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3 years ago

If your automobile runs out of fuel while driving, the engine stops. You don't come to an abrupt stop due to

Physics

1 answer:

anastassius [24]3 years ago

3 0

If your automobile runs out of fuel while driving, the engine

stops. You don't come to an abrupt stop due to the concept of

inertia.

<h3>What is Inertia?</h3>

This is referred to a resistance in an object due to its change in

velocity. The object doesn't come to a sudden state of rest.

When the fuel runs out and the engine stops, there is no abrupt

stop as a result of the inertia(resistance) to the change in

velocity.

Read more about Inertia here brainly.com/question/1140505

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If the time of the impact in a collision is extended by 4 times ,how much does the force of impact change

Rama09 [41]

We know, I = F.Δt
As Δt is increased to 4 times, then, F would decrease to 4 times, in order to keep that impulse constant.

In short, Your force will change to 1/4th of it's initial value

Hope this helps!

7 0

3 years ago

Two coils are wound around the same cylindrical form. When the current in the first coil is decreasing at a rate of -0.245 A/s ,

Airida [17]

Answer:

Complete question:

c.If the current in the second coil increases at a rate of 0.365 A/s , what is the magnitude of the induced emf in the first coil?

a. $M= 6.53\times10^{-3} H$

b.flux through each turn = Ф = $4.08\times10^{-4} Wb$

c.magnitude of the induced emf in the first coil = e= $2.38\times10^{-3} V$

Explanation:

a. rate of current changing = $\frac{di}{dt}=[tex]M=\frac{1.60\times10^{-3} V}{0.240\frac{A}{s} }}$ [/tex]

Induced emf in the coil =e= $1.60\times10^{-3} V$

For mutual inductance in which change in flux in one coil induces emf in the second coil given by the farmula based on farady law

$e=-M\frac{di}{dt}$

$M=\frac{e}{\frac{di}{dt} }$

$M=\frac{1.60\times10^{-3} }{-0.245}$

$M= 6.53\times10^{-3} H$

Flux through each turn=?

Current in the first coil =1.25 A

Number of turns = 20

using MI = NФ

flux through each turn = Ф = $\frac{6.53\times10^{-3}\times1.25}{20}$

flux through each turn = Ф = $4.08\times10^{-4} Wb$

second coil increase at a rate = 0.365 A/s

magnitude of the induced emf in the first coil =?

using $e=-M\frac{di_{2} }{dt}$

$e= 6.53\times10^{-3} \times 0.365$

magnitude of the induced emf in the first coil = e= $2.38\times10^{-3} V$

4 0

3 years ago

Is anyone good at science I need help with 2 tests

Katyanochek1 [597]

Answer:

i am!

Explanation:

7 0

3 years ago

Two protons (each with q = 1.60 x 10-19)

otez555 [7]

Answer:

230.4 N

Explanation:

From the question given above, the following data were obtained:

Charge (q) of each protons = 1.6×10¯¹⁹ C

Distance apart (r) = 1×10¯¹⁵ m

Force (F) =?

NOTE: Electric constant (K) = 9×10⁹ Nm²/C²

The force exerted can be obtained as follow:

F = Kq₁q₂ / r²

F = 9×10⁹ × (1.6×10¯¹⁹)² / (1×10¯¹⁵)²

F = 9×10⁹ × 2.56×10¯³⁸ / 1×10¯³⁰

F = 2.304×10¯²⁸ / 1×10¯³⁰

F = 230.4 N

Therefore, the force exerted is 230.4 N

5 0

3 years ago

A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of

uysha [10]

Answer:

q = 4.04×10^-12C

n = 2.53×10^7

Explanation:

1. Electrostatic force on charge due to electrical field E is F = qE

Gravitational force due to weight of drop of mass M is F = mg, when net force is zero we have to find the charges on the drop.

2. Charge on drop: since gravitational force is acting downward, electric field is acting upward and we equate both equations.

Therefore, Fg = Fe

qe = mg

q = mg/E

q = 3.50×10^-9kg×9.8m/s²/8480N/c

q = 4.04 × 10^-12

3. Number of protons by quantitization law

n = q/e

n = 4.04× 10^-12/8480

n = 2.53× 10^7

8 0

3 years ago