All categories

3 years ago

Why is cancer so horrible

Physics

2 answers:

____ [38]3 years ago

7 0

Answer:

Because it kills people. I have had 6 relatives who have died from it in the past 2 years. It's horrible.

Explanation:

If you're wondering what causes cancer, it is rapid and uncontrollable cell regeneration. The majority of cancers result from random mutations arising during DNA replication in the normal stem cells required during development and tissue maintenance.

erica [24]3 years ago

4 0

Explanation:

Cancer kills by invading key organs such as

the intestines, lungs, brain, liver, and kidneys. cancer interfers with body functions that are necessary to live.

You might be interested in

A spaceship left earth to collect samples from mars? Which statement is true about the strength of the earths gravity on moving

NNADVOKAT [17]

Answer:

The first one

Explanation:

6 0

4 years ago

Which statements describe elliptical galaxies? *Select the two correct answers.*

Alecsey [184]

Elliptical galaxies are galaxies that are shaped like an ellipse. An ellipse is oval in shape, which means elliptical galaxies are oval in shape.

The two correct statements that describe elliptical galaxies are:

B. They are the most common galaxy type.

C. They typically contain very little gas and dust.

Elliptical galaxies are galaxies where you will find little or no gas and dust. The galaxies that are numerous in number are the elliptical galaxies.

In the elliptical galaxy, you can't find any spiral component in it.

Therefore, from the above explanation, we can see that option B and C is the correct option.

To learn more, visit the link below:

brainly.com/question/13645005

6 0

3 years ago

Read 2 more answers

A circular loop of radius r carries a current i. at what distance along the axis of the loop is the magnetic field one-half its

lana [24]

At r = 0.766 R the magnetic field intensity will be half of its value at the center of the current carrying loop.

We have a circular loop of radius ' r ' carrying current ' i '.

We have to find at what distance along the axis of the loop is the magnetic field one-half its value at the center of the loop.

<h3>What is the formula to calculate the Magnetic field intensity due to a current carrying circular loop at a point on its axis?</h3>

The formula to calculate the magnetic field intensity due to a current carrying ( i ) circular loop of radius ' R ' at a distance ' x ' on its axis is given by -

$B(x) = \frac{\mu_{o} iR^{2} }{2(x^{2} +R^{2})^{\frac{3}{2} } }$

Now, for magnetic field intensity at the center of the loop can calculated by putting x = 0 in the above equation. On solving, we get -

$B(0) = \frac{\mu_{o} i}{2R}$

Let us assume that the distance at which the magnetic field intensity is one-half its value at the center of the loop be ' r '. Then -

$\frac{\mu_{o} iR^{2} }{2(r^{2} +R^{2})^{\frac{3}{2} } } = \frac{1}{2} \frac{\mu_{o}i }{2R}$

$2R^{3} = (r^{2} +R^{2} )^{\frac{3}{2} }$

$4R^{6} = (r^{2} +R^{2} )^{3}$

$r^{2} =0.587R^{2}$

r = 0.766R

Hence, at r = 0.766 R - the magnetic field intensity will be half of its value at the center of the current carrying loop.

To solve more questions on magnetic field intensity, visit the link below-

brainly.com/question/15553675

#SPJ4

6 0

2 years ago

Radioactive nuclei decay to become unstable. Is this true or false?

Damm [24]

The unstable nuclei undergo radioactive decay. The nucleus decay in form of emitting the radiations or changing into the different chemical element.

Thus, the nucleus decay takes place till the nuclei become stable.

Hence, given statement is false.

5 0

1 year ago

For Part A, Sebastian decided to use the copper cylinder. How would the magnitude of his q and ∆H compare if he were to redo Par

Vitek1552 [10]

The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

The given parameters;

initial temperature of metals, = $T_m$
initial temperature of water, = $T_i$
specific heat capacity of copper, $C_p$ = 0.385 J/g.K
specific heat capacity of aluminum, $C_A$ = 0.9 J/g.K
both metals have equal mass = m

The quantity of heat transferred by each metal is calculated as follows;

Q = mcΔt

For copper metal, the quantity of heat transferred is calculated as;

$Q_p = (m_wc_w + m_pc_p)(T_m - T_i)\\\\Q_p = (T_m - T_i)(m_wc_w ) + (T_m - T_i)(m_pc_p)\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m_p(T_m - T_i)\\\\m_p = m\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m(T_m - T_i)\\\\let \ (T_m - T_i)(m_wc_w ) = Q_i, \ \ \ and \ let \ (T_m- T_i) = \Delta t\\\\Q_p = Q_i + 0.385m\Delta t$

The change in heat energy for copper metal;

$\Delta H = Q_p - Q_i\\\\\Delta H = (Q_i + 0.385m \Delta t) - Q_i\\\\\Delta H = 0.385 m \Delta t$

For aluminum metal, the quantity of heat transferred is calculated as;

$Q_A = (m_wc_w + m_Ac_A)(T_m - T_i)\\\\Q_A = (T_m -T_i)(m_wc_w) + (T_m -T_i) (m_Ac_A)\\\\let \ (T_m -T_i)(m_wc_w) = Q_i, \ and \ let (T_m - T_i) = \Delta t\\\\Q_A = Q_i \ + \ m_Ac_A\Delta t\\\\m_A = m\\\\Q_A = Q_i \ + \ 0.9m\Delta t$

The change in heat energy for aluminum metal ;

$\Delta H = Q_A - Q_i\\\\\Delta H = (Q_i + 0.9m\Delta t) - Q_i\\\\\Delta H = 0.9m\Delta t$

Thus, we can conclude that the magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

Learn more here:brainly.com/question/15345295

6 0

3 years ago