Which of the following are not units of frequency: Hertz. m/s. waves/s

On a winter day with a temperature of -10°C, 500g of snow (water ice) is brought inside where the temperature is 18 °C. The snow

Serjik [45]

Answer:

Explanation:

Mass of ice m = 500g = .5 kg

Heat required to raise the temperature of ice by 10 degree

= mass of ice x specific heat of ice x change in temperature

= .5 x 2093 x 10 J

10465 J

Heat required to melt the ice

= mass of ice x latent heat

0.5 x 334 x 10³ J

167000 J

Heat required to raise its temperature to 18 degree

= mass x specific heat of water x rise in temperature

= .5 x 4182 x 18

=37638 J

Total heat

=10465 +167000+ 37638

=215103 J

7 0

3 years ago

What is the speed of a wave that has a frequency of 125 Hz and a wavelength of 1.25 meters?

kirill115 [55]

Using the "v = f. λ" <span>equation...

Your "v" or </span>velocity = 156.25 meters/second

5 0

3 years ago

A satellite’s velocity is 30000m/s. After 60 secs, it’s velocity shows to 15000m/a. What is the satellite’s acceleration?

Orlov [11]

Answer:

Acceleration = 9 × 10^5 m/s^2 ( deceleration )

Explanation:

From the first equation of motion:

V = u + at

15000 = 30000 + 60a

a = ( 15000-30000)/60

a = 9 × 10^5 m/s^2

4 0

2 years ago

What is the upper block's acceleration if the coefficient of kinetic friction between the block and the table is 0.13?

IgorLugansk [536]

Let us assume that pulley is mass less.

Let the tension produced at both ends of the pulley is T.

We are asked to calculate the acceleration of the block.

Let the masses of two bodies are denoted as $m_{1} \ and\ m_{2}\ respectively$

$Let\ m_{1} =1 kg\ and\ m_{2} =2 kg$

As per this diagram, the body having mass 1 kg is moving downward and the body having mass 2 kg is moving on the surface of the table.

Let the acceleration of each block is a .

For body having mass 1 kg:

The net force acting on 1 kg body will be-

$m_{1} g-T=m_{1} a$ [1]

Here tension in the rope will be vertically upward and weight of the body will be in vertical downward direction.

For body having mass 2 kg:

The coefficient of kinetic friction $[\mu]=0.13$

$Hence\ the\ frictional\ force\ F=\mu N$

$F=\mu m_{2} g$

Hence the net force acting on the body having mass 2 kg-

$T-\mu m_{2} g=m_{2} a$ [2]

Here the tension of the rope is towards right i.e along the direction of motion of the 2 kg block and frictional force is towards left.

Combining 1 and 2 we get-

$m_{1} g-T=m_{1}a$ [1]

$T-\mu m_{2}g= m_{2} a$ [2]

---------------------------------------------------

$[m_{1} -\mu m_{2} ]g=[m_{1} +m_{2} ]a$

$a=\frac{m_{1}-\mu m_{2}} {m_{1}+ m_{2}}*g$

$a=\frac{1-[2*0.13]}{1+2} *9.8\ m/s^2$

$a=\frac{0.74}{3} *9.8\ m/s^2$

$a=2.417 m/s$ [ans]

6 0

3 years ago

Read 2 more answers

If a student has a weight of 360 n on earth, what is the student’s weight on the moon?

Monica [59]

Use w=m*g. The value of g is 1.67m/s^2

3 0

2 years ago