All categories

3 years ago

Which of the following is an example of uniform circular motion?

Physics

2 answers:

Alex787 [66]3 years ago

7 0

Mmm i think it’s B I lose the quiz so don’t trust me :(

Allisa [31]3 years ago

3 0

Answer:

C. A rocket traveling to the Moon

You might be interested in

If I release a golf ball at a height of 31 cm and a length of 31 cm the mass is equal to 45.93 g what is theKinetic energy

prohojiy [21]

When a ball is at a particular height it possess potential energy which is given as,

$U=\text{mgh}$

The potential energy is converted into kinetic energy of the ball. According to conservation of energy,

$K=U$

which can be further given as,

$K=\text{mgh}$

Substitute the known values,

$\begin{gathered} K=(45.93\text{ g)(}\frac{1\text{ kg}}{1000\text{ g}})(9.8m/s^2)(31\text{ cm)(}\frac{1\text{ m}}{100\text{ cm}})(\frac{1\text{ J}}{1kgm^2s^{-2}^{}}) \\ =0.140\text{ J} \end{gathered}$

Thus, the kinetic energy of the ball is 0.140 J.

5 0

2 years ago

A charge of -3.40 nC is placed at the origin of an xy-coordinate system, and a charge of 2.45 nC is placed on the y axis at y =

gavmur [86]

Answer:

The magnitude of this force is $1.866\times10^{-4}\ N$

The direction of this force is 55.69°.

Explanation:

Given that,

Charge at origin $q_{1}= -3.40\ nC$

Charge at y axis $q_{2}= 2.45\ nC$

Distance on y axis = 4.25 cm

Third charge $q_{3}= 5.00\ nC$

Distance on x axis = 2.90 cm

(a). We need to calculate the force F₁₃

Using formula of force

$F_{13}=\dfrac{kq_{1}q_{3}}{r^2}$

Put the value into the formula

$F_{13}=\dfrac{9\times10^{9}\times(-3.40\times10^{-9})\times5.00\times10^{-9}}{(2.90\times10^{-2})^2}$

$F_{13}=-0.00018192\ N$

$F_{13}=-1.82\times10^{-4}\ N$

We need to calculate the force F₁₂

Using formula of force

$F_{12}=\dfrac{kq_{1}q_{2}}{r^2}$

Put the value into the formula

$F_{12}=\dfrac{9\times10^{9}\times(-3.40\times10^{-9})\times2.45\times10^{-9}}{(4.25\times10^{-2})^2}$

$F_{12}=-0.00004150\ N$

$F_{12}=-4.15\times10^{-5}\ N$

The magnitude of this force is

The total force exerted on this charge by the other two charges.

$F=\sqrt{F_{13}^2+F_{12}^2+2F_{13}F_{12}\cos \theta}$

Here, $\theta=90$

$F=\sqrt{F_{13}^2+F_{12}^2}$

Put the value into the formula

$F=\sqrt{(-1.82\times10^{-4})^2+(-4.15\times10^{-5})^2}$

$F=0.0001866\ N$

$F=1.866\times10^{-4}\ N$

(c). We need to calculate the direction of this force

Using formula of direction

$\tan\theta=\dfrac{y}{x}$

Put the value into the formula

$\theta=\tan^{-1}(\dfrac{4.25}{2.90})$

$\theta=55.69^{\circ}$

Hence, The magnitude of this force is $1.866\times10^{-4}\ N$

The direction of this force is 55.69°.

3 0

4 years ago

Two balloons are charged with an identical quantity and type of charge: -6.25 x 10-6 C. They are held apart at a separation dist

steposvetlana [31]

Answer:

The electric force between them is 878.9 N

Explanation:

Given:

Identical charge $q = -6.25 \times 10^{-6}$ C

Separation between two charges $r = 0.02$ m

For finding the electrical force,

According to the coulomb's law

$F = \frac{k q^{2} }{r^{2} }$

Here, force between two balloons are repulsive because both charges are same.

Where $k = 9 \times 10^{9}$

$F = \frac{9 \times 10^{9} \times (-6.25 \times 10^{-6} )^{2} }{4 \times 10^{-4} }$

$F = 878.9$ N

Therefore, the electric force between them is 878.9 N

8 0

3 years ago

Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm

mihalych1998 [28]

Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm.?

Point a is at the center of the square, and point b is at the empty corner closest to q2. Take the electric potential to be zero at a distance far from both charges.

(a) What is the electric potential at point a due to q1 and q2?

(b) What is the electric potential at point b?

(c) A point charge q3 = -6.00 μC moves from point a to point b. How much work is done on q3 by the electric forces exerted by q1 and q2?

Answer:

a) the potential is zero at the center .

Explanation:

a) since the two equal-magnitude and oppositely charged particles are equidistant

b)(b) Electric potential at point b, v = Σ kQ/r

r = 5cm = 0.05m

k = 8.99*10^9 N·m²/C²

Q = -2 microcoulomb

v= (8.99*10^9) * (2*10^-6) * (1/√2m - 1) / 0.0500m