I tried to experiment of dropping candy into a liter of soda and caused a big explosion of foamy soda

How much energy must be absorbed by water with a mass of 0.5 kg in order to raise the temperature from 30°C to 65°C? Note: Water

kvasek [131]

Answer:

73325J

Explanation:

Given parameters:

Mass of water = 0.5kg

Initial temperature = 30°C

Final temperature = 65°C

Specific heat capacity = 4190J/kg°C

Unknown:

Amount of energy absorbed = ?

Solution:

The amount of energy absorbed can be derived using the expression below;

H = m c Δt

H is the amount of energy

m is the mass

c is the specific heat

Δt is the change in temperature

H = 0.5 x 4190 x (65 - 30 )

H = 73325J

6 0

2 years ago

A straight wire of length 0.62 m carries a conventional current of 0.7 amperes. What is the magnitude of the magnetic field made

anyanavicka [17]

Answer:

Magnetic field at point having a distance of 2 cm from wire is 6.99 x 10⁻⁶ T

Explanation:

Magnetic field due to finite straight wire at a point perpendicular to the wire is given by the relation :

$B=\frac{\mu_{0}I }{2\pi R }\times\frac{L}{\sqrt{L^{2}+R^{2} } }$ ......(1)

Here I is current in the wire, L is the length of the wire, R is the distance of the point from the wire and μ₀ is vacuum permeability constant.

In this problem,

Current, I = 0.7 A

Length of wire, L = 0.62 m

Distance of point from wire, R = 2 cm = 2 x 10⁻² m = 0.02 m

Vacuum permeability, μ₀ = 4π x 10⁻⁷ H/m

Substitute these values in equation (1).

$B=\frac{4\pi\times10^{-7}\times 0.7 }{2\pi \times0.02 }\times\frac{0.62}{\sqrt{(0.62)^{2}+(0.02) ^{2} } }$

B = 6.99 x 10⁻⁶ T

3 0

2 years ago

A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre

vredina [299]

Answer:

5080.86m

Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

$y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}$

$v_1=v_{01}+a_1t$

We must consider that it's launched from the ground ( $y_{01}=0m$ ) and from rest ( $v_{01}=0m/s$ ), with an upwards acceleration $a_{1}=28m/s^2$ that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

$y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m$

And the velocity achieved in part 1:

$v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s$

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 ( $y_{02}=1317.26m$ ) and its initial velocity is the one achieved in part 1 ( $v_{02}=271.6m/s$ ), now in free fall, which means with a downwards acceleration $a_{2}=-9,8m/s^2$ . For the data we have it's faster to use the formula $v_f^2=v_0^2+2ad$ , where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

$v_{02}^2+2a_2(y_2-y_{02})=v_2^2=0m/s$

Then, to get $y_2$ , we do:

$2a_2(y_2-y_{02})=-v_{02}^2$

$y_2-y_{02}=-\frac{v_{02}^2}{2a_2}$

$y_2=y_{02}-\frac{v_{02}^2}{2a_2}$

And we substitute the values:

$y_2=y_{02}-\frac{v_{02}^2}{2a_2}=(1317.26m)-\frac{(271.6m/s)^2}{2(-9.8m/s^2)}=5080.86m$

3 0

2 years ago

Which theory of the origin of the moon would be most supported if the moon had an irregular shape, like an asteroid? impact theo

Fiesta28 [93]

<span>Its the impact theory.
It suggests that the moon resulted from the collision of two protoplanets, or embryonic worlds. One of those was the just-forming Earth, and the other was a Mars-size object called Theia. The moon then coalesced from the debris, thus giving it its irregular shape.</span>

3 0

2 years ago

Read 2 more answers

A cyclist and his bicycle have a combined mass of 88 kg and a combined

SpyIntel [72]

Answer:D

Explanation:

Mark brainly

5 0

2 years ago