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Julli [10]
3 years ago
9

You have a series circuit powered by a 9V battery. If you double the amount of

Physics
2 answers:
choli [55]3 years ago
8 0
I think it get 2x weaker
ss7ja [257]3 years ago
5 0
It gets 2 times weaker
You might be interested in
A particle with positive charge q = 9.61 10-19 C moves with a velocity v = (5î + 4ĵ − k) m/s through a region where both a unifo
user100 [1]

Answer:

\vec{F}=(1.537*10^{-17}\hat{i}-9.61*10^{-19}\hat{j}-2.883*10^{-18}\hat{k})N

Explanation:

The total force on the particle is given by

\vec{F}=q\vec{v}\ X\ \vec{B}+q\vec{E}

Then, by replacing we have:

q\vec{v}\ X \vec{B}=q[7\hat{k}-9\hat{j}-\hat{k}]\\\\q\vec{E}=q[5\hat{i}-\hat{j}-2\hat{k}]\\\\\vec{F}=(9.61*10^{-19}C)[(7+9)\hat{i}+(-9-1)\hat{j}+(-1-2)\hat{k}]\\\\\vec{F}=(1.537*10^{-17}\hat{i}-9.61*10^{-19}\hat{j}-2.883*10^{-18}\hat{k})N

where the cross product can be made with the determinant method.

Hope this helps!!

3 0
4 years ago
Read 2 more answers
a 4kg metal block absorbs 5000j of energy and increases to a temperature of 22°c. the metal has a specific heat capacity of 250j
e-lub [12.9K]

Answer:

17 °C

Explanation:

From specific Heat capacity.

Q = cm(t₂-t₁)................. Equation 1

Where Q = Heat absorb by the metal block, c = specific heat capacity of the metal block, m = mass of the metal block, t₂ = final temperature, t₁ = Initial temperature.

make t₁ the subject of the equation

t₁ = t₂-(Q/cm)............... Equation 2

Given: t₂ = 22 °C, Q = 5000 J, m = 4 kg, c = 250 J/kg.°c

Substitute into equation 2

t₁ = 22-[5000/(4×250)

t₁ = 22-(5000/1000)

t₁ = 22-5

t₁ = 17 °C

6 0
3 years ago
he atomic radii of Li and O2- ions are 0.068 and 0.140 nm, respectively. (a) Calculate the force of attraction in newtons betwee
Crazy boy [7]

Answer:

a

 F =  -1.07 *10^{-8} \  N

b

F_r  =  1.07 *10^{-8} \  N

Explanation:

Generally the force of attraction between this two irons is mathematically represented as

F =  \frac{k *  [Q_{Li}  ] * [Q_{O}  ]  }{ r^2}

Here k is known as the proportionality constant with value k = 2.31  *  10^ {-28} J \cdot m

substituting -2 for Q_{O} i.e the charge on oxygen , +1 for Q_{Li} i.e the charge on Lithium and [0.140 + 0.068 ] nm= 0.208 nm =  0.208*10^{-9} for r

So

F =  \frac{ 2.31  *  10^ {-28}*  +1   * -2   }{ ( 0.208*10^{-9} )^2   }

F =  -1.07 *10^{-8} \  N

Generally the force of repulsion will be the magnitude but different direction to the force o attraction

So Force of repulsionn is

F_r  =  1.07 *10^{-8} \  N

6 0
4 years ago
Which statement describes the effects of forces on an object
lukranit [14]

Answer:

Sorry this isn’t going to be any help. You don’t have any statement that I’m able to see.

Explanation:

8 0
3 years ago
What are the three important safety factors for surgical technologists to consider when exposed to ionizing radiation?
Gemiola [76]

Answer:

Exposure time limitation, shielding and distance.

Explanation:

  • Limitation of exposure time, since the dose received is directly proportional to the exposure time, so that, at a shorter time, lower dose. For this reason, planning is suggested, to reduce time.
  • Use of shields. This allows a reduction in the dose received by the technician when filtered by the barrier (screen). There are two types of shields or screens, the primary barriers (attenuate the radiation of the primary beam) and the secondary barriers (avoid diffuse radiation).
  • Distance to the radioactive source. The dose received is inversely proportional to the square of the distance to the radioactive source. Therefore, if the distance is doubled, the dose received will decrease by a quarter. Reason for this, it is advisable to use devices or remote controls whenever possible.
7 0
3 years ago
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