In which part of the scientific method do you sum up your results

Martina has a sample of an unknown substance. She measures the substance. It’s mass is 13.5 grams, and it’s volume is 5 cm3 .Whi

Dominik [7]

Cm^3 is same as mL

13.5 g / 5 mL = 2.7 g/mL
look up densities of metals
aluminum has a density of 2.7 g/mL

6 0

3 years ago

If the voltage across a circuit of constant resistance is doubled, the power dissipated by that circuit will

densk [106]

Answer:

The voltage will quadruple

Explanation:

The power dissipated in a circuit is given by

$P=\frac{V^2}{R}$

where

V is the voltage

R is the resistance

In this problem, the voltage across the circuit is doubled:

V' = 2V

So the new power dissipated is

$P'=\frac{V'^2}{R}=\frac{(2V)^2}{R}=4\frac{V^2}{R}=4 P$

so, the power dissipated will quadruple.

6 0

3 years ago

What is the magnitude of the electric force of attraction between two point charges + 2.8 mc and -1.2 mc if the distance

Anit [1.1K]

Answer:

a

Explanation:

it just a

8 0

2 years ago

An aircraft with a mass of 10,000 kg starts from rest at sea level and takes off, then flies to a cruising speed of 620 km/h and

Natasha_Volkova [10]

Answer:

The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

Explanation:

Given that,

Mass of aircraft = 10000 kg

Speed = 620 km/h = 172.22 m/s

Altitude = 10 km = 1000 m

We calculate the change in potential energy

$\Delta P.E=mg(h_{2}-h_{1})$

$\Delta P.E=10000\times9.8\times(10000-0)$

$\Delta P.E=10000\times9.8\times10000$

$\Delta P.E=980000000\ J$

$\Delta P.E=980\ MJ$

For g = 10 m/s²,

The change in potential energy will be 1000 MJ.

We calculate the change in kinetic energy

$\Delta K.E=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)$

$\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2-0^2)$

$\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2)$

$\Delta K.E=148298642\ J$

$\Delta K.E=148.3\ MJ$

For g = 10 m/s²,

The change in kinetic energy will be 150 MJ.

Hence, The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

7 0

3 years ago

A descending elevator of mass 1,000 kg is uniformly decelerated to rest over a distance of 8 m by a cable in which the tension i

Stolb23 [73]

The speed $V_{i}$ of the elevator at the beginning of the 8 m descent is nearly 4 m/s. Hence, option A is the correct answer.

We are given that-

the mass of the elevator (m) = 1000 kg ;

the distance the elevator decelerated to be y = 8m ;

the tension is T = 11000 N;

let us determine the acceleration 'a' by using Newton's second law of motion.

∑Fy = ma

W - T = ma

(1000kg x 9.8 m/s² ) - 11000N = 1000 kg x a

9800 - 11000 = 1000

a = - 1.2 m/s²

Using the equation of kinematics to determine the initial velocity.

$V_{f}$ ² = $V_{i}$ ² + 2ay

$V_{i}$ = √ ( 2 x 1.2m/s² x 8 m )

$V_{i}$ = √19.2 m²/s²

$V_{i}$ = 4.38 m/s ≈ 4 m/s

Hence, the initial velocity of the elevator is 4m/s.

Read more about the Equation of kinematics:

brainly.com/question/12351668

#SPJ4

8 0

1 year ago