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zhannawk [14.2K]
3 years ago
13

The diagrams below show forces acting on a toy car as it moves to the right which diagram shows the toy car that will most likel

y accelerate to the right based on these conditions?

Physics
2 answers:
choli [55]3 years ago
6 0

Answer:

Option 1

Explanation:

aniked [119]3 years ago
5 0

Answer:

Option 1.

Explanation:

To know which option is correct, let us calculate the net force in each case to know which will move to the right direction.

Option 1:

Force to the right (Fᵣ) = 8 N

Force to the left (Fₗ) = 6 N

Net force (Fₙ) =?

Fₙ = Fᵣ – Fₗ

Fₙ = 8 – 6

Fₙ = 2 N to the right

Option 2:

Force to the right (Fᵣ) = 8 N

Force to the left (Fₗ) = 10 N

Net force (Fₙ) =?

Fₙ = Fₗ – Fᵣ

Fₙ = 10 – 8

Fₙ = 2 N to the left

Option 3:

Force to the right (Fᵣ) = 10 N

Force to the left (Fₗ) = 10 N

Net force (Fₙ) =?

Fₙ = Fᵣ – Fₗ

Fₙ = 10 – 10

Fₙ = 0 (no movement)

Option 4:

Force to the right (Fᵣ) = 10 N

Force to the left (Fₗ) = 13 N

Net force (Fₙ) =?

Fₙ = Fₗ – Fᵣ

Fₙ = 13 – 10

Fₙ = 3 N to the left

From the illustrations made above, only option 1 has a net force toward the right direction. Therefore, only option 1 will have an acceleration towards the right direction.

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1. Juan wants to measure the current traveling through the straight wire in his experiment. He
olga2289 [7]

Answer:

galvanometer

Explanation:

  • A galvanometer is an instrument that can detect currents.
  • Juan wants to measure the current traveling through the straight wire in his experiment. He uses a galvanometer as his tool. He places the wire in between two permanent magnets to measure the current created by the magnetic field.
5 0
3 years ago
A commuter train blows its 200-Hz horn as it approaches a crossing. The speed of sound is 335 m/s. An observer waiting at the cr
pogonyaev

Answer:

346m/s

Explanation:

v1=335m/s

f1=200hz

f2=207hz

v2=?

v1/f1=v2/f2

335/200=v2/207

v2=335*207/200

v2=346m/s

4 0
3 years ago
1. A 2,000-turn solenoid is 65 cm long and has cross-sectional area 30 cm2. What rate of change of current will produce a 600 Vo
JulsSmile [24]

Answer:

\frac{dI}{dt} = 2.59\ x\ 10^4\ A/s

Explanation:

First, we will calculate the inductance of the solenoid by using the following formula:

L = \frac{\mu_o AN^2}{l}

where,

L = self-inductance of solenoid = ?

μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²

A = Cross-sectional area = 30 cm² = 3 x 10⁻³ m²

N  = No. of turns = 2000

l = length = 65 cm = 0.65 m

Therefore,

L = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(3\ x\ 10^{-3}\ m^2)(2000)^2}{0.65\ m}\\\\L =  0.0232\ H

Now, we will use Faraday's law to calculate the rate of change of current:

emf = L\frac{dI}{dt}\\\\ \frac{dI}{dt} =\frac{emf}{L} \\\\ \frac{dI}{dt} =\frac{600\ V}{0.0232\ H}\\\\  \frac{dI}{dt} = 2.59\ x\ 10^4\ A/s

5 0
3 years ago
If the potential in a region is given by the function V = 2 x − y 2 − cos(z), what is the y-component of the electric field at t
Bess [88]

Answer:

2y

Explanation:

Electric field in terms of Electric potential is given as:

E = dV/dr(x, y, z)

Where r(x, y, z) = position in x, y, z plane

The y component of the Electric field will be:

Ey = -dV/dy

Given that

V = 2x - y² - cos(z)

dV/dy = -2y

=> E = - (-2y)

E = 2y

8 0
4 years ago
To stretch a spring 8.00cm from its unstretched length, 16.0J of work must be done.A)What is the force constant of this spring?B
ad-work [718]

A) 5000 N/m

The force constant of the spring can be found by using the expression for the elastic potential energy stored in the spring (which is equal to the work done on it):

W=U=\frac{1}{2}kx^2

where

k is the spring constant

x is the stretching/compression of the spring

In this problem, we have

W = 16.0 J is the work done

x = 8.00 cm = 0.08 m is the stretching

Substituting into the formula and re-arranging it, we find

k=\frac{2W}{x^2}=\frac{2(16.0 J)}{(0.08 m)^2}=5000 N/m

B) 400 N

The magnitude of the force needed to stretch the spring by x = 8.00 cm = 0.08 m is given by Hook's law:

F=kx

where k=5000 N/m as we found previously. Substituting x=0.08 m, we find:

F=(5000 N/m)(0.08 m)=400 N

C) 4 J

The work done to compress the spring by x=4.00 cm=0.04 m is given by the same formula used for part A:

W=\frac{1}{2}kx^2

where in this case, k=5000 N/m and x=0.04 m. Substituting, we find

W=\frac{1}{2}(5000 N/m)(0.04 m)^2=4 J

D) 200 N

As we did in part B), the force needed to stretch this distance is given by Hook's law:

F=kx

where in this case, k=5000 N/m and x=0.04 m. Substituting, we find

F=(5000 N/m)(0.04 m)=200 N

8 0
3 years ago
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