1. The matter a wave travels through is called a

6 An object can never have

MAVERICK [17]

Answer:

B

Explanation:

Speed is the magnitude of the velocity vector, so it can never be negative.

6 0

3 years ago

Why is weathering slow in cold, dry places?

Alexxandr [17]

Rate of weathering depends on temperature and moisture. Cold dry places have less water to weather things

Hope this helped!

Don't forget: Mark Brainliest!

Have a good day :)

6 0

4 years ago

17. A volleyball weighs about 300 grams.

atroni [7]

Answer:

PE = 44.1 J

Explanation:

Ok, to have the specific data, the first thing we must do is convert from grams to kilograms. Since mass must always be in kilograms (kg)

We have:

1 kilograms = 1000 grams.

We convert it using a rule of 3, replacing, simplifying units and solving:

$\boxed{\bold{x=\frac{gr*1\ kg}{1000\ gr}=\frac{300\ gr*1\ kg}{1000\ gr}=\frac{300\ kg}{1000}=\boxed{\bold{0.3\ kg}}}}$

==================================================================

Earth's gravity is known to be 9.8 m/s², so we have:

Data:

m = 0.3 kg
g = 9.8 m/s²
h = 15 m
PE = ?

Use formula of potencial energy:

$\boxed{\bold{PE=m*g*h}}$

Replace and solve:

$\boxed{\bold{PE=0.3\ kg*9.8\frac{m}{s^{2}}*15\ m}}$

$\boxed{\boxed{\bold{PE=44.1\ J}}}$

Since the decimal number, that is, the number after the comma is less than 5, it cannot be rounded, then we have this result.

The potential energy of the volleyball is <u>44.1 Joules.</u>

Greetings.

8 0

3 years ago

Walking across a carpet is an example of charge being transferred by

Sati [7]

static electricity and friction

8 0

4 years ago

A 5-g lead bullet traveling in 20°C air at 300 m/s strikes a flat steel plate and stops.

densk [106]

To solve this problem it is necessary to apply the concepts related to the Kinetic Energy and the Energy Produced by the heat loss. In mathematical terms kinetic energy can be described as:

$KE = \frac{1}{2} mv^2$

Where,

m = Mass

v = Velocity

Replacing we have that the Total Kinetic Energy is

$KE = \frac{1}{2} mv^2$

$KE = \frac{1}{2} (5*10^{-3})(300)^2$

$KE = 225J$

On the other hand the required Energy to heat up t melting point is

$Q_1 = mC_p \Delta T$

$Q_2 = L_f m$

Where,

m = Mass

$C_p =$ Specific Heat

$\Delta T =$ Change at temperature

$L_f =$ Latent heat of fussion

Heat required to heat up to melting point,

$Q = Q_1+Q_2$

$Q = mC_p \Delta T+L_f m$

$Q = 5*0.128*(327-20) + 5*24.7$

$Q = 310J$

The energy required to melt is larger than the kinetic energy. Therefore the heat of fusion of lead would be 327 ° C: The melting point of lead.

4 0

3 years ago