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Masteriza
7 days ago
14

Two identical carts travel at the same speed toward each other, and then a collision occurs. The graphs show the momentum of eac

h cart as a function of time, both before and after the collision.
Which graph shows a perfectly inelastic collision?

Physics
1 answer:
inna [987]7 days ago
5 0

Explanation:

The term 'collision' refers to the interaction between two objects. There are two distinct types of collisions: elastic and inelastic.

In this scenario, two identical carts are heading towards each other at the same speed, resulting in a collision. In an inelastic collision, the momentum is conserved before and after the incident, but kinetic energy is lost.

After the event, both objects combine and move together at a single velocity.

The graph representing a perfectly inelastic collision is attached, illustrating that both carts move together at the same speed afterward.

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A solid conducting sphere carrying charge q has radius a. It is inside a concentric hollow conducting sphere with inner radius b
Softa [913]

Response:

Clarification:

Refer to the diagram indicating the charges on the specified sphere (see attachment).

The electric field at the stated positions is

E(r) = 0 for r≤a.  Equation 1

E(r) = kq/r² for a<r<b.   Equation 2

E(r) = 0 for b<r<c.      Equation 3

E(r) = kq/r² for r>c.    Equation 4.

We understand that electric potential correlates with the electric field through

V = Ed

A. To compute the potential at the outer surface of the hollow sphere (r=c), we determine that the electric field there is

E = kQ / r²

Then,

V = Ed,

At d = r = c

Thus,

Vc = (kQ / c²) × c

Vc = kQ / c

As a result, the total charge Q consists of +q, -q, and +q

Hence, Q = q - q + q = q

V = kq / c

B. To calculate the potential at the inner surface of the hollow sphere (r=b), we have

V = kQ/r

V = kQ / b,   noting that r = b

So, Q = q

V = kq / b

C. At r = a

Following from equation 1:

E(r) = 0 for r≤a.  Equation 1

The electric field at the surface of the solid sphere is 0, E = 0N/C

Thus,

V = Ed = 0 V

Consequently, the electric potential at the solid sphere's surface is 0.

D. At r = 0

The electric potential can be determined by

V = kq / r

As r approaches 0,

V = kq / 0

V approaches infinity.

8 0
10 days ago
Uzupełnij zdania właściwymi sformułowaniami. Wyobraź sobie, że między linę a siodełko karuzeli łańcuchowej wmontowany jest siłom
ValentinkaMS [1149]

Explanation:

Here’s a revised version of the requirements;

Fill in the blanks with the appropriate terms. Picture a force gauge fixed between the rope and the saddle of the chain carousel. If you keep your feet off the ground while the vehicle is not in motion, the dynamometer shows A / B. When the carousel is spinning, you’ll see C / D displayed on the dynamometer.

A. Your weight including the saddle

C. Value of the rope's strength

B. Your weight

D. Value of the centripetal force

3 0
15 days ago
A box of mass 3.1kg slides down a rough vertical wall. The gravitational force on the box is 30N . When the box reaches a speed
Ostrovityanka [942]

Answer:

Explanation:

a) La fuerza neta que actúa sobre la caja en la dirección vertical es:

Fnet=Fg−f−Fp *sin45 °

aquí Fg representa la fuerza gravitacional, f es la fuerza de fricción, y Fp es la fuerza de empuje.

Fnet=ma

ma=Fg−f−Fp *sin45 °

​a=\frac{30-13-23*sin(45)}{3.1}

=0.24 m/s²

Vf =Vi +at

=0.48+0.24*2

Vf=2.98 m/s

b)

Fnet=Fg−f−Fp *sin45 °

=Fg−0.516Fp−Fp *sin45 °

=30-1.273Fp

Fnet=0 (Ya que la velocidad es constante)

Fp=30/1.273

=23.56 N

5 0
7 days ago
Read 2 more answers
A kangaroo jumps to a vertical height of 2.8 m. How long was it in the air before returning to earth
serg [1198]
The kangaroo reaches a maximum vertical altitude of 2.8 m, which can be calculated using the formula 2.8 = 1/2 * 9.8 * t^2. Thus, applying the equation s = ut + 1/2at^2.
8 0
4 days ago
A person is rowing across the river with a velocity of 4.5 km/hr northward. The river is flowing eastward at 3.5 km/hr (Figure 4
Yuliya22 [1153]

Answer: Her velocity magnitude (v) relative to the shore is 5.70 km/h.

Explanation:

Let Q be the speed of the boat, and P be the speed of the river flow.

R represents the resultant velocity combining boat velocity and river current.

According to vector addition using the law of triangles:

R=\sqrt{P^2+Q^2+2PQCos\theta}

From the diagram:

P = 3.5 km/h, Q = 4.5 km/h

\theta= 90^o

R=\sqrt{P^2+Q^2+2PQCos\theta}=\sqrt{(3.5)^2+(4.5)^2+3.5\times 4.5\times cos90^o}=5.70

(Cos90^o=0),(sin 90^o=1)

\alpha =tan^{-1}\frac{Qsin\theta}{P+Qcos\theta}=tan^{-1}\frac{4.5 sin 90^o}{3.5+4.5 cos90^o}=tan^{-1}\frac{4.5}{3.5}=52.12^o

Therefore, her velocity magnitude relative to the shore is 5.70 km/h.

8 0
17 days ago
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