Can someone please help?

Which of the following is equal to impulse?<br> A. Ft<br> B. m/s<br> C. pt<br> D. mv

tankabanditka [31]

Answer: A (Ft)

Explanation: The impulse experienced by the object equals the change in momentum of the object. In equation form, F • t = m • Δ v

5 0

3 years ago

The second-order decomposition of hi has a rate constant of 1.80 x 10−3 m−1 s−1. How much hi remains after 45.6 s if the initial

zlopas [31]

Answer:

2.9 M

Explanation:

The concentration-time equation for a second order reaction is:

1/[A] = kt + 1/[A°]

Where,

A = concentration remaining at time, t

A° = initial concentration

k = rate constant

1/[A] = (1.80 x 10^-3) * (45.6) + 1/3.81

1/[A] = 0.345

= 1/0.345

= 2.9 M.

6 0

2 years ago

You are holding a positive charge and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east. Wh

lara [203]

If I hold a positive charge in my hand and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east then the direction of the force on the charge I am holding is towards the north-east direction.

Reasoning:

It is given that there is a positive charge in my hand. There are two more positive charges with the same magnitude. One is 1 mm far towards the east, and the other one is 1 mm far towards the north. It is required to find the direction of the force acting on the charge in my hand.

Let the magnitude of the charge in my hand is Q, and the magnitude of the other charges is q.

Thus the electric force applied on the charge in my hand due to each other is,

$F=\frac{kQq}{r^2}$

Here k is the Coulomb constant, and r is the distance between the charges.

It is also known that the force on a positive charge due to another positive charge is acted outwards.

Thus, the force on the charge due to the charge on the east is,

$\vec{F_1}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{i}$

And the force on the charge due to the charge on the north is,

$\vec{F_2}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{j}$

As the forces are equal in magnitude and one is perpendicular to the other, thus the net force will be acted at an angle of $45^\circ$ from the north or from the north direction.

Thus the net force is acting in the north-east direction.

Learn more about the direction of the force here,

brainly.com/question/2037071

#SPJ4

3 0

1 year ago

Imagine a place in the cosmos far from all gravitational and frictional influences. Suppose that you visit that place (just supp

xz_007 [3.2K]

The rock will continue to travel in a straight line with a constant velocity for ever... The reason is, once it leaves your hand there is no force acting on the rock, so it will just continue to move in a natural motion which is constant velocity.

6 0

2 years ago

A plastic rod 1.3 m long is rubbed all over with wool, and acquires a charge of -3e-08 coulombs. we choose the center of the rod

Anestetic [448]

(a) The plastic rod has a length of L=1.3m. If we divide by 8, we get the length of each piece:
$L/8=1.3m/8=0.1625 m$

(b) The center of the rod is located at x=0. This means we have 4 pieces of the rod on the negative side of x-axis, and 4 pieces on the positive side. So, starting from x=0 and going towards positive direction, we have: piece 5, piece 6, piece 7 and piece 8. Each piece is 0.1625 m long. Therefore, the center of piece 5 is at 0.1625m/2=0.0812 m. And the center of piece 6 will be shifted by 0.1625m with respect to this:
$c_6 = 0.0812m+0.1625m=0.2437 m$

(c) The total charge is $Q=-3 \cdot 10^{-8}C$ . To get the charge on each piece, we should divide this value by 8, the number of pieces:
$Q/8=-3\cdot 10^{-8}C/8=-3.75\cdot 10^{-9}C$

(d) We have to calculate the electric field at x=0.7 generated by piece 6. The charge on piece 6 is the value calculated at point (c):
$q= -3.75\cdot 10^{-9}C$
If we approximate piece 6 as a single charge, the electric field is given by
$E=k_e \frac{q}{d^2}$
where $k_e=8.99\cdot 10^9Nm^2C^{-2}$ and d is the distance between the charge (center of piece 6, located at 0.2437m) and point a (located at x=0.7m). Therefore we have
$E= 8.99\cdot 10^9 Nm^2C^{-2} \frac{-3.75\cdot 10^9 C}{(0.2437m-0.7m)^2} =-161.9 V/m$
poiting towards the center of piece 6, since the charge is negative.

(e) missing details on this question.

5 0

3 years ago