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2 years ago

The velocity of 200kg object changes from 10m/s to 22m/s in 12 sec.what is the force on the accelerated object?

Physics

1 answer:

Vikentia [17]2 years ago

4 0

Initial velocity (u) = 10 m/s
Final velocity (v) = 22 m/s
Time (t) = 12 s
Mass (m) = 200 Kg
Let the acceleration be a.
By using the equation of motion,

v = u + at, we have

22 m/s = 10 m/s + 12 s × a
or, 22m/s - 10 m/s = 12 s × a
or, 12 m/s = 12 s × a
or, a = 1 m/s^2
Let the force be F.
We know, F = ma
Therefore, the force on the accelerated object (F)
= ma
= (200 × 1) N
= 200 N

Answer:

b) 200 N

Hope you could understand.

If you have any query, feel free to ask.

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A circular loop of flexible iron wire has an initial circumference of 165cm , but its circumference is decreasing at a constant

ArbitrLikvidat [17]

Answer:

emf induced is 0.005445 V and direction is clockwise because we can see area is decrease and so that flux also decrease so using right hand rule direction of current here clockwise

Explanation:

Given data

initial circumference = 165 cm

rate = 12.0 cm/s

magnitude = 0.500 T

tome = 9 sec

to find out

emf induced and direction

solution

we know emf in loop is - d∅/dt ........1

here ∅ = ( BAcosθ)

so we say angle is zero degree and magnetic filed is uniform here so that

emf = - d ( BAcos0) /dt

emf = - B dA /dt ..............2

so area will be

dA/dt = d(πr²) / dt

dA/dt = 2πr dr/dt

we know 2πr = c,

r = c/2π = 165 / 2π

r = 26.27 cm

c is circumference so from equation 2

emf = - B 2πr dr/dt ................3

and

here we find rate of change of radius that is

dr/dt = 12/2π = 1.91 $10^{-2}$ cm/s

so when 9.0s have passed that radius of coil = 26.27 - 191 (9)

radius = 9.08 $10^{-2}$ cm

so now from equation 3 we find emf

emf = - (0.500 ) 2π(9.08 $10^{-2}$ ) 1.91 $10^{-2}$

emf = - 0.005445

and magnitude of emf = 0.005445 V

emf induced is 0.005445 V and direction is clockwise because we can see area is decrease and so that flux also decrease so using right hand rule direction of current here clockwise

4 0

3 years ago

QUESTION 3

Alisiya [41]

The force of frictions is opposed to relative motion.

The acceleration of the crate is approximately 2.937 m/s².

Reason:

The given parameters are;

The mass of the wood, m = 100 kg

The force which can move the wood, F = 588 N

Wood on wood static friction, $\mu_s$ = 0.5

Wood on wood kinetic friction, $\mu_k$ = 0.3

Solution;

The force of friction, $F_f$ , acting when the crate is moving is given as

follows;

$F_f$ = m × g × $\mu_k$

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

Therefore, we have;

$F_f$ = 100 × 9.81 × 0.3 = 294.3

The force of friction, $F_f$ = 294.3 N

The force with which the crate moves, F = 588 - 294.3 = 293.7

The force with which the crate moves, F = 293.7 N

Force = Mass, m × Acceleration, a

$a = \dfrac{F}{m}$

Therefore;

$a = \dfrac{293.7 \ N}{100 \ kg} = 2.937$

The acceleration of the crate, a ≈ 2.937 m/s².

Learn more about friction here:

brainly.com/question/94428

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.Explanation:

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3 years ago

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solniwko [45]

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3 years ago

Read 2 more answers

In quantum mechanics, the fundamental constant called Planck's constant, h, has dimensions of [ML^2T^-1 ]. Construct a quantity

Mice21 [21]

Answer:

h/(m*c)

Explanation:

Hi!

Lets denote the units of X as [X]

Since the dimentions of h are:

$[h] = \frac{ML^{2}}{T}$