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oee [108]
3 years ago
8

A person tosses a ball from the ground up into the air at an initial speed of 10 m/sec and an initial angle of 43◦ off the groun

d. after the ball is released, what is the total acceleration vector acting on the ball when the ball is at the top of its arc? 1. 9.8 m/s2 , down 2. none of these 3. 9.8 m/s2 , up 4. zero 5. 9.8 m/s2 , in the horizontal direction
Physics
1 answer:
Feliz [49]3 years ago
7 0
Just try your best best friend everyone
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In a typical badminton swing the racket is in contact with the birdy for about 0.0010 seconds. If the 0.045kg birdy acquires a s
zepelin [54]

Answer:

3015 N

Explanation:

From Newton's second law, we know that;

F.t = mv

F = force on the ball= ?

m= mass of the ball

v= velocity

F= mv/t

F= 0.045 × 67/0.0010

F= 3.015/0.0010

F= 3015 N

3 0
3 years ago
Throughout the problem, take the speed of sound in air to be 343 m/s . Part A Consider a pipe of length 80.0 cm open at both end
White raven [17]

Answer:

the lowest frequency f of the sound wave is 214.375 Hz

Explanation:

The computation of the lowest frequency f of the sound wave is shown below;

Length = L= 80 cm

= 0.8 m

V = 343 m/s (sound speed in air )

Now

V1 = n V ÷ 2 L

= 1 × 343 ÷ 2 × 0.8

V1 = 214.375 Hz

Hence, the lowest frequency f of the sound wave is 214.375 Hz

We simply applied the above formula so that the correct value could come

And, the same is to be considered  

6 0
3 years ago
Particles of atoms
vredina [299]
Answer:

Protons , neutrons and elections
3 0
3 years ago
A person is spinning an object around on a circular path on the end of a string of length 0.96 m. The object has a mass of 0.34
Rama09 [41]
I’m so sorry I needed points but I hope you get it right
8 0
3 years ago
A bullet of mass 6.20 10-3 kg, moving at 1320 m/s impacts a tree stump and penetrates 11.00 cm into the wood before coming to re
sladkih [1.3K]

Answer:

  F = -49.1   10³ N

Explanation:

Let's use the kinematics to find the acceleration the acceleration of the bullet that they tell us is constant

   v_{f}² = v₀² + 2 a x

Since the bullet is at rest, the final speed is zero

    x = 11.00 cm (1 m / 100 cm) = 0.110 m

    0 = v₀² + 2 a x

   a = -v₀² / 2 x

   a = -1320²/(2 0.110)

   a = -7.92 10⁶ m / s²

With Newton's second law we find the force

   F = m a

   F = 6.20 10⁻³ (-7.92 10⁶)

   F = -49.1   10³ N

The sign means that it is the force that the tree exerts to stop the   bullet

8 0
3 years ago
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