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2 years ago

A water heater warms 35 l of water from a temperature of 22.7 c to a temperature of 83.7

1 answer:

8 0

The amount of energy needed to increase the temperature of a substance by $\Delta T$ is given by
$Q=m C_s \Delta T$
where
m is the mass of the substance
$C_s$ is its specific heat capacity
$\Delta T$ is the increase in temperature

The water volume is $V=35 L= 35 dm^3 = 0.035 m^3$ , since its density is $d=1000 kg/m^3$ , the mass of this sample of water is
$m=dV=(1000 kg/m^3)(0.035 m^3)=35 kg$

The water specific heat capacity is $C_s = 4.18 kJ/kg ^{\circ}C$

and the increase in temperature is $\Delta T=83.7 ^{\circ}C-22.7 ^{\circ}C=61^{\circ}C$

Therefore, the amount of energy needed is
$Q=mC_s \Delta T=(35 kg)(4.18 kJ/kg ^{\circ}C)(61^{\circ}C)=8924 kJ = 8.92 \cdot 10^6 J$

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Jules is conducting an experiment involving friction. He is measuring the temperature of various objects and surfaces after quic

Semenov [28]

Answer:

The correct prediction will be:

The temperature of the surface of the ball bearing when rubbed over glass will be the least.

The incorrect prediction:

The tennis ball over the linoleum floor will have no friction, as the temperatures will not change.

6 0

2 years ago

A rod 150 cm long and of diameter 2.0 cm is subjected to an axial pull of 20 kn. if the modulus of elasticity of the material of

Scilla [17]

Given:
rod of circular cross section is subjected to uniaxial tension.
Length, L=1500 mm
radius, r = 10 mm
E=2*10^5 N/mm^2
Force, F=20 kN = 20,000 N
[note: newton (unit) in abbreviation is written in upper case, as in N ]

From given above, area of cross section = π r^2 = 100 π =314 mm^2
(i) Stress,
σ
=force/area
= 20000 N / 314 mm^2
= 6366.2 N/mm^2
= 6370 N/mm^2 (to 3 significant figures)

(ii) Strain
ε
= ratio of extension / original length
= σ / E
= 6366.2 /(2*10^5)
= 0.03183
= 0.0318 (to three significant figures)

(iii) elongation
= ε * L
= 0.03183*1500 mm
= 47.746 mm
= 47.7 mm (to three significant figures)

5 0

2 years ago

What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then

Darya [45]

Answer:

$y = -8.37 cm$

Explanation:

As we know that the equation of SHM is given as

$y = A cos(\omega t)$

here we know that

$\omega = \frac{2\pi}{T}$

here we have

$T = 0.66 s$

now we have

$\omega = \frac{2\pi}{0.66}$

$\omega = 3\pi$

now we have

$y = (8.8 cm) cos(3\pi t)$

now at t = 2.3 s we have

$y = (8.8 cm) cos(3\pi \times 2.3)$

$y = -8.37 cm$

6 0

2 years ago

Read 2 more answers

To prevent collisions and violations at intersections that have traffic signals, use the _____ to ensure the intersection is cle

Ann [662]

The answer should be:
To prevent collisions and violations at intersections that have traffic signals, use the delayed acceleration technique to ensure the intersection is clear before you enter it.
Delayed acceleration technique refers to waiting to go through an intersection until you have a chance to scan for other vehicles.

3 0

2 years ago

Rotational dynamics about a fixed axis: A person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface

FrozenT [24]

Answer:

I = 2 kgm^2

Explanation:

In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:

$\tau=I\alpha$ (1)

I: moment of inertia of the door

α: angular acceleration of the door = 2.00 rad/s^2

τ: torque exerted on the door

You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:

$\tau=Fd$ (2)

F: force = 5.00 N

d: distance to the hinges = 0.800 m

You replace the equation (2) into the equation (1), and you solve for α:

$Fd=I\alpha\\\\I=\frac{Fd}{\alpha}$

Finally, you replace the values of all parameters in the previous equation for I:

$I=\frac{(5.00N)(0.800m)}{2.00rad/s^2}=2kgm^2$

The moment of inertia of the door around the hinges is 2 kgm^2

3 0

2 years ago