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fiasKO
4 days ago
9

Theresa is swinging a 4.0 kg ball on a 1.2 m long rope with a tangential speed of 1.8 m/s. What is the centripetal force exerted

on the rope? Round to the tenths place.
Physics
2 answers:
Keith_Richards [1K]4 days ago
8 0
Rounding the number gives us a result of 10.8N.
ValentinkaMS [1.1K]4 days ago
5 0

In this scenario, the centripetal force is expressed as:

F = m * (\frac{v^2}{r})

Where,

m: object mass

v: speed along a tangent

r: radius of the rope

By substituting the values into the equation, we find:

F = (4.0) * (\frac{1.8^2}{1.2})

Next, we perform the necessary calculations:

F = 10.8 N

Result:

The force acting on the rope is:

F = 10.8 N

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4 0
4 days ago
Jack pulls a sled across a level field by exerting a force of 110 n at an angle of 30 with the ground. what are the parallel and
Softa [913]
<span>A force of 110 N is applied at an angle of 30</span>°<span> to the horizontal. Because the force does not align directly either vertically or horizontally with the sled, it can be broken down into two components based on sine and cosine.

For the component parallel to the ground:
x = rcos</span>β
<span>x = 110cos30</span>°
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3 0
6 days ago
Read 2 more answers
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
A champion athlete can produce one horsepower (746 W) for a short period of time. The number of 16-cm-high steps a 70-kg athlete
Sav [1105]

Answer:

407 steps

Explanation:

Based on the question,

P = mgh/t........... Equation 1

Where P stands for power, m is mass, g denotes gravity, h is height, and t represents time.

Rearranging the equation to solve for h, we have:

h = Pt/mg............. Equation 2

Providing values: P = 746 W, t = 1 minute = 60 seconds, m = 70 kg.

Given constant: g = 9.8 m/s²

By substituting into equation 2

h = 746(60)/(70×9.8)

h = 44760/686

h = 65.25 m

h = 6525 cm

Calculating number of steps: 6525/16

The resulting number of steps = 407 steps

6 0
9 days ago
For a projectile, which of the following quantities are constant during the flight: x, y, vx, vy, v, ax, ay? Check all that appl
Yuliya22 [1153]

Response:

C. vx

F. ax

G. ay

Clarification:

The projectile follows a curved trajectory toward the ground, causing changes in x and y positions.

Since there is no external force acting in the x-direction, the acceleration in x remains at zero. Consequently, ax and vx remain unchanged.

The projectile is subject to the force of gravity, directed downwards, leading to an increase in its velocity due to the rise in its y-component.

Meanwhile, the y-component of acceleration remains constant due to gravitational acceleration.

5 0
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