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Andreas93 [3]
3 years ago
14

And 8 kg bowling ball is rolling along the frictionless alley

Physics
1 answer:
VLD [36.1K]3 years ago
8 0
It will stop eventually
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Pulling up on a rope, you lift a 4.25 kg bucket of water from a well with an acceleration of 1.80 m/s2 . What is the tension in
sattari [20]

Answer:

49.3 N

Explanation:

Given that Pulling up on a rope, you lift a 4.25 kg bucket of water from a well with an acceleration of 1.80 m/s2 . What is the tension in the rope?

The weight of the bucket of water = mg.

Weight = 4.25 × 9.8

Weight = 41.65 N

The tension and the weight will be opposite in direction.

Total force = ma

T - mg = ma

Make tension T the subject of formula

T = ma + mg

T = m ( a + g )

Substitutes all the parameters into the formula

T = 4.25 ( 1.8 + 9.8 )

T = 4.25 ( 11.6 )

T = 49.3 N

Therefore, the tension in the rope is 49.3 N approximately.

8 0
2 years ago
Determine the slit width that produces a diffraction pattern with the 2nd dark fringe at 6.2mm from the central fringe. The scre
Elanso [62]

Answer:

d= 0.242 mm

Explanation:

Slit width (d ) = ?

Screen distance ( D ) = 1.25 m

Wave length of light λ = 600 nm

Distance of n the dark fringe from centre

= n λ D / d

Here n = 2

so

6.2\times10^{-3}=\frac{2\times600\times10^{-9}}{d}

d=\frac{1500\times10^{-6}}{6.2}

d= 0.242 mm

4 0
3 years ago
A substance that Connor be broken down into simplersubstance by ordinary chemical means
Nana76 [90]

Answer:

An element

Explanation:

8 0
2 years ago
Please answer I will give good rating and mark brainlest
lina2011 [118]
The answer is C. You must divide your wavelength and your frequency to get your answer.
8 0
3 years ago
(a) what is the acceleration of two falling sky divers (mass 132 kg including parachute) when the upward force of air resistance
Komok [63]

As per the question the mass of two falling sky drivers is 132 kg.

First we have to calculate their acceleration.

Whenever a body falls freely under gravity,its acceleration is acceleration due to gravity i.e g whose value is 9.8 m/s^2.

The earth pulls the object with a force equal to the weight of the body.

Hence the force gravity  F=W= mg   [ here m is mass of the body]

Here m =132 kg.

Hence force of gravity F= mg

                                        =132 kg ×9.8 m/s^2

                                        =1293.6 kg m/s^2

                                         =1293.6 N      [ here N[newton] is the unit of force.]

As per the question the air resistance is one fourth of weight of the bodies.

Hence air resistance F' =1/4 mg

                                       =\frac{1}{4} *1293.6N

                                        =323.4 N

Here F acts in vertically downward direction while F' acts in vertically upward direction.

Hence the net force acting on the particle is F-F'.

                                                 F_{net} =1293.6N -323.4N

                                                         =970.2 N

From Newton's second law of motion we know that net force is the product of mass and acceleration i.e  

                                   F_{net} =ma  [Here a is the acceleration]

                                             a =\frac{F_{net} }{m}

                                                  = \frac{970.2}{132} m/s^2

                                                  =7.35 m/s^2

In the second question it has been told that they descend with uniform speed.hence acceleration of the two bodies will be zero.

 we know that F= ma

                           =m×0

                            =0 N

Hence they will not get any force when they will descend with a uniform speed.


4 0
2 years ago
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