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tensa zangetsu [6.8K]

2 years ago

10

The particle starts from rest at t=0. What is the magnitude p of the momentum of the particle at time t? Assume that t>0. Exp

ress your answer in terms of any or all of m, F, and t.

1 answer:

olganol [36]2 years ago

6 0

Answer:

Ft

Explanation:

We are given that

Initial velocity=u=0

We have to find the magnitude of p of the momentum of the particle at time t.

Let mass of particle=m

Applied force=F

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

Final velocity , $v=a+ut$

Substitute the values

$v=0+\frac{F}{m}t=\frac{F}{m}t$

We know that

Momentum, p=mv

Using the formula

$p=m\times \frac{F}{m}t=Ft$

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A point charge with charge q1 is held stationary at the origin. A second point charge with charge q2 moves from the point (x1, 0

Scilla [17]

Answer:

$W=kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})$

Explanation:

Position of charge q₁ is (0,0)

Position of charge q₂ is (x₁,0)

So, the electric potential energy between the charges is given by :

$U_1=k\dfrac{q_1q_2}{x_1}$

Now the position of charge q₂ has been changes from (x₁,0) to (x₂,y₂). Now, electric potential energy between the charges is :

$U_2=k\dfrac{q_1q_2}{\sqrt{x_2^2+y_2^2}}$

We know form the work energy theorem that, the change in potential energy is equal to the work done. Mathematically, it is given by :

$W=-\Delta U$

$W=-(U_2-U_1)$

$W=(U_1-U_2)$

$W=(k\dfrac{q_1q_2}{x_1}-k\dfrac{q_1q_2}{\sqrt{x_2^2+y_2^2}})$

$W=kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})$

Hence, the work done by the electrostatic force on the moving point charge is $kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})$ . Hence, this is the required solution.

3 0

2 years ago

The mass of the Sun is 2 × 1030 kg, and the distance between Neptune and the Sun is 30 AU. What is the orbital period of Neptune

Veronika [31]

Kepler's third law states that, for a planet orbiting around the Sun, the ratio between the cube of the radius of the orbit and the square of the orbital period is a constant:
$\frac{r^3}{T^2}= \frac{GM}{4 \pi^2}$ (1)
where
r is the radius of the orbit
T is the period
G is the gravitational constant
M is the mass of the Sun

Let's convert the radius of the orbit (the distance between the Sun and Neptune) from AU to meters. We know that 1 AU corresponds to 150 million km, so
$1 AU = 1.5 \cdot 10^{11} m$
so the radius of the orbit is
$r=30 AU = 30 \cdot 1.5 \cdot 10^{11} m=4.5 \cdot 10^{12} m$

And if we re-arrange the equation (1), we can find the orbital period of Neptune:
$T=\sqrt{ \frac{4 \pi^2}{GM} r^3} = \sqrt{ \frac{4 \pi^2}{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2} )(2\cdot 10^{30} kg)}(4.5 \cdot 10^{12} m)^3 }= 5.2 \cdot 10^9 s$

We can convert this value into years, to have a more meaningful number. To do that we must divide by 60 (number of seconds in 1 minute) by 60 (number of minutes in 1 hour) by 24 (number of hours in 1 day) by 365 (number of days in 1 year), and we get
$T=5.2 \cdot 10^9 s /(60 \cdot 60 \cdot 24 \cdot 365)=165 years$

3 0

2 years ago

Moving water, like that of a river, carries sediment as it moves along its bed. The faster the water flows, the more sediment th

katovenus [111]

Correct option: A

An object remains at rest until a force acts on it.

As the water moves faster, it applies greater force on the sediment, which over comes the frictional forces between the bed and the sediment. So, when the river flows faster, more and larger sediment particles are carried away. When the flow slows down, the river couldn't apply enough force on the larger sediments which can overcome the frictional force between the sediment and the river bed. So, the net force on the heavier particles become zero. Hence, the heavier particles of the load will settle out.

3 0

2 years ago

Read 2 more answers

A particle has a velocity of v→(t)=5.0ti^+t2j^−2.0t3k^m/s.

Makovka662 [10]

Answer:

a) $a=5 i+2t j - 6\ t^2k$

b) $a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2$

Explanation:

Given that

v(t) = 5 t i + t² j - 2 t³ k

We know that acceleration a is given as

$a=\dfrac{dv}{dt}$

$\dfrac{dv}{dt}=5 i+2t j - 6\ t^2k$

$a=5 i+2t j - 6\ t^2k$

Therefore the acceleration function a will be

$a=5 i+2t j - 6\ t^2k$

The acceleration at t = 2 s

a= 5 i + 2 x 2 j - 6 x 2² k m/s²

a=5 i + 4 j -24 k m/s²

The magnitude of the acceleration will be

$a=\sqrt{5^2+4^2+24^2}\ m/s^2$

a= 24.83 m/s²

The direction of the acceleration a is given as

$a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2$

a) $a=5 i+2t j - 6\ t^2k$

b) $a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2$

5 0

2 years ago

In the sport of parasailing, a person is attached to a rope being pulled by a boat while hanging from a parachute-like sail. A r

scoray [572]

Answer:

W = 506.75 N

Explanation:

tension = 2300 N

Rider is towed at a constant speed means there no net force acting on the rider.

hence taking all the horizontal force and vertical force in consideration.

net horizontal force:

F cos 30° - T cos 19° = 0

F cos 30° = 2300 × cos 19°

F = 2511.12 N

net vertical force:

F sin 30° - T sin 19°- W = 0

W = F sin 30° - T sin 19°

W = 2511.12 sin 30° - 2300 sin 19°

W = 506.75 N

8 0

2 years ago