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11111nata11111 [884]

3 years ago

15

A coffee machine transfers 12 kj of energy in 15 seconds. It uses 230 v mains supply. Use this information to work out the curre

nt through the coffee machine

1 answer:

Otrada [13]3 years ago

6 0

Answer:

3.5 A

Explanation:

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A photon of wavelength 192 nm strikes an aluminum surface along a line perpendicular to the surface and releases a photoelectron

alex41 [277]

Answer:

$KE=3.529\times10^{−27}\ J$

Explanation:

Given that

Wavelength λ=192 nm

So energy of photon,E

$E=\dfrac{hC}{\lambda }$

Now by putting the values

$h=6.6\times 10^{-34}\ m^2.kg/s$

$C=3\times 10^{8}\ m/s$

$E=\dfrac{6.6\times 10^{-34}\times 3\times 10^{8}}{192\times 10^{-9} }$

$E=1.03\times 10^{-18} J$

We know that

Kinetic energy given as

$KE=\dfrac{P^2}{2m}$

$KE=\dfrac{E^2}{2mC^2}$

$KE=\dfrac{(1.03\times 10^{-18})^2}{2\times 1.67\times 10^{-27}(3\times 10^8)^2}$

$KE=3.529\times10^{−27}\ J$

5 0

3 years ago

An Object, Start from rest w Confront Aiceleration 8m/s2 along a

shutvik [7]

Answer:

A) v = 40 m / s, B) v_average = 20 m / s

Explanation:

For this exercise we will use the kinematics relations

A) the final velocity for t = 5 s and since the body starts from rest its initial velocity is zero

v = vo + a t

v = 0 + 8 5

v = 40 m / s

B) the average velocity can be found with the relation

v_average = vf + vo / 2

v-average = 0+ 40/2

v_average = 20 m / s

6 0

2 years ago

A 3 5m container is filled with 900 kg of granite (density of 2400 3 kg m/ ). The rest of the volume is air, with density equal

Elina [12.6K]

Complete question:

A 5-m³ container is filled with 900 kg of granite (density of 2400 kg/m3). The rest of the volume is air, with density equal to 1.15 kg/m³. Find the mass of air and the overall (average) specific volume.

Answer:

The mass of the air is 5.32 kg

The specific volume is 5.52 x 10⁻³ m³/kg

Explanation:

Given;

total volume of the container, $V_t$ = 5 m³

mass of granite, $m_g$ = 900 kg

density of granite, $\rho _g$ = 2,400 kg/m³

density of air, $\rho_a$ = 1.15 kg/m³

The volume of the granite is calculated as;

$V_g = \frac{m_g}{ \rho_g}\\\\V_g = \frac{900 \ kg}{2,400 \ kg/m^3} \\\\V_g = 0.375 \ m^3$

The volume of air is calculated as;

$V_a = V_t - V_g\\\\V_a = 5 \ m^3 \ - \ 0.375 \ m\\\\V_a = 4.625 \ m^3$

The mass of the air is calculated as;

$m_a = \rho_a \times V_a\\\\m_a = 1.15 \ kg/m^3 \ \times \ 4.625 \ m^3\\\\m_a = 5.32 \ kg$

The specific volume is calculated as;

$V_{specific} = \frac{V_t}{m_g \ + \ m_a} = \frac{5 \ m^3}{900 \ kg \ + \ 5.32\ kg} = 5.52 \times 10^{-3} \ m^3/kg$

4 0

3 years ago

How is the acceleration of a car traveling on an elevated air track related to the angle of elevation and the height of elevatio

vitfil [10]

on a given inclined we know that net force is given by

$F_{net} = - mgsin\theta$

here we know that

$F_{net} = ma$

so here we have

$ma = - mg sin\theta$

$a = - gsin\theta$

so here acceleration depends directly on angle of inclination

now we also know that if height of the inclined is H and its length is L

then we can write

$sin\theta = \frac{H}{L}$

so the acceleration is given as

$a = - g*\frac{H}{L}$

so acceleration also depends directly on height of the inclined plane

4 0

3 years ago

A negative charge of -0.550 m exerts an upward <a href="/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="5868766e

almond37 [142]

Answer:

a. +10.9μC

b. 0.600N and downward

Explanation:

To determine the magnitude of the charge, we use the force rule that exist between two charges which us expressed as

F=(kq₁q₂)/r²

since q₁=-0.55μC and the force it applied on the charge above it is upward,we can conclude that the second charge is +ve, hence we calculate its magnitude as

q₂=Fr²/kq₁

q₂=(0.6N*0.3²)/(9*10⁹*0.55*10⁻⁶)

q₂=0.054/4950

q₂=1.09*10⁻⁵c

q₂=10.9μC.

Hence the second charge is +10.9μC

b. From the rule of charges which state that like charges repel and unlike charges attract, we can conclude that the two above charges will attract since they are unlike charges. Hence the direction of the force will be downward into the second charge and the magnitude of the force will remain the same as 0.600N

8 0

3 years ago