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Tatiana [17]
3 years ago
8

Three men are pulling on a tree attached with ropes to keep it from falling down. If one man is pulling the rope with a force of

370N acting at 30 degrees east of north and the second man is pulling with a force of 280N acting at 70 degrees west of south. What force should the third man pull with and in what direction? PLEASE HELP AND EXPLAIN THANK YOU SO MUCH
Physics
1 answer:
slamgirl [31]3 years ago
3 0
The correct answer to this question is <span>48.8N</span><span> on a bearing of </span><span>134.2 degrees.

In order to solve this, first we need to find the resultant force of the men pulling in the north and south directions:

F = 40 - 6 = 34N [due south - 180]

Then, find the resultant force.

R^2 =34^2 + 35^2 
R^2 = 2381
R = sqrt(2381)
R = 44.8N

The angle from the vertical is given by:
tan(x) = 35/34
tan(x) = 1.0294
angle = 45.8 degrees

Taking N as zero degrees, then it bears 134.2 degrees.


</span>
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Explanation:

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F = ma

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The drag on a pitched baseball can be surprisingly large. Suppose a 145 g baseball with a diameter of 7.4 cm has an initial spee
kupik [55]

Answer:

<h2>Part A)</h2><h2>Acceleration of the ball is 10.1 m/s/s</h2><h2>Part B)</h2><h2>the final speed of the ball is given as</h2><h2>v_f = 35.3 m/s</h2>

Explanation:

Part a)

As we know that drag force is given as

F = \frac{C_d \rho A v^2}{2}

C_d = 0.35

A = \frac{\pi d^2}{4}

A = \frac{\pi(0.074)^2}{4}

A = 4.3 \times 10^{-3} m^2

v = 40.2 m/s

so we have

F = \frac{0.35\times 1.2 (4.3 \times 10^{-3})(40.2)^2}{2}

F = 1.46 N

So acceleration of the ball is

a = \frac{F}{m}

a = \frac{1.46}{0.145}

a = 10.1 m/s^2

Part B)

As per kinematics we know that

v_f^2 - v_i^2 = 2 a d

v_f^2 - 40.2^2 = 2(-10.1)(18.4)

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4 0
3 years ago
Am i correct? If not then which one
cupoosta [38]

Answer:

Yes, it's correct

Explanation:

Newton's second Law states that the acceleration of an object is proportional to the net force applied on it, according to the equation:

F=ma

where

F is the net force on the object

m is the mass of the object

a is the acceleration of the object

We can re-arrange the previous equation in order to solve explicitely for a, the acceleration, and we find:

F=ma\\\frac{F}{m}=\frac{ma}{m}\\\frac{F}{m}=a\\a=\frac{F}{m}

So, we see that the acceleration is proportional to the net force and inversely proportional to the mass of the object.

4 0
3 years ago
URGENT
Advocard [28]

Answer:

give \\ mass(m) = 1350kg  \\ acceleration(a) = 1ms {}^{ - 2}  \ \\ sln\ \\  from \: our \: formula \\  \: f = ma \\1350kg \times 1ms {}^{ - 2}  \\ f = 1350newton

the force you applied to your car =1350N

5 0
3 years ago
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