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likoan [24]
3 years ago
12

4. A 75 kg bobsled is pushed along a horizontal surface by two athletes. After the

Physics
1 answer:
vlada-n [284]3 years ago
8 0

The net force on the sled is 300 N

Explanation:

First of all, we start by finding the acceleration of the bobsled, by using the suvat equation:

v^2-u^2=2as

where:

v = 6.0 m/s is the final velocity of the sled

u = 0 is the initial velocity

a is the acceleration

s = 4.5 m is the displacement of the sled

Solving for a, we find

a=\frac{v^2-u^2}{2s}=\frac{6.0^2-0}{2(4.5)}=4 m/s^2

Now we can find the net force on the sled by using Newton's second law:

F = ma

where

F is the net force

m = 75 kg is the mass of the sled

a=4 m/s^2 is the acceleration

Solving the equation, we find the net force:

F=(75)(4)=300 N

Learn more about acceleration and Newton laws here:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

#LearnwithBrainly

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pashok25 [27]

Answer:

22.11 m / s

Explanation:

The falcon catches the prey from behind means both are flying in the same direction ( suppose towards the left )

initial velocity of falcon = 28 cos 35 i - 28 sin 35 j  

( falcon was flying in south east direction making 35 degree from the east )

momentum = .9 ( 28 cos 35 i - 28 sin 35 j  )

= 20.64 i - 14.45 j

initial velocity of pigeon

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initial momentum = .325 x 7i

= 2.275 i

If final velocity of composite mass of falcon and pigeon be V

Applying law of conservation of momentum

( .9 + .325) V = 20.64 i - 14.45 j +2.275 i

V = ( 22.915 i - 14.45 j ) / 1.225

= 18.70 i - 11.8 j

magnitude of V

= √ [  (18.7 )² + ( 11.8 )²]

= 22.11 m / s

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Which organization functions to privide clinics and surgeries i developing nations
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Doctors without boarders.
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In the following atomic model, where does the strong nuclear force happen? mc006-1.jpg outside A between A and B between B and C
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The answer is the third option
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3 years ago
The free body diagram shows a box being pulled to the left up a 25-degree incline
Greeley [361]
The question is incomplete but still I answer to assume your thinking.
The picture is attached below!.
Here,
F is the force with which you pull up the incline.
N is the normal force.
w is the weight acting downward.
Axis are mentioned in the attached picture.
Concept:
You can see there is no movement of object in the y-direction that means acceleration is zero in y-direction, sum of all the forces in y-direction equal to zero.
According to newton second law,
<span>∑ F = ma
</span>As, acceleration is zero in y-direction, so right hand side is zero in the above equation.
<span>∑ F = 0</span>
N-wcosθ=0 
N= m*g*cos25°
N= m*(9.8)*(0.9063)
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Hopefully, this is the answer of your question.


5 0
4 years ago
Read 2 more answers
1. Calcula la fuerza de atracción electrostática entre dos cuerpos de cargas q1 = -18 C y q2 = +5 mC, separados entre sí por una
romanna [79]

Answer:

A) F=-20.16×10⁹N

B) if the distance doubles, force is 4 times smaller.

Explanation:

q1=-28C

q2=5mC=0.005C

d=25cm=0.25m

Electrostatic force between charges: F=k×q1×q2/d², where k is a coefficient that has the value k=9 × 10⁹ N⋅m²⋅C^(-2) for air.

Thus:

F=9×10⁹×(-28)×0.005/0.25²

F=-20.16×10⁹N

The minus sign indicates attraction.

If distance doubles, d1=2×d, then we have 4d² at the denominator and the force is 4 times smaller.

6 0
3 years ago
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