All categories

3 years ago

1.2 Define the following terms and in each case give the symbol and the unit: 1.2.1 wavelength (4) ·

Physics

1 answer:

Alex777 [14]3 years ago

8 0

Wavelength is represented by lambda( $\lambda$ )

$\\ \bull\tt\longmapsto \lambda=\dfrac{c}{v}$

C is speed of light in air.
v is the frequency.

It has unit as meter(m)

You might be interested in

a 46 kilogram student climbs 11 meter up a rope at a constant speed if the student power output is 230 watts how long in seconds

Bezzdna [24]

230×46=10580÷11=961 second

3 0

3 years ago

NGSS Physics

vovangra [49]

Answer: 3.5 seconds

EXPLANATION:

Using the formula:
v = u + at
And taking the upwards direction as positive, we have the following information:

u = 35 m/s
a = -10m/s^2 (this is acceleration due to gravity)

At the top of its path, the apple will have a velocity of 0 m/s, therefore:

v = 0m/s

Once you substitute everything into the formula, you get:

0 = 35 + (-10)t

Therefore, t = 35/10 = 3.5 seconds

6 0

3 years ago

The iron nail’s mass is 16 grams and its temperature drops 650 C when dropped into the water. How much heat energy did the iron

Mice21 [21]

The heat energy transferred by the iron nail is 4680 J

Explanation:

The thermal energy transferred by a substance to another substance is given by the equation

$Q=mC\Delta T$

where

m is the mass of the substance

C is its specific heat capacity

$\Delta T$ is its change in temperature

For the iron nail in this problem, we have:

m = 16 g

$C=0.450 J/g^{\circ}C$

$\Delta T = -650^{\circ}C$

So, the amount of heat energy given off by the nail is

$Q=(16)(0.450)(-650)=-4680 J$

where the negative sign indicates that the heat is given off.

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

5 0

3 years ago

You pull straight up on the string of a yo-yo with a force 0.35 N, and while your hand is moving up a distance 0.16 m, the yo-yo

jarptica [38.1K]

Answer:

a) 0.138J

b) 3.58m/S

c) (1.52J)(I)

Explanation:

a) to find the increase in the translational kinetic energy you can use the relation

$\Delta E_k=W=W_g-W_p$

where Wp is the work done by the person and Wg is the work done by the gravitational force

By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

$Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J$

the change in the translational kinetic energy is 0.138J

b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

$\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2$

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

$v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}$

the new speed is 3.58m/s

c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

$\frac{E_{fr}}{E_{fr}}=\frac{1/2I\omega_f^2}{1/2I\omega_o^2}=\frac{\omega_f^2}{\omega_o^2}\\\\\omega_f=\frac{v_f}{r}\\\\\omega_o=\frac{v_o}{r}\\\\\frac{E_{fr}}{E_{fr}}=\frac{v_f^2}{v_o^2}=\frac{(3.58m/s)}{(2.9m/s)^2}=1.52J$