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3 years ago

15

How long does it take the students’ kettle to come to the boil if 3 kg of ice at 0 °C is put in it? The latent heat of fusion of

ice is 334 000 J/kg.

1 answer:

Alborosie3 years ago

3 0

Answer:

Explanation:

111.333

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A spring of spring constant k is attached to a support at the bottom of a ramp that makes an angle θ with the horizontal. A bloc

Nikitich [7]

Answer:

$x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}$

Explanation:

From the law of conservation of energy

Energy lost by the spring, W=Kinetic energy gained, KE+Potential energy gained, PE+Work done by friction, Fr

$0.5kd^{2}=0.5mv^{2}+mgLsin\theta+\mu_{k}mgcos\theta x$

$x(mgsin\theta+\mu_{k}mgcos\theta)=0.5kd^{2}$

$x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}$

The required distance from A to B is $x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}$

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3 years ago

A force does work on an object if a component of the force is called

ivanzaharov [21]

Explanation:

The work done by an object is given by :

$W=Fd\cos \theta$

Here,

F is force

d is displacement

$\theta$ is the angel between F and d.

If the angle between force and displacement is 0, then the work done is equal to, $W=Fd$ .

So, a force does work on an object if a component of the force is parallel to the displacement of the object.

4 0

3 years ago

A child observes a caterpillar walking on a window sill. The caterpillar walks 18 cm to the left, then 6 cm to

horrorfan [7]

Answer: 34 cm

Explanation:

18+6+10=34 cm

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4 years ago

A negative charge of -6.0x10-6 C exerts an

Ray Of Light [21]

The magnitude of the second charge given that the first is –6×10¯⁶ C and is located 0.05 m away is +3.0×10¯⁶ C

<h3>Coulomb's law equation </h3>

F = Kq₁q₂ / r²

Where

F is the force of attraction
K is the electrical constant
q₁ and q₂ are two point charges
r is the distance apart

<h3>How to determine the second charge </h3>

Charge 1 (q₁) = –6×10¯⁶ C
Electric constant (K) = 9×10⁹ Nm²/C²
Distance apart (r) = 0.05 m
Force (F) = 65 N

Charge 2 (q₂) =?

F = Kq₁q₂ / r²

Cross multiply

Fr² = Kq₁q₂

Divide both side by Kq₁

q₂ = Fr² / Kq₁

q₂ = (65 × 0.05²) / (9×10⁹ × 6×10¯⁶)

q₂ = +3.0×10¯⁶ C (since the force is attractive)

Learn more about Coulomb's law:

brainly.com/question/506926

7 0

2 years ago

A man pulls a wagon with a handle that is at an angle of 42° with the ground. If the man pulls with 330 N of force, how much for

Fed [463]

Answer:

245N

Explanation:

We just need to multiply the force the man pulls the wagon by the cosine of the angle with the ground.

F = 330N * cos(42°) = 245.2N

8 0

4 years ago

Read 2 more answers