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3 years ago

What is the magnitude of the Box's Acceleration?

Physics

1 answer:

mojhsa [17]3 years ago

4 0

The Box's Acceleration : g sin θ

<h3>Further explanation </h3>

Newton's 2nd law explains that the acceleration produced by the resultant force on an object is proportional and in line with the resultant force and inversely proportional to the mass of the object

∑F = m. a

F = force, N

m = mass = kg

a = acceleration due to gravity, m / s²

We plot the forces acting on the block (picture attached) according to the y-axis and the x-axis.

Because the motion of the block is in the same direction as the x-axis, ignoring the friction force with the inclined plane, then

$\tt \sum F_x=m.a\\\\W.sin\theta=m.a\\\\mgsin\theta=m.a\\\\a=gsin\thet\theta$

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Answer:

(4) full moon.

Explanation:

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2 years ago

A projectile is launched in the horizontal direction. It travels 2.050 m horizontally while it falls 0.450 m vertically, and it

kompoz [17]

Answer:

a) 0.303 s b) 6.77 m/s c) - 2.97 m/s d) 7.39 m/s

Explanation:

horizontal distance travel = 2.050 m, vertical distance travel = 0.45 m

Using equation of linear motion

Sy = Uy t + 1/2 gt² Uy is the inital vertical component of the velocity, t is the time taken for the vertical motion in seconds, and S is the vertical distance traveled, taken downward vertical motion as negative

-0.45 = 0 - 0.5 × 9.81×t²

0.45 / (0.5 × 9.81) = t²

t = √0.0917 = 0.303 s

b) horizontal distance traveled = horizontal velocity (Ux) × t

Ux = horizontal distance traveled / t = 2.050 / 0.303 = 6.77 m/s

c) Velocity in the vertical direction can be calculated using

Vy = Uy - gt where g is negative since initial U is zero

Vy = - 9.81 × 0.303 = - 2.97 m/s

d) the magnitude of the velocity = resultant of the the two velocity

using Pythagoras theorem

the magnitude of the velocity = √ ( 6.77² + 2.97²) = 7.39 m/s

at angle = tan^-1 ( 2.97 / 6.77) = 23.69⁰

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3 years ago

How would Newton's Three laws apply to a Ferris Wheel?

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The Central pedal force of the wheels center is opposed by an equal force from the objects being spun around by the wheel

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3 years ago

A 0.060-kg tennis ball, moving with a speed of 5.50m/s , has a head-on collision with a 0.090-kg ball initially moving in the sa

Yuliya22 [10]

Answer:

Explanation:

For elastic collision , the formula for velocity of .06 kg is

v₁ = (m₁-m₂)/(m₁+m₂) u₁ + 2m₁m₂/(m₁+m₂) u₂

=( .06-.09 / .06+.09 ) 5.5 + (2 x .06 x .09 / .06+.09 ) 3.4

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Its direction will be - opposite direction

the formula for velocity of .09kg is

v₂ = (m₂-m₁)/(m₁+m₂) u₂ + 2m₁m₂/(m₁+m₂) u₁

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3 years ago

Jason and Guy are throwing watermelons straight up from the ground. (They’re making fruit salad.) Jason throws his watermelon at

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Answer:2t

Explanation:

Given

Jason throwing watermelon at speed v and it spend t time in air

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$v_f=u_i+at'$

0=v-gt'

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and time to reach bottom is also t'

t=2t'

$t=2\frac{v}{g}$

For guy throws his watermelon at speed 2v

So total time in air will be

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$t''=4\frac{v}{g}$

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3 years ago

What is the magnitude of the Box's Acceleration?​

What is the magnitude of the Box's Acceleration?