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Alexxandr [17]
1 year ago
15

What is the work done by the 200.-N tension shown if it is used to drag the 150-N crate 25 m across the floor at a constant spee

d?
Physics
1 answer:
WINSTONCH [101]1 year ago
8 0

Answer:

0 J

Explanation:

Work equals force times distance, but the force is zero because the crate being dragged will have zero acceleration. Force equals mass times acceleration and since acceleration is zero, force has to equal zero as well. Since the force is zero, the work required also has to be zero.

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A very long, straight wire has charge per unit length 3.50×10^−10 C/m . At what distance from the wire is the electricfield magn
Dafna11 [192]

Answer:

r= 2.17 m

Explanation:

Conceptual Analysis:

The electric field at a distance r from a charge line of infinite length and constant charge per unit length is calculated as follows:

E= 2k*(λ/r) Formula (1)

Where:

E: electric field .( N/C)

k: Coulomb electric constant. (N*m²/C²)

λ: linear charge density. (C/m)

r : distance from the charge line to the surface where E calculates (m)

Known data

E= 2.9  N/C

λ = 3.5*10⁻¹⁰ C/m

k= 8.99 *10⁹ N*m²/C²

Problem development

We replace data in the formula (1):

E= 2*k*(λ/r)

2.9= 2*8.99 *10⁹*(3.5*10⁻¹⁰/r)

r =( 2*8.99 *10⁹*3.5*10⁻¹⁰) / (2.9)

r= 2.17 m

5 0
2 years ago
If one replaces the conducting cube with one that has positive charge carriers, in what direction does the induced electric fiel
Grace [21]

Answer:

There will be no change in the direction of the electric field .

Explanation:

The direction will remain the same because the sign of the charges has no effect on it.

When one replaces the conducting cube with one that has positive charge carriers there will be no change in the direction of the field as there is no defined relationship between the direction of the electric field and sign of the charge.

3 0
2 years ago
A spaceship is travelling at 20,000.0 m/s. After 5.0 seconds, the rocket thrusters are turned on. At the 55.0 second mark, the s
tankabanditka [31]

Answer:

80 m/s^2

Explanation:

The acceleration of an object is given by:

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time interval it takes for the velocity to change from u to v

For the rocket in this problem,

u = 20,000 m/s

v = 24,000 m/s

t = 55.0 - 5.0 = 50.0 s

Substituting,

a=\frac{24000-20000}{50}=80 m/s^2

7 0
1 year ago
Calculate the amount of work done to draw a current of 8A from a point at 100V to a point at 120V in 2 seconds?
Morgarella [4.7K]
Given:
I=8A
t=2second
Potential difference,V=120-100=20volt
Workdone=V×i×t
=20×8×2
=320 joule.
3 0
2 years ago
A 0.305 kg book rests at an angle against one side of a bookshelf. The magnitude and direction of the total force exerted on the
tankabanditka [31]

Answer

given,

F_L= 1.52\ N

\theta_L= 31^0

mass of book = 0.305 Kg

so, from the diagram attached  below

F_L cos {\theta_L} + F_b sin {\theta_b} = m g

1.52 times cos {31^0} + F_b sin {\theta_b} = 0.305 \times 9.8

F_b sin {\theta_b} = 2.989 -1.303

F_b sin {\theta_b} = 1.686

computing horizontal component

F_b cos {\theta_b} = F_L sin {\theta_L}

cos {\theta_b} = \dfrac{F_L sin {\theta_L}}{F_b}

cos {\theta_b} = \dfrac{1.52 \times sin {31^0}}{1.686}

cos {\theta_b} = 0.464

θ = 62.35°

5 0
2 years ago
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