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2 years ago

PLEASE HELP ME , it’s my exam

Physics

1 answer:

tatyana61 [14]2 years ago

5 0

Answer:

I only know how to answer 22

which is 250 J

Explanation:

KE: 1/2 mv^2

m- 5kg

v- 10 m/s

KE: 1/2 × 5 × 10^2

= 250 J

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The origin of an x axis is placed at the center of a nonconducting solid sphere of radius R that carries a charge +qsphere distr

MA_775_DIABLO [31]

Answer:

$q=49Q/64$

and

$x =16R/15$

Explanation:

See attached figure.

$E_{Q}=$ E due to sphere

$E_{q}=$ E due to particule

$E_{total}=E_{Q}-E_{q}=0$ (1)

according to the law of gauss and superposition Law:

$E_{Q}=E_{1}+E_{2}=E_{2}$ ; electric field due to the small sphere with r1=R/4

$E_{Q}=kq_{2}/(r_{1}^{2})=$

$q_{2}=density*4/3*pi*r_{1}^{3}=Q/(4/3*pi*R^{3})*4/3*pi*r_{1}^{3}=Q*r_{1}^{3}/R^{3}$

then: $E_{Q}=kq_{2}/(r_{1}^{2})=k*Q*r_{1}^{3}/(R^{3}*r_{1}^{2}) = kQ/(4*R^{2})$ (2)

on the other hand, for the particule:

$E_{q}=kq/(r_{p}^{2})$

$r_{p}=2R-R/4=7R/4$ ⇒ $E_{q}=16kq/(49R^{2})$ (3)

We replace (2) y (3) in (1):

$E_{total}=E_{Q}-E_{q}=0=kQ/(4*R^{2}) - 49kq/(16R^{2})$

$q=49Q/64$

--------------------

if R<x<2R AND $E_{total}=E_{Q}-E_{q}=0$

$E_{total}=E_{Q}-E_{q}=0=kQ/(x^{2}) - kq/(2R-x^{2})$

remember that $q=49Q/64$

then:

$Q(2R-x^{2})=49/64*x^{2}$

solving:

$x_{1} =16R/15$

$x_{2} =16R$

but: R<x<2R

so : $x =16R/15$

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"Without forces there can be no movement"- Do you agree with this statement? Why or Why not?

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Answer: Yes.

Explanation: It is clearly stated in Newton’s first law of physics that an object will not change its motion unless a force acts on.

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3 years ago

A balloon filled with helium occupies 20.0 l at 1.50 atm and 25.0◦

bija089 [108]

At stp (standard temperature and pressure), the temperature is T=0 C=273 K and the pressure is p=1.00 atm. So we can use the ideal gas law to find the number of moles of helium:
$pV=nRT$
where p is the pressure (1.00 atm), V the volume (20.0 L), n the number of moles, T the temperature (273 K) and $R=0.082 atm L K^{-1} mol^{-1}$ the gas constant. Using the numbers and re-arranging the formula, we can calculate n:
$n= \frac{pV}{RT}= \frac{(1.00atm)(20.0L)}{(0.082 LatmK^{-1}mol^{-1})(273 K)}=0.89 mol$

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3 years ago

The near point of a farsighted person's uncorrected eyes is 80 cm. what power contact lens should be used to move the near point

makvit [3.9K]

To solve this problem, we will get f and then we will use it to calculate the power.

So, for this farsighted person,
do = 25 cm and di = -80
Therefore:
1/f = (1/25) + (1/-80) = 0.00275 = 0.275 m

Power = 1/f = 1/0.275 = +3.6363 Diopeters.
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2 years ago

A neutron consists of one "up" quark of charge +2e/3 and two "down" quarks each having charge -e/3. If we assume that the down q

Anon25 [30]

Answer:

The magnitude of the electrostatic force is 120.85 N

Explanation:

We can use Coulomb's law to find the electrostatic force between the down quarks.

In scalar form, Coulomb's law states that for charges $q_1$ and $q_2$ separated by a distance d, the magnitude of the electrostatic force F between them is: