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10 months ago

Hi im really confused:

Physics

1 answer:

Vitek1552 [10]10 months ago

3 0

The energy required to raise the temperature of the air in a room by 5.0°C is 336 kJ

U = $c_{p}$ m ΔT

U = Energy

$c_{p}$ = Specific heat

m = Mass

ΔT = Change in temperature

ρ = Density

V = Volume

ρ = 1000 g / m³ (Dry air )

$c_{p}$ = 1 J / g K

ΔT = 5 °C

V = 4 * 4 * 3

V = 48 m³

m = ρ V

m = 1000 * 48

m = 48000 g

U = 1 * 48000 * 7

U = 336000 J

U = 336 kJ

Therefore, the energy required to raise the temperature of the air in a room by 5.0°C is 336 kJ

To know more about energy required to raise temperature

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The diagram below represents a flashlight that has been turned on. Which form of energy is being converted to electrical energy

Orlov [11]

Answer:

Chemical energy to electrical energy

Explanation:

In nature, there are several types of energy.

In this example (a flashlight being turned on), we have a conversion of energy from chemical energy to electrical energy. In fact:

- Chemical energy is the energy stored in the chemical bonds of the molecules of the substances used inside the battery. When the chemical reaction inside the battery occurs, this energy is liberated, and it is used to "push" the electrons along the circuit connected to the battery

- Electric energy is the energy associated to the motion of the electrons along the circuit of the flashlight; it is the energy associated to an electric current.

Moreover, in the flashlight the electric energy is then converted into two more types of energy: light energy (since the bulb in the flashlight produces light) and heat energy (because the flashlight also produces heat, so thermal energy).

3 0

2 years ago

The electron's velocity at that instant is purely horizontal with a magnitude of 2 \times 10^5 ~\text{m/s}2×10 5 m/s then ho

Lesechka [4]

Complete question:

At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10⁵ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.

[Assume that this experiment takes place in deep space so that the effect of gravity is negligible.]

Answer:

The time it will take the particle to pass through point (P) again is 1.639 ns.

Explanation:

F = qvB

Also;

$F = \frac{MV}{t}$

solving this two equations together;

$\frac{MV}{t} = qVB\\\\t = \frac{MV}{qVB} = \frac{M}{qB}$

where;

m is the mass of electron = 9.11 x 10⁻³¹ kg

q is the charge of electron = 1.602 x 10⁻¹⁹ C

B is the strength of the magnetic field = 3.47 x 10⁻³ T

substitute these values and solve for t

$t = \frac{M}{qB} = \frac{9.11 *10^{-31}}{1.602*10^{-19}*3.47*10^{-3}} = 1.639 *10^{-9} \ s \ = 1.639 \ ns$

Therefore, the time it will take the particle to pass through point (P) again is 1.639 ns.

5 0

3 years ago

How high can a 40 N force move a load, when 395 J of work is done?

Juliette [100K]

Answer:

9.875

Explanation:

w=f×s

395=40×s

make s the subject of the formula

s=395/40

=9.875

7 0

3 years ago

A ball thrown by Ginger is moving upward through the air. Diagram A shows a box with a downward arrow. Diagram B shows a box wit

vichka [17]

As the ball is moving in air as well as we have to neglect the friction force on it

So we can say that ball is having only one force on it that is gravitational force

So the force on the ball must have to be represented by gravitational force and that must be vertically downwards

So the correct FBD will contain only one force and that force must be vertically downwards

So here correct answer must be

Diagram A shows a box with a downward arrow.

8 0

2 years ago

wo lacrosse players collide in midair. Jeremy has a mass of 120 kg and is moving at a speed of 3 m/s. Hans has a mass of 140 kg

Julli [10]

2.71 m/s fast Hans is moving after the collision.

Explanation:

Given that,

Mass of Jeremy is 120 kg ( $M_J$ )

Speed of Jeremy is 3 m/s ( $V_J$ )

Speed of Jeremy after collision is ( $V_{JA}$ ) -2.5 m/s

Mass of Hans is 140 kg ( $M_H$ )

Speed of Hans is -2 m/s ( $V_H$ )

Speed of Hans after collision is ( $V_{HA}$ )

Linear momentum is defined as “mass time’s speed of the vehicle”. Linear momentum before the collision of Jeremy and Hans is

= $=\mathrm{M}_{1} \times \mathrm{V}_{\mathrm{J}}+\mathrm{M}_{\mathrm{H}} \times \mathrm{V}_{\mathrm{H}}$

Substitute the given values,

= 120 × 3 + 140 × (-2)

= 360 + (-280)

= 80 kg m/s

Linear momentum after the collision of Jeremy and Hans is

= $=\mathrm{M}_{\mathrm{J}} \times \mathrm{V}_{\mathrm{JA}}+\mathrm{M}_{\mathrm{H}} \times \mathrm{V}_{\mathrm{HA}}$

= 120 × (-2.5) + 140 × $V_{HA}$

= -300 + 140 × $V_{HA}$

We know that conservation of liner momentum,

Linear momentum before the collision = Linear momentum after the collision

80 = -300 + 140 × $V_{HA}$

80 + 300 = 140 × $V_{HA}$

380 = 140 × $V_{HA}$

380/140= $V_{HA}$

$V_{HA}$ = 2.71 m/s

2.71 m/s fast Hans is moving after the collision.

4 0

2 years ago