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mariarad [96]
3 years ago
5

An object, initially at rest, is subject to an acceleration of 34 m/s2. How long will it take for that

Physics
2 answers:
SIZIF [17.4K]3 years ago
6 0

Answer:

If it is moving 34 m/s it will take 100 seconds, or 1:40 to reach 3400 meters.

Explanation:

I found this answer by dividing 3400 by 34 and converting seconds to minutes

nevsk [136]3 years ago
6 0

Answer:

14 s

Explanation:

Given:

v₀ = 0 m/s

a = 34 m/s²

Δx = 3400 m

Find: t

Δx = v₀ t + ½ at²

(3400 m) = (0 m/s) t + ½ (34 m/s²) t²

t = 14.1 s

Rounded to two significant figures, it takes 14 seconds.

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According to Lenz's law, the induced current in a circuit always flows to oppose the external magnetic field through the circuit
Makovka662 [10]

Answer:

True.

Explanation:

According to Lenz's law, the induced current in a circuit always flows to oppose the external magnetic field through the circuit. This statement is true.

The Faraday's law of induction is given by :

\epsilon=-\dfrac{d\phi}{dt}

Here, negative sign shows that the direction of induced emf is such that opposes the changing current that is its cause.

Hence, the statement is true.

3 0
3 years ago
A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calcu
Alekssandra [29.7K]

Complete question:

A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

a) 60 ⁰

b) 90 ⁰

c) 120 ⁰

Answer:

(a) When the angle, θ = 60 ⁰,  force = 4.07 N

(b) When the angle, θ = 90 ⁰,  force = 4.7 N

(c) When the angle, θ = 120 ⁰,  force = 4.07 N

Explanation:

Given;

length of the wire, L = 2.8 m

current carried by the wire, I = 5.6 A

magnitude of the magnetic force, F = 0.3 T

The magnitude of the magnetic force is calculated as follows;

F = BIl \ sin(\theta)

(a) When the angle, θ = 60 ⁰

F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(60)\\\\F = 4.07 \ N

(b) When the angle, θ = 90 ⁰

F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(90)\\\\F = 4.7 \ N

(c) When the angle, θ = 120 ⁰

F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(120)\\\\F = 4.07 \ N

6 0
2 years ago
A cannonball is fired horizontally at the same time a ball being dropped from the same height How do the times it takes them to
ipn [44]

Answer:

The cannonball and the ball will both take the same amount of time before they hit the ground.

Explanation:

For a ball fired horizontally from a given height, there is only a vertical acceleration on it towards the ground. This acceleration is equal to the acceleration due to gravity (g = 9.81 m/s^2). A ball dropped from a height will also only experience the same vertical acceleration downwards which is also equal to g = 9.81 m/s^2.

Therefore both the cannonball and the ball will take the same amount of time to hit the ground if they are released/fired from the same height.

3 0
3 years ago
On your first day at work as an electrical technician, you are asked to determine the resistance per meter of a long piece of wi
Lostsunrise [7]

Answer:

0.06\Omega/m

Explanation:

Firstly, when you measure the voltage across the battery, you get the emf,

E = 13.0 V

In order to proceed we have to assume that the voltmeter offers no loading effect, which is a valid assumption since it has a very high resistance.

Secondly, the wires must be uniform. So the resistance per unit length is constant (say z). Now, even though the ammeter has very little resistance it cannot be ignored as it must be of comparable value/magnitude when compared to the wires. This is can seen in the two cases when currents were measured. Following Ohm's law and the resistance of a length of wire being proportional to it's length, we should have gotten half the current when measuring with the 40 m wire with respect to the 20 m wire (I=\frac{V}{R}). But this is not the case.

Let the resistance of the ammeter be r

Hence, using Ohm's law we get the following 2 equations:

\frac{13}{20z+r} =7.6   .......(1)

\frac{13}{40z+r} =4.5     ......(2)

Substituting the value of r from (2) in (1), we have,

13=152z+7.6\times\frac{13-180z}{4.5}

which simplifying gives us, z=0.0589\Omega/m\approx0.06\Omega/m (which is our required solution)

putting the value of z in either (1) or (2) gives us, r = 0.5325 \Omega

3 0
3 years ago
4. What is the electric field strength 1.4 nm from a charge of 4.7 cC?
pentagon [3]

The electric field strength is 2.16\cdot 10^{26} N/C

Explanation:

The strength of the electric field produced by a single point charge is given by:

E=k\frac{q}{r^2}

where

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q is the magnitude of the charge

r is the distance from the charge at which the field strength is calculated

For the charge in the problem, we have:

q=4.7 cC = 0.047 C is the charge

r=1.4 nm = 1.4\cdot 10^{-9} m

Therefore, the electric field strength is

E=(8.99\cdot 10^9)\frac{0.047}{(1.4\cdot 10^{-9})^2}=2.16\cdot 10^{26} N/C

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

5 0
3 years ago
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