What is the acceleration of a 1000kg car subject to a 550N net force?

The graph in the figure shows the variation of the electric potential V(x) (in arbitrary units) as a function of the position x

dmitriy555 [2]

The best and most correct answer among the choices provided by your question is the second choice or letter b.

Ex is negative from x = -2 to x = 0, and positive from x = 0 to x = 2 <span>correctly describes the orientation of the x-component of the electric field along the x-axis.</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

7 0

3 years ago

10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ?????

KATRIN_1 [288]

Answer: Tension = 47.8N, Δx = 11.5× $10^{-6}$ m.

Tension = 95.6N, Δx = 15.4× $10^{-5}$ m

Explanation: A speed of wave on a string under a tension force can be calculated as:

$|v| = \sqrt{\frac{F_{T}}{\mu} }$

$F_{T}$ is tension force (N)

μ is linear density (kg/m)

Determining velocity:

$|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }$

$|v| = \sqrt{0.00874 }$

$|v| =$ 0.0935 m/s

The displacement a pulse traveled in 1.23ms:

$\Delta x = |v|.t$

$\Delta x = 9.35.10^{-2}*1.23.10^{-3}$

Δx = 11.5× $10^{-6}$

With tension of 47.8N, a pulse will travel Δx = 11.5× $10^{-6}$ m.

Doubling Tension:

$|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }$

$|v| = \sqrt{2.0.00874 }$

$|v| = \sqrt{0.01568}$

|v| = 0.1252 m/s

Displacement for same time:

$\Delta x = |v|.t$

$\Delta x = 12.52.10^{-2}*1.23.10^{-3}$

$\Delta x =$ 15.4× $10^{-5}$

With doubled tension, it travels $\Delta x =$ 15.4× $10^{-5}$ m

4 0

3 years ago

Two long, parallel wires separated by 3.50 cm carry currents in opposite directions. The current in one wire is 1.55 A, and the

vaieri [72.5K]

Answer:

Therefore,

The magnitude of the force per unit length that one wire exerts on the other is

$\dfrac{F}{l}=2.79\times 10^{-5}\ N/m$

Explanation:

Given:

Two long, parallel wires separated by a distance,

d = 3.50 cm = 0.035 meter

Currents,

$I_{1}=1.55\ A\\I_{2}=3.15\ A$

To Find:

Magnitude of the force per unit length that one wire exerts on the other,

$\dfrac{F}{l}=?$

Solution:

Magnitude of the force per unit length on each of @ parallel wires seperated by the distance d and carrying currents I₁ and I₂ is given by,

$\dfrac{F}{l}=\dfrac{\mu_{0}\times I_{1}\times I_{2}}{2\pi\times d}$

where,

$\mu_{0}=permeability\ of\ free\ space =4\pi\times 10^{-7}$

Substituting the values we get

$\dfrac{F}{l}=\dfrac{4\pi\times 10^{-7}\times 1.55\times 3.15}{2\pi\times 0.035}$

$\dfrac{F}{l}=2.79\times 10^{-5}\ N/m$

Therefore,

The magnitude of the force per unit length that one wire exerts on the other is

$\dfrac{F}{l}=2.79\times 10^{-5}\ N/m$

7 0

3 years ago

Read 2 more answers

A cart for hauling ore out of a gold mine has a mass of 435 kg, including its load. the cart runs along a straight stretch of tr

marta [7]

What are the answer choices

4 0

3 years ago

A and B start walking in the same direction at the same time around a circular park of diameter 4200 m. If A walks 10 m more tha

Liono4ka [1.6K]

Answer:

Va = 5000 m / 3600 s = 1.39 m/s

(Va - Vb) 60 = 10

Vb = Va - .167 = 1.22 m/s

(Va - Vb) T = 4200 Π where T is time for A to complete 1 more lap

.17 T = 4200 Π

T = 24700 Π time for A to again catch B

N = 1.39 * 24700 Answer:

Va = 5000 m / 3600 s = 1.39 m/s

(Va - Vb) 60 = 10

Vb = Va - .167 = 1.22 m/s

(Va - Vb) T = 4200 Π where T is time for A to complete 1 more lap

.17 T = 4200 Π

T = 24700 Π time for A to again catch B

N = 1.39 * 24700 Π / (4200 Π) = 8.2 laps

A will make 8 but not 9 rounds before catching B

6 0

2 years ago