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likoan [24]
4 years ago
5

Two common terms for a decrease in velocity are

Physics
2 answers:
Colt1911 [192]4 years ago
6 0

deceleration or rėtardation i’m pretty sure (it won’t let me say the second word but it’s correct)

polet [3.4K]4 years ago
5 0

-- negative acceleration

-- deceleration

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A student lifts a box of books 2 meters with a force of 45 N. He then carries the box 10 meters to the living room. What is the
Alisiya [41]

Answer:

90J

Explanation:

The only time work is being done is when he lifts the box off the ground. Therefore, using the work formula, 2 x 45, you get 90J. Hope this helps someone.

5 0
3 years ago
Given a double slit apparatus with slit distance 2 mm, what is the theoretical maximum number of bright spots that I would see w
katen-ka-za [31]

Answer:

The values is  m_{max} = 8001 \  bright \ spots

Explanation:

From the question we are told that

    The slit distance is  d  =  2 \ mm  =  2*10^{-3} \ m

    The  wavelength is  \lambda =  500 \ nm =  500 *10^{-9} \ m

At the first half of the screen from the central maxima

   The number of bright spot according to the condition for constructive interference is  

          n  =  \frac{d *  sin (\theta )}{\lambda}

For maximum number of spot \theta =  90^o

So  

       n  =  \frac{2*10^{-3} *  sin (90 )}{500 *10^{-9}}

        n  =4000

Now for the both sides plus the central maxima  we have

      m_{max} = 2 * n  + 1

substituting values

       m_{max} = 2 *  4000 + 1

       m_{max} = 8001 \  bright \ spots

   

6 0
4 years ago
A block of mass M slides down an inclined plane that makes an angle θ with the horizontal. The coefficient of kinetic friction b
Llana [10]

Answer:

Mg Cosθ

Explanation:

mass of block = M

Angle of inclination = θ

coefficient of friction = μk

The force of gravity acts of the block is Mg

there are two components of the weight

the component parallel to the inclined plane is Mg Sinθ

the component perpendicular to the plane of inclined is Mg Cosθ

So, the normal force exerted by the inclined is Mg Cosθ

5 0
4 years ago
A stationary 0.750kg ball is thrown up by doing 2.50J of work on it. What is the velocity of the ball?
Lena [83]

When a ball is thrown up by doing work, the velocity of the ball will be 2.6 m/s.

<h3>What is Work energy theorem?</h3>

It states that the Work done in moving a body is equal to the change in kinetic energy of the body

Kinetic energy = 1/2 mv²

Given is a ball of mass m = 0.750 kg and the work done on ball W = 2.50 J

The ball is initially at rest. So, initial velocity is zero. Then, change in kinetic energy will be

W= ΔK.E = K.Ef - K.Ei

According to work energy theorem, work done is

W = 2.5J  = 1/2 x 0.750 x (v)² -0

v =2.6 m/s

Thus, the velocity of the ball is 2.6 m/s

Learn more about work energy theorem.

brainly.com/question/10063455

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6 0
2 years ago
The radius of a lead atom is 175 pm. how many lead atoms would have to be laid side by side to span a distance of 6.11 mm?
podryga [215]

The number of atoms needed to be laid side by side to span a distance of 6.11 mm is 17,457,143.

<h3>What is Unit conversion?</h3>

Unit conversion is a way of converting some common units into another without changing their real value. for, example, 1 centimeter is equal to 10 mm, though the real measurement is still the same the units and numerical values have been changed.

Given that the radius of a lead atom is 175 picometers. Therefore, the radius of the lead atom in mm can be written as,

1 picometer = 10⁻⁹ millimeter

175 picometer = 175×10⁻⁹ millimeter

The radius of the lead atom = 175×10⁻⁹ millimeter

Since the atoms are needed to be arranged side by side, therefore, the diameter of one lead atom will join with the other, thus, the diameter of the lead atom is,

Diameter of lead atom = 2 × 175 × 10⁻⁹ millimeter = 350 ×10⁻⁹ millimeter

Now, let the number of lead atoms be represented by n. Therefore, the number of atoms needed to be laid side by side to span a distance of 6.11 mm is,

6.11 mm = Number of atoms × Diameter of atoms

6.11 mm = n × 350 ×10⁻⁹ mm

Solving for n,

n = 6.11 mm / 350 ×10⁻⁹ mm

n = 17,457,142.86 ≈ 17,457,143

Hence, the number of atoms needed to be laid side by side to span a distance of 6.11 mm is 17,457,143.

Learn more about Units conversion here:

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7 0
2 years ago
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