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Vilka [71]
3 years ago
6

An archer puts a 0.30 kg arrow to the bowstring. An average force of 201 N is exerted to draw the string back 1.3 m.a. Assuming

that all the energy goes into the arrow, with what speed does the arrow leave the bow? (41.7 m/s)b. If the arrow is shot straight up, how high does it rise? (88.9 m)
Physics
1 answer:
Vlad [161]3 years ago
4 0

Answer:

Explanation:

Given

mass of archer m=0.3\ kg

Average force F_{avg}=201\ N

extension in arrow x=1.3\ m

Work done to stretch the bow with arrow

W=F\cdot x

W=201\times 1.3=261.3\ m

This work done is converted into kinetic Energy of arrow

W=\frac{1}{2}mv^2

where v= velocity of arrow

261.3=\frac{1}{2}\times 0.3\times v^2

v=\sqrt{1742}

v=41.73\ m/s

(b)if arrow is thrown vertically upward then this energy is converted to Potential energy

W=mgh

261.3=0.3\times 9.8\times h

h=\frac{261.3}{0.3\times 9.8}

h=88.87\ m

   

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Gain in decibels is given by;

Gain db = 10*log (Po/Pi), where Po = Power output, Pin = Power input

Substituting;

Gain in db = 10 * log (50/5) = 10 db
6 0
3 years ago
A person throws a ball straight up into the air with a speed of 7.3m/s. How high does the ball go?
Colt1911 [192]

Answer:

2.6645m

Explanation:

applying motion equations we can find the answer,

v^{2}=u^{2}  +2as

Let assume ,

u = starting speed(velocity)

v = Final speed (velocity)

s =  distance traveled

a = acceleration

by the time of reaching the highest point subjected to the gravity , the speed should be equal to zero

we consider the motion upwards , in this case the gravitational acceleration should be negative in upwards (assume g =10 m/s2 downwards)

that is,

v^{2}=u^{2}  +2as\\0=7.3^{2} -2*10*s\\s=2.6645m

4 0
4 years ago
Starting velocity: 50 m/s
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6 0
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Complete the sentence to explain when waves interact.
rewona [7]

Answer:

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7 0
3 years ago
As soon as a traffic light turns green, a car speeds up from rest to its cruising speed 22.3 m/sec with constant acceleration 4.
Ganezh [65]

Answer:  (a) The bicycle is ahead of the car for 4 s.

               (b) The bicycle leads the car by the maximum distance of 55 m.

Explanation:

(a)

Use the equation of the motion to calculate the time taken by the car.

v=u+at  

As it is given in the problem, a car speeds up from rest to its cruising speed 22.3 m/sec with constant acceleration 4.02 m/sec2 .

Put u=0, v=22.3 m/sec and a=4.02 m/sec^2.

22.3=0+4.02t

t=\frac{22.3}{4.02}

t= 5.5 s

Use the equation of the motion to calculate the time taken by the  bicycle.

v=u+at_{1}

As it is given in the problem, a cyclist speeds up from rest to its cruising speed 8.94 m/sec with constant acceleration 5.81 m/sec^2.

Put u=0, v=8.94 m/sec and a=5.81 m/sec^2.

8.94=0+5.81t_{1}

t_{1}=\frac{8.94}{5.81}[tex]t_{1}=1.5 s

Calculate the time interval for which the bicycle is ahead of the car.

t-t_{1}= 5.5 s - 1.5s

t-t_{1}= 4s

Therefore, the bicycle is ahead of the car for 4 s.

(b)

Use the equation motion to calculate the distance covered by the car.

S=ut+\frac{1}{2}at^{2}

As it is given in the problem, a car speeds up from rest to its cruising speed 22.3 m/sec with constant acceleration 4.02 m/sec^2 .

Put t= 5.5 s, u=0 s and a=4.02 m/sec^2.

S=(0)t+\frac{1}{2}(4.02)(5.5)^{2}

S= 60.8 m

Use the equation motion to calculate the distance covered by the bicycle.

S_{1}=ut+\frac{1}{2}at^{2}

As it is given in the problem, a cyclist speeds up from rest to its cruising speed 8.94 m/sec with constant acceleration 5.81 m/sec^2.

Put t= 1.5 s, u=0 s and a=5.81 m/sec^2.

S_{1}=(0)t+\frac{1}{2}(5.18)(1.5)^{2}

S_{1}= 5.8 m

Calculate the maximum distance covered by the bicycle to lead the car.

S-S_{1}=60.8-5.8=55m

Therefore, the bicycle leads the car by the maximum distance of 55 m.

6 0
3 years ago
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