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elena55 [62]
4 years ago
13

Before there was central heating, hot water bottles were used to keep people warm at night. These flat containers were filled wi

th hot water and placed at the foot of the bed, under the blankets. How was the heat energy transferred from the hot water bottle into the space under the blankets?
Physics
1 answer:
mestny [16]4 years ago
8 0
The answer is D radiation because the heat from the hot water bottle is radiating the space around the hot water bottle. Hope this helps
You might be interested in
A string that is restricted at both ends has a length of 1.50 m. what is the wavelength of the string’s fundamental frequency?
lara31 [8.8K]
Since it is restricted at both ends, λ/2 = length of string

λ/2 = 1.5m
λ = 1.5*2 = 3m
6 0
3 years ago
Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant
Reika [66]

Answer:

v = 21 m / s

Explanation:

We can solve this exercise with the kinematics equations, let's start by finding the acceleration of the train with the initial data

            v = v₀ + a t

the initial speed is the speed within the city 6 m / s, the final speed is v = 11 m / s and the time is t = 8 s

             a = (v-v₀) / t

             a = (11 - 6) / 8

             a = 0.625 m / s²

when it leaves the city with speed vo = 11 m / s it accelerates for t = 16 s

            v = v₀ + a t

            v = 11 + 0.625 16

             v = 21 m / s

4 0
4 years ago
alayn h uderstands the half life of a certain substance is 90days. how long would it take for 80lbs of her substance to be reduc
yan [13]

Answer:

135 days...............

7 0
3 years ago
A wire carries 3.7 A of current. A second wire is placed parallel to the first 8.0 cm away. What is the current flowing through
IgorC [24]

Answer:

The current in second wire is 5.0 A.

(B) is correct option.

Explanation:

Given that,

Current in first wire = 3.7 A

Distance = 8.0 cm

We need to calculate the magnetic field due to the current carrying wire

Using formula of magnetic field

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

Where, I = current

r = distance

Put the value into the formula

For first wire

B = \dfrac{\mu_{0}\times3.7}{2\pi \times3.4\times10^{-2}}...(I)

For second wire,

The distance is 8-3.7 =  4.3 cm

B' = \dfrac{\mu_{0}\times I'}{2\pi \times(8-3.4)\times10^{-2}}...(II)

The magnetic field in both the wires,

From equation (I) and (II)

\dfrac{\mu_{0}\times3.7}{2\pi \times3.4\times10^{-2}}= \dfrac{\mu_{0}\times I'}{2\pi \times(8-3.4)\times10^{-2}}

I'=\dfrac{3.7\times4.3\times10^{-2}}{3.4\times10^{-2}}

I'=4.68\ A\ approx = 5.0\ A

Hence, The current in second wire is 5.0 A.

8 0
4 years ago
A 25kg mass is suspended at the end of a horizontal, massless rope that extends from a wall on the left and from the end of a se
patriot [66]
<h2>Answer:20.97g N,32.63g N</h2>

Explanation:

We consider the forces at the knot.

The vertical forces are

T_{2}Sin(50^{0}}) is the vertical component of tension T_{2} at the knot.

-25g is the weight of the mass 25Kg acting downwards.

The horizontal forces are

-T_{1} is the tension in the rope acting left.

T_{2}Cos(50^{0}) is the horizontal component of tension T_{2} acting towards right.

Since the knot has no mass,it is always in equilibrium.

So,the sum of forces acting on it will be zero.

Balancing vertical forces gives,

T_{2}Sin(50^{0}})-25g=0

T_{2}=32.63gN

Balancing horizontal forces gives,

T_{2}Cos(50^{0})-T_{1}=0

T_{1}=32.63g\times Cos(50^{0})=20.97gN

7 0
3 years ago
Read 2 more answers
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