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2 years ago

What are 2 ways electromagnets are used

1 answer:

lisabon 2012 [21]2 years ago

4 0

Electromagnets are used for various purposes but I fathom in this instance, the questioner is asking about how electromagnetics can be used to attraction or repulsion.
Example, electromagnets are used for attraction in cranes which attach them to containers in order to lift them.
Meanwhile, Maglev trains use electromagnets repulsive properties.

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The semi-major axis of this ellipse is 8.8 cm, and the distance from one of the foci to the

scoundrel [369]

The eccentricity is 0.5

8 0

2 years ago

If a substance has a density of 2.7g/cm3 and a mass of 86.4g, what is its volume?

Gennadij [26K]

Answer:

32cm³

Explanation:

Given parameters:

Density of substance = 2.7g/cm³

Mass of substance = 86.4g

Unknown:

Volume of substance = ?

Solution:

Density is the mass per unit volume of a substance.

Density = $\frac{mass}{volume}$

Since the unknown is volume we solve for it;

mass = density x volume

86.4 = 2.7 x volume

volume = $\frac{86.4}{2.7}$ = 32cm³

7 0

2 years ago

Water is boiled in a pan on a stove at sea level. During 10 min of boiling, it is observed that 200 g of water has evaporated. T

diamong [38]

Answer:

Rate of heat is 45.1 kJ/min

Explanation:

Heat required to evaporate the water is given by

Q = mL

here we know that

$L = 2.25 \times 10^6 J/kg$

now we have

$Q = (0.200)(2.25 \times 10^6 J/kg)$

$Q = 452.1 kJ$

now the power is defined as rate of energy

$P = \frac{Q}{t}$

$P = \frac{452.1 kJ}{10}$

$P = 45.1 kJ/min$

6 0

3 years ago

What is the weight of a person with a massa of 80kg

Archy [21]

784 Newtons or 176.37 lbs

5 0

2 years ago

A 62.0-kg athlete leaps straight up into the air from a trampoline with an initial speed of 9.6 m/s. The goal of this problem is

pochemuha

Answer:

2856.96 J

$\frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f$

6.78822 m/s

Explanation:

$v_i$ = Initial velocity = 9.6 m/s

g = Acceleration due to gravity = 9.81 m/s²

h = Height

The athlete only interacts with the gravitational potential energy. Air resistance is neglected.

At height y = 0

Kinetic energy

$K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 62\times 9.6^2\\\Rightarrow K=2856.96\ J$

At height y = 0 the potential energy is 0 as

$P=mgy\\\Rightarrow P=mg0=0$

At maximum height her velocity becomes 0 so the kinetic energy becomes zero.

As the the potential and kinetic energy are conserved

The general equation

$K_i+P_i=K_f+P_f\\\Rightarrow \frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f$

Half of maximum height

$\\\Rightarrow mgh_i+\frac{1}{2}mv_f^2=mg\frac{h_i}{2}+\frac{1}{2}mv^2\\\Rightarrow gh_i=g\frac{h_i}{2}+\frac{1}{2}v^2\\\Rightarrow g\frac{h_i}{2}=\frac{1}{2}v^2\\\Rightarrow v=\sqrt{gh}$

$h_i=\frac{v_i^2}{2g}$

$v=\sqrt{gh}\\\Rightarrow v=\sqrt{g\times \frac{v_i^2}{2g}}\\\Rightarrow v=\sqrt{\frac{v_i^2}{2}}\\\Rightarrow v=\sqrt{\frac{9.6^2}{2}}\\\Rightarrow v=6.78822\ m/s$

The velocity of the athlete at half the maximum height is 6.78822 m/s

8 0

2 years ago