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3 years ago

What type of charge does an proton have?

Physics

2 answers:

Serga [27]3 years ago

7 0

Answer:

positive change

Explanation:

Whitepunk [10]3 years ago

4 0

A proton has a positive (1+) charge

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Suppose you have two small pith balls that are 5.5 cm apart and have equal charges of -29 nc?

zysi [14]

The question is missing, however, I guess the problem is asking for the value of the force acting between the two balls.

The Coulomb force between the two balls is:
$F= k_e \frac{ q_1 q_2}{r^2}$
where $k_e=8.99\cdot10^9~N m^2 C^{-2}$ is the Coulomb's constant, $q_1=q_2=29~nC=29\cdot 10^{-9}~C$ is the intensity of the two charges, and $r=5.5~cm=0.055~m$ is the distance between them.

Substituting these numbers into the equation, we get
$F=2.5~10^{-3}~N$

The force is repulsive, because the charges have same sign and so they repel each other.

6 0

3 years ago

In the Hunger Games movie, Katniss Everdeen fires a 0.0200-kg arrow from ground level to pierce an apple up on a stage. The spri

egoroff_w [7]

Answer:

a) $v=99.8584\ m.s^{-1}$

b) $v'=99.366\ m.s^{-1}$

Explanation:

Given:

mass of the arrow, $m=0.02\ kg$

stiffness constant of the bow, $k=330\ N,m^{-1}$

distance of pulling back the arrow on the bow from its mean position, $\Delta x=0.55\ m$

height of the apple targeted, $h=5\ m$

<u>Force on the arrow due to the stiffness of the bow:</u>

$F=k.\Delta x$

$F=330\times 0.55$

$F=181.5\ N$

Now the acceleration of the arrow upwards:

$a=\frac{F}{m}$

$a=\frac{181.5}{0.02}$

$a=9075\ m.s^{-2}$

a) For the course of motion when the arrow leaves the bow after the stretch is relaxed we consider that the arrow left the bow after its string goes to the mean position. During this phase the arrow also faces gravity in the downward direction.

Using the equation of motion:

$v^2=u^2+2(a-g).\Delta x$

where:

$v=$ velocity with which the arrow leaves the bow

$u=$ initial velocity of the arrow after it left

$v^2=0^2+2\times (9075-9.81)\times 0.55$

$v=99.8584\ m.s^{-1}$

b) Now when the arrow travels up then it is under a constant gravitational force acting opposite to the motion.

<u>Using eq. of motion:</u>

$v'^2=v^2-2\times g.h$

where:

$v'=$ final velocity when the arrow hits the target

$v=$ initial velocity after the arrow has been launched

$v'^2=99.8584^2-2\times 9.81\times 5$

$v'=99.366\ m.s^{-1}$

3 0

3 years ago

dalvyx [7]

Answer:

$4.5\ \text{N}$

Explanation:

$F_1$ = Gravitational force between the objects = $18\ \text{N}$

$r_1$ = Initial distance between the two objects

$r_2$ = Final distance between the two objects = $2r_1$

Gravitational force between two objects is given by

$F=\dfrac{Gm_1m_2}{r^2}$

$F\propto \dfrac{1}{r^2}$

$\dfrac{F_2}{F_1}=\dfrac{r_1^2}{r_2^2}\\\Rightarrow \dfrac{F_2}{F_1}=\dfrac{r_1^2}{(2r_1)^2}\\\Rightarrow \dfrac{F_2}{F_1}=\dfrac{1}{4}\\\Rightarrow F_2=\dfrac{F_1}{4}\\\Rightarrow F_2=\dfrac{18}{4}\\\Rightarrow F_2=4.5\ \text{N}$

The new force of attraction between the objects is $4.5\ \text{N}$ .

3 0

2 years ago

Which element is required for the combustion of fossil fuels? nitrogen hydrogen oxygen carbon

Svetradugi [14.3K]

Hydrogen fuel is a zero emission that uses electrochemical cells or combustion in internal engines to propel vehicles and electrical devices. It is also used in the propulsion of spacecraft and can potentially be mass produced and marketed for use in land vehicles of passengers and aircraft.

Hydrogen is located in the first group and the first period of the periodic table is the first element of the periodic table, making it the lightest element in the universe. Hydrogen is neither a metal nor a non-metal but it is still considered a non-metal. It acts like a metal when it is compressed at high densities.

Since hydrogen gas is so light, it rises in the atmosphere and therefore is rarely found in its pure form.

8 0

3 years ago

Read 2 more answers

Three cables AB, AC, and AD are supporting a 70-m tall tower. The magnitude of force in each cable is 2 kN. Find the total force

Mrac [35]

Explanation:

The coordinates of points are:

A (0, 70, 0), B (40, 0, 0), C (40, 0, 40) D (60, 0, 60).

The position vectors for corresponding cables are:

$r_{AD} =(-60-0)i+(0-70)j+(-60-0)k\\r_{AD} =-60i-70j-60k\\r_{AC} =(-40-0)i+(0-70)j+(40-0)k\\r_{AC} =-40i-70j+40k$