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NeTakaya
3 years ago
9

An arrow, starting from rest, leaves the bow with a speed of25

Physics
1 answer:
xz_007 [3.2K]3 years ago
5 0

Answer:

  • The speed will be v_f= 43.30 \frac{m}{s}

Explanation:

We can use the following kinematics equation

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

where v_f is the final speed, v_i its the initial speed, a is the acceleration, and d the distance.

The force will be tripled, the force is:

\vec{F} = m \vec{a}

in 1D

F = m a

Now, for the original problem, we have

(25 \frac{m}{s})^2-(0 \frac{m}{s})^2=2 \ a' \ d'

(25 \frac{m}{s}))^2=2 \ a' \ d'

625 \frac{m^2}{s^2}=2 \ a' \ d'

For the second problem, we have

(v_f})^2-(v_i)^2=2 \ a'' \ d''

starting from the rest, we have the same initial velocity.

(v_f})^2-(0\frac{m}{s})^2=2 \ a'' \ d''

(v_f})^2=2 \ a'' \ d''

As the force is tripled, we have:

F'' = 3 F'

m'' \ a'' = 3 m' \ a'

But the mass its the same,  so

m' \ a'' = 3 m' \ a'

a'' = 3 \ a'

So the acceleration its also tripled.

(v_f})^2=2 \ (3 * a') \ d''

(v_f})^2=3 ( 2 \ ( a' \ d'' )

As the distance traveled by the arrow must also be the same, we have:

(v_f})^2=3 ( 2 \ ( a' \ d' )

(v_f})^2=3 (625 \frac{m^2}{s^2})

v_f= \ sqrt{3 (625 \frac{m^2}{s^2})}

v_f= \ sqrt{3} 25 \frac{m}{s}

v_f= 43.30 \frac{m}{s}

And this will be the speed from the arrow leaving the bow.

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In traveling a distance of 2.5 km between points A and D, a car is driven at 99 km/h from A to B for t seconds and 48 km/h from
Andreyy89

Answer:

d = 1.954 Km

Explanation:

given,

total distance, D = 2.5 Km

in stretch A to B =

speed = 99 Km/h = 99 x 0.278 = 27.22 m/s     time =t

in stretch B to C

time = 3.4 s

In stretch C to D

speed = 48 Km/h = 48 x 0.278 = 13.34 m/s     time =t      

we know,

distance = speed x time

distance of BC

using equation of motion

v = u + a t

27.22 = 13.34 - a x 3.4

a = 4.08 m/s²

uniform deceleration is equal to 4.08 m/s²

distance traveled in BC

s = ut + \dfrac{1}{2}at^2

s = 13.34\times 3.4 + \dfrac{1}{2}\times 4.08 \times 3.4^2

s = 68.94 m

3000 = 99 \times \dfrac{1000\ t}{3600}+ 68.94 + 48\times \dfrac{1000\ t}{3600}

3000 = 27.5 t + 68.94 + 13.33 t

40.83 t = 2931.06

t = 71.79 s

distance travel in AB

distance = s x t

d = 27.22 x 71.79

d = 1954 m

d = 1.954 Km

distance between A and B is equal to 1.954 Km.

4 0
3 years ago
An electric drill transfers 200 J of energy into a useful kinetic energy store. It also transfers 44 J of energy by sound and 48
kramer

Answer:

Er = 108 [J]

Explanation:

To solve this problem we must understand that the total energy is 200 [J]. Of this energy 44 [J] are lost in sound and 48 [J] are lost in heat. In such a way that these energy values must be subtracted from the total of the kinetic energy.

200 - 44 - 48 = Er

Where:

Er = remaining energy [J]

Er = 108 [J]

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3 years ago
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Mumz [18]
Acc= vf-vi/time=10-0/20= 0.5 m/s/s
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3 years ago
If a basketball travels a distance of 4 meters in 5 seconds, what is its average speed?
aliya0001 [1]

Average speed = (distance covered) / (time to cover the distance)


Average speed = (4 meters) / (5 seconds)


Average speed = (4/5) (meters/seconds)


Average speed = 0.8 m/s

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3 years ago
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Answer:

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Explanation:

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