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yulyashka [42]
3 years ago
15

20 Point Offer for Full Answer.

Physics
1 answer:
Fantom [35]3 years ago
4 0
I think it is number 4 because he is putting force on the chair but the chair is not exerting force on him
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A light plane must reach a speed of 33 m/s per take off. How long a runway is needed if the constant acceleration is 3.0 m/s^2.
aleksley [76]
Given the final velocity (Vf) and the acceleration (a), the distance that should be traveled by the plane is calculated through the equation,
                            d = (Vf² - Vi²) / 2a
V1 should be zero because the light plane started the motion from rest. Substituting the given values,
                          d = ((33 m/s)² - 0)) / 2(3 m/s²)
The distance is therefore equal to 181.5 meters. 
3 0
3 years ago
Compare and contrast the CONFLICT (choose one) in the short story you read with the elements appearing in The Watsons Go to Birm
forsale [732]

Answer:

they were in two places in flint and Birmingham and in Birmingham it is hot but flint of cold the Simi is they both have Sunday school for Joetta

Explanation:

use in your own words teachers know when your not trust me.

4 0
3 years ago
A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s
Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g=4.67\times 10^{-3} kg

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1,m_1=1177g=1.177kg

Mass of block 2,m_2=1626 g=1.626 kg

Velocity of block 1,v_1=0.681m/s

(a)

Let velocity of the second block  after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

mv=m_1v_1+(m+m_2)v_2

4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2

1.66719=0.801537+1.63067v_2

1.66719-0.801537=1.63067v_2

0.865653=1.63067v_2

v_2=\frac{0.865653}{1.63067}

v_2=0.531m/s

Hence, the  velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

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

k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)

k_i=297.59 J

Final kinetic energy after collision

K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2

K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2

K_f=0.5028 J

Now, he ratio of the total kinetic energy after the collision to that before the collision

=\frac{k_f}{k_i}=\frac{0.5028}{297.59}

=0.00169

5 0
3 years ago
What is the volume of a bar of soap that is 9 cm long, 5 cm wide, and 2 cm high?
GalinKa [24]

Answer:

90cm

Explanation:

2x5=10

10x9=90

7 0
3 years ago
Read 2 more answers
Two parakeets sit on a swing with their combined center of mass 10.0 cm below the pivot. At what frequency do they swing?
Oksanka [162]

Answer:

1.58 Hz

Explanation:

The frequency of the simple pendulum is given by

f = 1/T

 = 1/2π√g/l  

In this problem, I = 10.0 cm = 0.1 m  

f = 1/2π√9.8/0.1

=  1.58 Hz

7 0
4 years ago
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