Calculate the wavelength of a wave if 5 complete waves occupy a length of 20m

How much heat is required to raise 100 grams of water (c= 4.18) by 5 degrees Celsius?

Andrei [34K]

Answer:

Heat capacity, Q = 2090 Joules.

Explanation:

Given the following data;

Mass = 100 grams

Specific heat capacity = 4.18 J/g°C.

Temperature = 5°C

To find the quantity of heat required;

Heat capacity is given by the formula;

$Q = mct$

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

t represents the temperature of an object.

Substituting into the formula, we have;

$Q = 100*4.18*5$

Heat capacity, Q = 2090 Joules.

7 0

3 years ago

In a laboratory, it is often convenient to make measurements in centimeters and grams, but SI units are needed for calculations.

zheka24 [161]

Answer:

(a) 0.92 cm= 0.092 m.

(b) 141.64 g=0.14164 kg.

(c) 15. 8 cm³=0.0000158 m³

(d) 63.6 g/cm³= 63600 kg/m³

Explanation:

The International System of Units, abbreviated S.I., also called the International System of Measurements is a system of measurements in which its units are based on fundamental physical phenomena. The units of the S.I. They are the international reference for the indications of all measuring instruments.

The International System of Units (SI) arose from the need to unify and give coherence to a great variety of unit subsystems.

The International System of Units consists of seven basic units, also called fundamental units, which define the corresponding fundamental physical quantities and which allow any physical quantity to be expressed in terms or as a combination of them. The fundamental physical quantities are complemented by two more physical quantities, called supplementary ones.

By combining the basic units, the other units are obtained, called units derived from the International System, and which allow defining any physical quantity.

(a) The SI unit of length is the meter. Being 1 cm = 0.01 m, then 0.92 cm= 0.092 m.

(b) The SI unit of mass is kg. Being 1 g = 0.001 kg, then 141.64 g=0.14164 kg.

(c) Being 1 cm³ = 0.000001 m³, then 15. 8 cm³=0.0000158 m³

(d) Being 1 g/cm³= 1000 kg/m³, then 63.6 g/cm³= 63600 kg/m³

3 0

3 years ago

Explain the difference between mass and weight (in at least 2 sentences)

Irina-Kira [14]

Mass= is how big something is.
Weight= is how heavy something is.
They are different things because weight is talking about heavy... not how big it is.

5 0

3 years ago

Read 2 more answers

Objects 1 and 2 attract each other with a gravitational force of 179 units. If the distance separating objects 1 and 2 is change

yaroslaw [1]

Explanation:

Fgravity = G*(mass1*mass2)/D²

G is the gravitational constant throughout the universe.

D is the distance between the 2 objects.

the distance is now quadrupled.

Fgravitynew = G*(mass1*mass2)/(4D)² =

= G*(mass1*mass2)/(16D²) =

= (G*(mass1*mass2)/D²) / 16 = Fgravity/16

the new gravitational force will be 179/16 = 11.1875 units

3 0

3 years ago

a ball kicked with a velocity of 8m/s at an angle of 30 degree to horizontal. calculate the time of flight of the ball. (g=10ms^

posledela

Answer:

Approximately $0.8\; \rm s$ (assuming that air resistance is negligible.)

Explanation:

Let $v_0$ denote the initial velocity of this ball. Let $\theta$ denote the angle of elevation of that velocity.

The initial velocity of this ball could be decomposed into two parts:

Initial vertical velocity: $v_0(\text{vertical}) = v_0 \cdot \sin(\theta)$ .
Initial horizontal velocity: $v_0(\text{vertical}) = v_0 \cdot \cos(\theta)$ .

If air resistance on this ball is negligible, $v_0(\text{vertical})$ alone would be sufficient for finding the time of flight of this ball.

Calculate $v_0(\text{vertical})$ given that $v_0 = 8 \; \rm m \cdot s^{-1}$ and $\theta = 30^\circ$ :

$\begin{aligned}& v_0(\text{vertical}) \\ &= v_0 \cdot \sin(\theta) \\ &= \left(8 \; \rm m \cdot s^{-1} \right) \cdot \sin\left(30^{\circ}\right) \\ &= 4\;\rm m \cdot s^{-1} \end{aligned}$ .

Assume that air resistance on this ball is zero. Right before the ball hits the ground, the vertical velocity of this ball would be exactly the opposite of the value when the ball was launched.

Since $v_0(\text{vertical}) = 4\; \rm m \cdot s^{-1}$ , the vertical velocity of this ball right before landing would be $v_1(\text{vertical}) = -4\; \rm m \cdot s^{-1}$ .

Calculate the change to the vertical velocity of this ball:

$\begin{aligned}& \Delta v(\text{vertical}) \\ & = v_1(\text{vertical}) - v_0(\text{vertical}) \\ &= -8\; \rm m \cdot s^{-1}\end{aligned}$ .

In other words, the vertical velocity of this ball should have change by $8\; \rm m \cdot s^{-1}$ during the entire flight (from the launch to the landing.)

The question states that the gravitational field strength on this ball is $g = 10\; \rm m \cdot s^{-2}$ . In other words, the (vertical) downward gravitational pull on this ball could change the vertical velocity of the ball by $10\; \rm m\cdot s^{-1}$ each second. What fraction of a second would it take to change the vertical velocity of this ball by $8\; \rm m \cdot s^{-1}$ ?

$\begin{aligned}t &= \frac{\Delta v(\text{initial})}{g} \\ &= \frac{8\; \rm m \cdot s^{-1}}{10\; \rm m \cdot s^{-2}} = 0.8\; \rm s\end{aligned}$ .

In other words, it would take $0.8\; \rm s$ to change the velocity of this ball from the initial velocity at launch to the final velocity at landing. Therefore, the time of the flight of this ball would be $0.8\; \rm s\!$ .

5 0

3 years ago