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1 year ago

An object is dropped from a 15 m ledge. How fast it is moving just before it hits the ground?

Physics

2 answers:

loris [4]1 year ago

7 0

Answer:

$v_f = 17.15 m/s$

Explanation:

Since the ball is dropped from the height

h = 15 m

so here initial speed of the ball is

$u = 0$

acceleration of the ball due to gravity is given as

$a = 9.8 m/s^2$

now the final speed of the ball is given as

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 0 = 2(9.8)(15)$

$v_f = 17.15 m/s$

Sergeeva-Olga [200]1 year ago

4 0

By v^2 = u^2 + 2gh
v^2 = 0 + 2 x 9.8 x 15
v = √294
v = 17.15 m/s

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1 year ago

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A cart starts from rest and accelerates at 4.0 m/s2 for 5.0 s, then maintains that velocity for 10 s, and then decelerates at th

zhannawk [14.2K]

Answer:

Final speed of car = 12 m/s

Explanation:

We have equation of motion v = u + at, where v is final velocity, u is initial velocity, a is acceleration and t is time.

a) A cart starts from rest and accelerates at 4.0 m/s² for 5.0 s

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t = 5 s

v = u + at = 0 + 4 x 5 = 20 m/s

b) Then maintains that velocity for 10 s

v = ?

u = 20 m/s

a = 0 m/s²

t = 10 s

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c) Then decelerates at the rate of 2.0 m/s² for 4.0 s

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3 0

1 year ago

for a given initial projectile speed, you observe that the projectile has a certain range R at a launch angle of a = 30. For wha

VLD [36.1K]

Answer:

The other angle is 30 degrees.

Explanation:

The range of projectile is given by :

$R=\dfrac{u^2\ \sin2\theta}{g}$

Here,

u is the speed of launch of projectile

Here, $\theta=30^{\circ}$

We need to find the other launch angle when the projectile have the same range, such that,

$\dfrac{u^2\ \sin(60)}{g}=\dfrac{u^2\ \sin2\alpha}{g}$

$\sin(60)=\sin2\alpha$

$\dfrac{\sqrt3}{2}=\sin2\alpha$

$\alpha =30^{\circ}$

So, the other angle is 30 degrees. Hence, this is the required solution.

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1 year ago

The boom hoisting sheave must have pitch diameters of no less than _______times the nominal diameter of the rope used.

alexira [117]

Answer:

18 times

Explanation:

According to the security purposes which is set under the rules and regulation OSHA, which describes all the rights to the worker.

In the boom hoist receiving system all the sheaves which are used should have a pitch diameter of rope not less than 18 times the diameter of the nominal rope which is used.

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1 year ago

the water behind hoover dam in nevada is 206 m higher than the colorado river below it. at what rate must water pass through the

Nitella [24]

Answer:

the required mass flow rate is 49484.37 kg/s

Explanation:

Given the data in the question;

we first determine the relation for mass flow rate of water that passes through the turbine;

so the relation for net work on the turbine due to the change in potential energy considering 100% efficiency is;

$W_{net}$ = m ( Δ P.E )

so we substitute (gh) for ( Δ P.E );

$W_{net}$ = m (gh)

m = $W_{net}$ / gh

so we substitute our given values into the equation

m = 100 MW / ( 9.81 m/s²) × 206 m

m = ( 100 MW × 10⁶W/MW) / ( 9.81 m/s²) × 206 m

m = 10 × 10⁷ / 2020.86

m = 49484.37 kg/s

Therefore, the required mass flow rate is 49484.37 kg/s

5 0

1 year ago