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3 years ago

15

20 Point Offer for Full Answer.

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