

d = distance

$B = \frac{V_{Hall}}{v_{d}sin\theta}$ (2)

(a) When $\theta = 90^{\circ}$

$B = \frac{0.069}{0.2\times sin90} = 0.345\ T$

(b) When $\theta = 60^{\circ}$

$B = \frac{0.069}{0.2\times sin60} = 0.398\ T$

5 0

3 years ago

One of the checks that you could do for problem 1) would be to check the output resistance of the Wheatstone bridge to make sure

Ray Of Light [21]

Answer:

Explanation:

Let the four resistances of th wheat stone bridge is

P, Q, R and S and the value of each is 350 ohm.

Here, P and Q are in series.

R' = P + Q = 350 + 350 = 700 ohm

Then R and S are in series

R' = R + S = 350 + 350 = 700 ohm

Now R' and R'' are in parallel.

So, the equivalent resistance is

Req = R' x R'' / ( R' + R'')

Req = 700 / 2 = 350 ohm

Thus, the reading of ohmmeter is 350 ohm.

6 0

3 years ago

4.2 mol of monatomic gas A interacts with 3.2 mol of monatomic gas B. Gas A initially has 9500 J of thermal energy, but in the p

Komok [63]

Answer:

14657.32 J

Explanation:

Given Parameters ;

Number of moles mono atomic gas A , n 1 = 4 .2 mol

Number of moles mono atomic gas B , n 2 = 3.2mol

Initial energy of gas A , K A = 9500 J

Thermal energy given by gas A to gas B , Δ K = 600 J

Gas constant R = 8.314 J / molK

Let K B be the initial energy of gas B.

Let T be the equilibrium temperature of the gas after mixing.

Then we can write the energy of gas A after mixing as

(3/2)n1RT = KA - ΔK

⟹ (3/2) x 4.2 x 8.314 x T = 9500 - 600

T = (8900 x 3 )/(2x4.2x8.314) = 382.32 K

Energy of the gas B after mixing can be written as

(3/2)n2RT = KB + ΔK

⟹ (3/2) x 3.2 x 8.314 x 382.32 = KB + 600

⟹ KB = [(3/2) x 3.2 x 8.314 x 382.32] - 600

⟹ KB = 14657.32 J

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4 years ago