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riadik2000 [5.3K]
3 years ago
10

An elevator is moving upward at 1.20 m/s when it experiences an acceleration of 0.31 downward over a distance of 0.75m. What wil

l its final velocity be?
Physics
2 answers:
iren2701 [21]3 years ago
5 0
<span>Info.: Vi = 1.20 m/s, a = 0.31 m/s^2, Δx = 0.75 m, Vf = ?

Vf = sqrt (Vi^2 + 2aΔx) = sqrt ((1.20 m/s)^2 + 2(0.31 m/s^2 * 0.75 m))
= 0.99 m/s</span>
Gemiola [76]3 years ago
5 0

Answer:

The final velocity is 0.99 m/s.

Explanation:

Given that,

Speed = 1.20 m/s

Acceleration = -0.31 m/s²

Distance = 0.75 m

We need to calculate the final velocity

Using equation of motion

v^2=u^2+2as

Where, v = final velocity

u = initial velocity

a = acceleration

s = distance

Put the value in the equation

v^2=1.20^2-2\times0.31\times0.75

v=\sqrt{0.975}

v=0.99\ m/s

Hence, The final velocity is 0.99 m/s.

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Running at 2 m/s, Bruce the 45 kg quarterback, collides with Biff, the 90 kg tackle, who is traveling at 7 m/s in the other dire
melomori [17]

Given data

*The mass of Bruce is m_1 = 45 kg

*The initial velocity of the Bruce is u_1 = 2 m/s

*The mass of the biff is m_2 = 90 kg

*The initial velocity of the Biff is u_2 = -7 m/s

*The final velocity of the first glider is v_2 = -1 m/s

According to the law of conservation of linear momentum, the total linear momentum of a system remains constant

Applying the law of conservation of momentum as

\begin{gathered} p_i=p_f \\ m_1u_1+m_2u_2=m_1v_1+m_2v_2 \\ v_1=\frac{m_1u_1+m_2u_2-m_2v_2_{}_{}_{}_{}}{m_1} \end{gathered}

Substitute the known values in the above expression as

\begin{gathered} v_1=\frac{(45)(2)+(90)(-7)-(90)(-1)}{45} \\ =-10\text{ m/s} \end{gathered}

Hence, the speed of the bruce knock backwards is v_1 = -10 m/s

3 0
2 years ago
A shark is cruising at 4 m/s when it sees a fish straight
Oksanka [162]

Answer:

Distance, S = 13m

Explanation:

Given the following data;

Initial velocity, u = 4m/s

Final velocity, v = 9m/s

Time, t = 2 seconds

To find the distance, S;

First of all, we would determine the acceleration of the shark.

Acceleration = (v - u)/t

Acceleration = (9 - 4)/2

Acceleration = 5/2

Acceleration = 2.5m/s²

Now, to find the distance we would use the second equation of motion

S = ut + ½at²

Substituting into the equation, we have

S = 4(2) + ½*2.5*2²

S = 8 + 1.25*4

S = 8 + 5

Distance, S = 13m

3 0
4 years ago
Where are the three seismographs used to find the epicenter of this earthquake located?
iogann1982 [59]
The seismographs are located in Minneapolis, Detroit, and Charleston.
3 0
3 years ago
Read 2 more answers
A sled and its rider are moving at a speed of 4.0 m/s along a horizontal stretch of snow. The snow exerts a kinetic frictional f
ehidna [41]

Answer:

The acceleration of the sled is "0.49 m/s²". A further explanation is given below.

Explanation:

The given values are,

Speed,

V = 4.0 m/s

Coefficient of kinetic friction,

μ = 0.05

As we know,

⇒  F \mu=\mu mg

and,

⇒  a = \mu g

On substituting the given values, we get

⇒     =0.0 5\times 9.8

⇒     =0.49 \ m/s^2

5 0
3 years ago
A tortoise and hare start from rest and have a race. As the race begins, both accelerate forward. The hare accelerates uniformly
Mnenie [13.5K]

Answer:

The acceleration of the hare once it begins to slow down is -0.68 m/s²

The acceleration of the tortoise is 0.28 m/s²

Explanation:

The equations that describe the position and velocity of the hare and the tortoise are the following:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

To find the acceleration of the hare once it begins to slow down, we have to find how much time the hare traveled during the deceleration and what was its initial speed.

First, the hare moves with constant acceleration for 4.7 s. Then, its velocity at  t = 4.7 s will be:

v = v0 + a · t    (v0 = 0 because the hare starts form rest)

v = a · t = 0.9 m/s² · 4.7 s = <u>4.2 m/s</u>

<u />

The distance traveled by the hare while accelerating can be calculated using the equation of the position:

x = x0 + v0 · t + 1/2 · a · t²      (x0 = 0 and v0 = 0)

x = 1/2 · a · t² = 1/2 · 0.9 m/s² · (4.7)² = <u>9.9 m</u>

<u />

Then, the hare runs at a constant speed of 4.2 m/s for 11.7 s. The distance traveled at constant speed will be:

x =  v · t

x = 4.2 m/s · 11.7 s = <u>49.1 m</u>

<u />

Then, the distance traveled by the hare while slowing down was:

Distance traveled while slowing down = 72 m - 49.1 m - 9.9 m = 13 m

Let´s find how much time it took the hare to come to stop, so we can calculate the acceleration. We know that when the position is 13 m, the velocity is 0.

v = v0 + a · t

0 = 4.2 m/s + a · t

-4.2 m/s / t = a

Replacing in the equation of the position:

x = v0 · t + 1/2 · a · t²      (considering x0 as the point at which the hare started to slow down)

13 m = 4.2 m/s · t - 1/2 · 4.2 m/s / t · t²

13 m = 4.2 m/s · t - 2.1 m/s · t

13 m = 2.1 m/s · t

t = 13 m / 2.1 m/s

t = 6.2 s

Then, the acceleration of the hare while slowing down will be:

-v0/t = a

-4.2 m/s / 6.2 s = a

a = -0.68 m/s²

The acceleration of the hare once it begins to slow down is -0.68 m/s²

The hare traveled 72 m in (6.2 s + 11.7 s + 4.7 s) 22.6 s. The tortoise reaches the final position of the hare at the same time, so, using the equation of the position we can calculate the acceleration of the tortoise:

x = x0 + v0 · t + 1/2 · a · t²     (x0 = 0 and v0 = 0)

x = 1/2 · a · t²

72 m = 1/2 · a · (22.6 s)²

144 m / (22.6 s)² = a

a = 0.28 m/s²

The acceleration of the tortoise is 0.28 m/s²

6 0
3 years ago
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