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OLga [1]
2 years ago
9

A 2300 kg truck has put its front bumper against the rear bumper of a 2500 kg suv to give it a push. with the engine at full pow

er and good tires on good pavement, the maximum forward force on the truck is 18,000 n.
Physics
2 answers:
devlian [24]2 years ago
8 0

Answer:

Acceleration = a = 3.75 m/s^2

Explanation:

Mass of truck1 = m1 = 2300 kg

Mass of truck2 = m2 = 2500 kg  

Total mass = m = m1 + m2 = 2300 + 2500 = 4800 kg

Force exerted by truck1 = F = 18000 N

As both trucks are joint together so, behaving as single object. The acceleration can be found by Newton’s second law of motion.

F = ma  

a = F/m = 18000/4800

a = 3.75 m/s^2

valentina_108 [34]2 years ago
7 0
F=ma
m=total mass = 2300kg+2500kg=4800
F=18000N
a=?
a=F/m
a=18000/4800
a=3.8m/s^2
Final answer
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At one point in the rescue operation, breakdown vehicle A is exerting a force of 4000 N and breakdown vehicle B is exerting a fo
lukranit [14]

Answer:

1.) Magnitude = 5596 N

2.) Direction = 60 degrees

Explanation: You are given that the breakdown vehicle A is exerting a force of 4000 N at angle 45 degree to the vertical and breakdown vehicle B is exerting a force of 2000 N

Let us resolve the two forces into X and Y component

Sum of the forces in the X - component will be 4000 × cos 45 = 2828.43 N

Sum of the forces in the Y - component will be 2000 + ( 4000 × sin 45 )

= 2000 + 2828.43

= 4828.43 N

The resultant force R will be

R = sqrt ( X^2 + Y^2 )

Substitutes the forces at X component and Y component into the formula

R = sqrt ( 2828.43^2 + 4828.43^2 )

R = sqrt ( 31313752.53 )

R = 5595.87 N

The direction will be

Tan Ø = Y/X

Substitute Y and X into the formula

Tan Ø = 4828.43 / 2828.43

Tan Ø = 1.707106

Ø = tan^-1( 1.707106 )

Ø = 59.64 degree

Therefore, approximately, the magnitude and direction of the resultant force on the truck are 5596 N and 60 degree respectively.

8 0
2 years ago
You are asked to design a spring that will give a 1160-kg satellite a speed of 2.50 m&gt;s relative to an orbiting space shuttle
Tju [1.3M]
Given:
m = 1160 kg
g = 9,81 m/s²
v = 2.5 m/s

Unknown:
k spring constant
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F = ma = m(5g) = |kx|
2. equation to use, the energy of the spring must equal the energy of the satellite:
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 Combinig the two equations:
\frac{1}{2} mv^2 =  \frac{1}{2}  m(5g)x \\ v^2 = 5gx \\ x =  \frac{v^2}{5g}  \\  \\ k =  \frac{m(5g)}{x} =  \frac{m(5g)^2}{v^2}

7 0
2 years ago
A 3-cm high object is in front of a thin lens. The object distance is 4 cm and the image distance is –8 cm. (a) What is the foca
xenn [34]

Answer:

a) Focal length of the lens is 8 cm which is a convex lens

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Explanation:

u = Object distance =  4 cm

v = Image distance = -8 cm

f = Focal length

Lens Equation

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c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.

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