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3 years ago

A 87 arrow is fired from a bow whose string exerts an average force of 105 on the arrow over a distance of 75 .

Physics

1 answer:

timofeeve [1]3 years ago

6 0

The solution would be like this for this specific problem:

V^2 = 2AS = 2FS/M

V = sqrt(2FS/M) = sqrt(2*105*.75/.087) = 44.52817783 = 42.5 mps

So the speed of the arrow as it leaves the bow is 42.5 mps.

I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

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For a standard production car, the highest road-tested acceleration ever reported occurred in 1993, when a Ford RS200 Evolution

Ann [662]

Answer:

a = 8.06 m/s²

Explanation:

The acceleration of this car can be found using the first equation of motion:

$v_f = v_i + at\\\\a = \frac{v_f-v_i}{t}$

where,

a = acceleration = ?

vf = final speed = 26.8 m/s

vi = initial speed = 0 m/s

t = time = 3.323 s

Therefore,

$a = \frac{26.8\ m/s-0\ m/s}{3.323\ s}$

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2 years ago

Which statement about stage 3 NREM is accurate?

cestrela7 [59]

Answer:

hope

that helps

Explanation:

D. The sleeper's heart rate, blood pressure, and breathing rate drop to their lowest levels.

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2 years ago

A 15 kg bucket of water is suspended by a very light ropewrapped around a solid uniform cylinder 0.300 m in diamter withmass 12.

matrenka [14]

Answer:

Part a)

$T = 42 N$

Part b)

$v_f = 11.8 m/s$

Part c)

$t = 1.7 s$

Part d)

$F = 159.7 N$

Explanation:

Part a)

While bucket is falling downwards we have force equation of the bucket given as

$mg - T = ma$

for uniform cylinder we will have

$TR = I\alpha$

so we have

$T = \frac{1}{2}MR^2(\frac{a}{R^2})$

$T = \frac{1}{2}Ma$

now we have

$mg = (\frac{M}{2} + m)a$

$a = \frac{mg}{(\frac{M}{2} + m)}$

$a = \frac{15 \times 9.81}{(6 + 15)}$

$a = 7 m/s^2$

now we have

$T = \frac{12 \times 7}{2}$

$T = 42 N$

Part b)

speed of the bucket can be found using kinematics

so we have

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

$v_f^2 - 0 = 2(7)(10)$

$v_f = 11.8 m/s$

Part c)

now in order to find the time of fall we can use another equation

$v_f - v_i = at$

$11.8 - 0 = 7 t$

$t = 1.7 s$

Part d)

as we know that cylinder is at rest and not moving downwards

so here we can use force balance

$F = T + Mg$

$F = 42 + (12 \times 9.81)$

$F = 159.7 N$

5 0

3 years ago

A car moving in a straight line starts at X=0 at t=0. It passesthe point x=25.0 m with a speed of 11.0 m/s at t=3.0 s. It passes

Agata [3.3K]

Answer:

Average velocity v = 21.18 m/s

Average acceleration a = 2 m/s^2

Explanation:

Average speed equals the total distance travelled divided by the total time taken.

Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)

Average acceleration equals the change in velocity divided by change in time.

Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)

Where;

v1 and v2 are velocities at time t1 and t2 respectively.

And x1 and x2 are positions at time t1 and t2 respectively.

Given;

t1 = 3.0s

t2 = 20.0s

v1 = 11 m/s

v2 = 45 m/s

x1 = 25 m

x2 = 385 m

Substituting the values;

Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)

v = (385-25)/(20-3)

v = 21.18 m/s

Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)

a = (45-11)/(20-3)

a = 2 m/s^2

8 0

3 years ago

What is the relationship between the strength of a bond (single vs double vs triple) and its wave-number on an IR spectrum? As t

Elanso [62]

<u>Explanation</u>

The relationship between the strength of a bond (single vs double vs triple) and its wave-number on an IR spectrum as the bond strength increases the wave number increases.

STRENGTH OF BONDS TRIPLE>DOUBLE>SINGLE

WAVE NUMBER SINGLE>DOUBLE>TRIPLE

wave number for single bond is greatest because it has greatest bond frequency among the three( more the frequency greater is the wave number).

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3 years ago