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3 years ago

When a force is applied to an object for an amount of time, it is known as __________.

Physics

2 answers:

stiks02 [169]3 years ago

8 0

Answer:

Impulse

Explanation:

I took the test and got it correct.

Hope this helps! :)

NikAS [45]3 years ago

7 0

The correct answer to the question is : D) Impulse

EXPLANATION:

Before going to answer this question, first we have to understand impulse.

Impulse of a body is defined as change in momentum or the product of force with time.

Mathematically impulse = F × t = m ( v - u ).

Here, v is the final velocity

u is the initial velocity

F is the force acting on the body for time t.

Hence, the perfect answer of this question is impulse m i.e the force multiplied with time is known as impulse.

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The main energy source during exercise of low to moderate intensity is

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Answer:

THUS,THE MAJOR SOURCES OF ENERGY DURING EXERCISE ARE CARBOHYDRATES AND FATS.

Explanation:

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3 years ago

Which statement is true about the relationship between carrying capacity and population size?

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Carry capacity determines maximum population size

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3 years ago

9) An object with a height of 18 cm is placed in front of a converging lens. The image has a

vovikov84 [41]

Answer:

Explanation:

a) Magnification = image height / object height = -9 / 18 = -0.5

b) Magnification = - image distance / object distance = -0.5

so image distance = 0.5 object distance

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object distance = 18.0 cm

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3 years ago

Read 2 more answers

At the instant a traffic light turns green, a car starts with a constant acceleration of 1.3 m/s2. At the same instant a truck,

Bas_tet [7]

Answer:

(a) Distance traveled = 75.3846 m

(b) Velocity of car at that instant will be 14 m/sec

Explanation:

We have given acceleration of the car $a=1.3m/sec^2$

Initial velocity of the cart u = 0 m/sec

(a) According to second equation of motion we know that $s=ut+\frac{1}{2}at^2$

So distance traveled by car $s_c=0\times t+\frac{1}{2}\times 1.3t^2=0.65t^2$

As the truck is moving with constant speed

So distance traveled by truck $s_t=ut=7t$

As the truck overtakes the car

So $s_c=s_t$

$0.65t^2=7t$

$t=10.769sec$

So distance traveled $s_c=s_t=7\times 10.769=75.3846m$

(b) From second equation of motion we know that v = u+at

So v = 0+1.3×10.769 = 14 m /sec

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3 years ago

Two concentric conducting spherical shells produce a radially outward electric field of magnitude49,000 N/C at a point 4.10 m fr

Korolek [52]

To solve this problem it is necessary to apply the concepts related to the electric field according to the definition of Coulomb's law.

The electric field is defined mathematically as a vector field that associates to each point in space the (electrostatic or Coulomb) force per unit of charge exerted on an infinitesimal positive test charge at rest at that point.

Mathematically this can be described as:

$E = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}$

Where,

$\epsilon_0 =$ permittivity of free space

r = Distance

q = Charge

E = Electric Field

Our values are given as,

$E= 49.000N/C$

$r= 4.1m$

$\epsilon_0=8.854*10{-12}C^2N^{-1}m^{-2}$

Replacing we have,

$E = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}$

$49000 = \frac{1}{4\pi (8.854*10{-12})}\frac{q}{(4.1)^2}$

$q= 9.16*10^{-5} C$

$q= 91.6\mu C$

Therefore the amoun of charge on the outer surface of the larger shell is $91.6 \mu C$

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3 years ago