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natta225 [31]
3 years ago
12

a custodian pulls a vacuum 13.5 m with a 33.9 N force at a 55 degree angle, against a 14.1 N friction force. Find the work done

by the custodian.
Physics
1 answer:
pshichka [43]3 years ago
5 0

Answer:

The answer is 190.35

Explanation:

Given values be put in the formula that is

W=F×D

W=14.1×13.5

W=190.35

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The gyri and sulci on the brain greatly increase the brain’s surface area. How do you think this increase in surface area relate
MrRa [10]

Answer:

Increase in surface area leads to more chances for connections between neurons hence more possible learning

Explanation:

The brain appears wrinkled due to the presence of gyri and sulci. Sulci are the grooves in the brain while gyri are the folds. The two major functions of gyri and sulci are to lead to formation of brain divisions and increase the surface area of cerebral cortex. The formation of brain divisions is done by creating boundaries between the brain lobes thereby leading to division of the brain into two hemispheres. Moreover, by increasing the surface area of the brain, the cortex processes more information since more neurons are formed.

8 0
4 years ago
Which scientist was the first to use the telescope in astronomy
Mashutka [201]

Answer:

Galileo Galilei

Explanation:

although Galileo was not the scientist who invented the telescope, he was the first to use it to observe celestial objects. he used the telescope in 1609. his discovery included more accurate information about the moon, the sun and some of the planets.

4 0
4 years ago
A 2.00 kg, frictionless block s attached to an ideal spring with force constant 550 N/m. At t = 0 the spring is neither stretche
gladu [14]

Answer:

a)    A = 0.603 m , b) a = 165.8 m / s² , c)  F = 331.7 N

Explanation:

For this exercise we use the law of conservation of energy

Starting point before touching the spring

    Em₀ = K = ½ m v²

End Point with fully compressed spring

    Em_{f} = K_{e} = ½ k x²

    Emo = Em_{f}

    ½ m v² = ½ k x²

    x = √(m / k)    v

    x = √ (2.00 / 550)   10.0

    x = 0.603 m

This is the maximum compression corresponding to the range of motion

     A = 0.603 m

b) Let's write Newton's second law at the point of maximum compression

    F = m a

    k x = ma

    a = k / m x

    a = 550 / 2.00 0.603

    a = 165.8 m / s²

With direction to the right (positive)

c) The value of the elastic force, let's calculate

    F = k x

    F = 550 0.603

   F = 331.65 N

5 0
3 years ago
PLEASE HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP
frozen [14]
A. Since it does not say the ball is moving your answer is A.

4 0
3 years ago
A 1350 kg uniform boom is supported by a cable. The length of the boom is l. The cable is connected 1/4 the
olchik [2.2K]

Answer:

Tension= 21,900N

Components of Normal force

Fnx= 17900N

Fny= 22700N

FN= 28900N

Explanation:

Tension in the cable is calculated by:

Etorque= -FBcostheta(1/2L)+FT(3/4L)-FWcostheta(L)= I&=0 static equilibrium

FTorque(3/4L)= FBcostheta(1/2L)+ FWcostheta(L)

Ftorque=(Fcostheta(1/2L)+FWcosL)/(3/4L)

Ftorque= 2/3FBcostheta+ 4/3FWcostheta

Ftorque=2/3(1350)(9.81)cos55° + 2/3(2250)(9.81)cos 55°

Ftorque= 21900N

b) components of Normal force

Efx=FNx-FTcos(90-theta)=0 static equilibrium

Fnx=21900cos(90-55)=17900N

Fy=FNy+ FTsin(90-theta)-FB-FW=0

FNy= -FTsin(90-55)+FB+FW

FNy= -21900sin(35)+(1350+2250)×9.81=22700N

The Normal force

FN=sqrt(17900^2+22700^2)

FN= 28.900N

4 0
4 years ago
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