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podryga [215]
3 years ago
5

An object of mass m is initially at rest. After a force of magnitude F acts on it for a time T, the object has a speed v. Suppos

e the mass of the object is doubled, and the magnitude of the force acting on it is quadrupled.In terms of T, how long does it take for the object to accelerate from rest to a speed v now?
Physics
1 answer:
adell [148]3 years ago
5 0

Answer:

It takes \frac{1}{2}T to accelerate the object from rest to the speed v.

Explanation:

From Newton's second law:

F=m\cdot a  (1)

and the definition of acceleration,

\displaystyle{a = \frac{\Delta v}{\Delta t}} (2)

we can solve this problem. Putting (2) in (1) we have:

\displaystyle{F = m\cdot \frac{\Delta v}{\Delta t}} and solving for \Delta t and considering the initial time as zero (t_0=0) and the initial velocity also zero (v_0=0) we have:

\displaystyle{T=\frac{mv}{F}}

Now, for a mass m^*= 2m and the F^*=4F we can wrtie the same equation:

\displaystyle{T^*=\frac{m^*v}{F^*}} and substituting m^* and F^*:

\displaystyle{T^*=\frac{2m\cdotv}{4F}=\frac{2}{4}T=\boxed{\frac{1}{2}T}}

So now, it only takes half the time to accelerate the object from rest to the speed v

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Vlad [161]

Answer:

(a) The package lands 682 meters horizontally ahead from the point the package was dropped from the plane

(b) The horizontal component = 39.0 m/s

The vertical component = 171.55 m/s

(c) The angle of impact is 77.19°

Explanation:

The parameters given are;

Velocity of the plane, vₓ = 39.0 m/s

Height of the plane above the ground, h = 1.50 × 10² m = 1,500 m

(a) The time, t, before the package hits the ground when dropped from the plane is given by the relation;

h = u·t + 1/2×g×t²

Where:

g = Acceleration due to gravity = 9.81 m/s²

u = Initial vertical velocity = 0 m/s

Hence;

1500 = 0×t + 1/2 × 9.81 × t² = 4.905·t²

∴ t = √(1500/4.905) = 17.49 s

The horizontal distance the package travels before landing = 17.49 × 39 ≈ 682 m

The package lands 682 meters horizontally ahead from the point the package was dropped from the plane

(b) The vertical velocity, v_y, of the package just before landing is given by the relation;

v_y² = u² + 2·g·h

u = 0 m/s

∴ v_y² = 0 + 2×9.81×1500 = 29430 m²/s²

v_y = √29430  = 171.55 m/s

Hence the horizontal component = 39.0 m/s

The vertical component = 171.55 m/s

(c) The angle of impact, θ, is given as follows;

tan \theta = \dfrac{v_y}{v_x}  = \dfrac{171.55}{39.0 } = 4.4

∴ θ = tan⁻¹(4.4) = 77.19°.

6 0
3 years ago
A crate with a mass of m = 450 kg rests on the horizontal deck of a ship. The coefficient of static friction between the crate a
Zielflug [23.3K]

Answer:F_{v} =\mu_{k} mg

Magnitude of the force is 2601.9 N

Explanation:

m = 450 kg

coefficient of static friction μs = 0.73

coefficient of kinetic friction is μk = 0.59

The force required to  start crate moving is F_{s} =\mu_{s} mg.

but once crate starts moving the force of friction is reduced  F_{v} =\mu_{k} mg.

Hence  to keep crate moving at constant velocity we have to reduce the  force pushing crate ie F_{v} =\mu_{k} mg.

Then the above pushing force will equal the frictional force due to kinetic friction and constant velocity is possible as  forces are balanced.

Magnitude of the force

F_{v} =\mu_{k} mg\\F_{v} =0.59 \times 450 \times 9.8\\F_{v} =2601.9  N

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Aleksandr-060686 [28]

Answer:

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Explanation:

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v = 27 m/s

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