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3 years ago

SOMEONE PLZ HELP , I HAVE 4 HW DUE TMR BC MY TEACHER JUST GIVE ME A VERY SHORT TIME TO DO THE WORK . 50POINTS

Physics

2 answers:

Artist 52 [7]3 years ago

4 0

d. Radio waves

e. Infrared waves

g. Infrared waves

h. X-ray waves

i. Gamma ray waves (UV)

6. Ultraviolet (UV) radiation

Alexxandr [17]3 years ago

3 0

Answer:

4. Violet light has the shortest wavelength of all the colours and the highest frequency. Red has the longest wavelength and the lowest frequency.

a. Radio waves

b. Radio waves

c. Radio waves

d. Radio waves

e. Infrared waves

f. Infrared waves

g. Infrared waves

h. X-ray waves

i. Gamma ray waves (UV)

6. Ultraviolet (UV) radiation

7. Premature aging of the skin and sun damage

(liver spots, keratosis, solar elastosis, etc.)

Hope it helps!

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Help me with the question b.

Morgarella [4.7K]

Answer:

a) The specific heat capacity means the amount of heat needed by a unit mass of a material to increase its temperature in one unit.

b) Liquid P - $Q = 3840\,J$ , Liquid Q - $Q = 5500\,J$ , Liquid R - $Q = 7800\,J$ , Liquid S - $Q = 2856\,J$

Explanation:

a) The specific heat capacity means the amount of heat needed by a unit mass of a material to increase its temperature in one unit.

b) Let suppose that heat transfer rates between liquids and surroundings are stable. The quantity of the heat released is determined by the following expression:

$Q = m\cdot c\cdot (T_{r} - T_{f})$ (1)

Where:

$m$ - Mass of the liquid, in kilograms.

$c$ - Specific heat capacity, in joules per kilogram-degree Celsius.

$T_{r}$ - Initial temperature of the sample, in degrees Celsius.

$T_{f}$ - Freezing point, in degrees Celsius.

Liquid P ( $m = 1\,kg$ , $c = 160\,\frac{J}{kg\cdot ^{\circ}C}$ , $T_{r} = 30\,^{\circ}C$ , $T_{f} = 6\,^{\circ}C$ )

$Q = (1\,kg)\cdot \left(160\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (30\,^{\circ}C - 6\,^{\circ}C)$

$Q = 3840\,J$

Liquid Q ( $m = 1\,kg$ , $c = 220\,\frac{J}{kg\cdot ^{\circ}C}$ , $T_{r} = 30\,^{\circ}C$ , $T_{f} = 5\,^{\circ}C$ )

$Q = (1\,kg)\cdot \left(220\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (30\,^{\circ}C - 5\,^{\circ}C)$

$Q = 5500\,J$

Liquid R ( $m = 1\,kg$ , $c = 300\,\frac{J}{kg\cdot ^{\circ}C}$ , $T_{r} = 30\,^{\circ}C$ , $T_{f} = 4\,^{\circ}C$ )

$Q = (1\,kg)\cdot \left(300\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (30\,^{\circ}C - 4\,^{\circ}C)$

$Q = 7800\,J$

Liquid S ( $m = 1\,kg$ , $c = 102\,\frac{J}{kg\cdot ^{\circ}C}$ , $T_{r} = 30\,^{\circ}C$ , $T_{f} = 2\,^{\circ}C$ )

$Q = (1\,kg)\cdot \left(102\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (30\,^{\circ}C - 2\,^{\circ}C)$

$Q = 2856\,J$

6 0

3 years ago

The amount of which element in the universe matches the predictions of scientists and helps support the big bang theory?

Elena-2011 [213]

The amount of URANIUM in the universe matches the predictions of scientists and helps support the big bang theory.

3 0

3 years ago

Read 2 more answers

A 1425 kg car located at < 266, 0, 0 > m has a momentum of < 46000, 0, 0 > kg·m/s. What is its location 11 s later?

True [87]

Answer:

<621, 0 , 0>

Explanation:

given,

mass of the car ,m = 1425 Kg

location of car = < 266, 0, 0 > m

momentum of the car = < 46000, 0, 0 > kg.m/s

M v_x = 46000

1425 x v_x = 46000

v_x = 32.28 m/s

v_y = 0 m/s

v_z = 0 m/s

Location of the car after 11 s

x = v_x t + 266

x = 32.28 x 11 + 266

x = 621 m

hence, the location of car after 11 s is equal to <621, 0 , 0>

4 0

3 years ago

An experimentalist fires a beam of electrons, creating a visible path in the air that can be measured. The beam is fired along a

gtnhenbr [62]

Answer:

Velocity of the electron in the beam.

Radius of the circulating electrons due to the magnetic field.

Explanation:

We have a Mathematical expression for the force on a moving charge in a magnetic field as:

$F=q.v.B.sin \theta$ ...........................(1)

where:

q= charge on the particle in coulomb

v= velocity of the charge projected into the magnetic field

B= intensity of the magnetic field in tesla

$\theta$ = angle between the velocity and direction of magnetic field

For the forces on rotating mass we have the formula:

$F=m.\frac{v^2}{r}$ ..........................................(2)

where:

m= mass of the charged particle

v= velocity of projection of charge into the magnetic field

r= radius of the path traced by the charge in the magnetic field

From eq. (1) and (2) we can calculate the magnetic field .

Now,

Using Ampere's Law we have:

$B = \frac{\mu_0 .I}{2 \pi r}$

where:

I= current in the wire

$\mu_0=$ The permeability of free space.

r= radial distance from the current carrying wire( in this case it is same as the radius of the circular path)

4 0

4 years ago

A car with a mass of 710 kg is traveling at 37 km/hr. It accelerates to a speed of 120 km/hr in 12.6 seconds. What is the net fo

guajiro [1.7K]

Answer: The net force acting on the car 1,299.3 N.

Explanation:

Mass of the car = 710 kg

Initial velocity of the car of the ,u= 37 km/h= 10.27 m/s $(1km\h=\frac{5}{18} m/s)$

Final velocity of the car,v = 120 km/h = 33.33 m/s

time taken b y car = 12.6 sec

v-u=at

$33.33m/s-10.27m/s=23.06 m/s=a(12.6 sec)$

$a = 1.83 m/s^2$

$Force=mass\times acceleration$

$Force=710 kg\times 1.83 m/s^2$

$Force=1,299.3 N$

The net force acting on the car 1,299.3 N.

8 0

3 years ago