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yawa3891 [41]
1 year ago
15

DeWayne meets a potential employer at a restaurant, who gives him a phone number. Because he doesn't have anything to write with

, he repeats the number several times to himself as a way of remembering the phone number. He is not very likely to remember this phone number later in the evening because of this strategy. True False
Physics
1 answer:
Ede4ka [16]1 year ago
5 0

Based on the fact that he didn't write the number down, the statement  "He is not very likely to remember this phone number later in the evening because of this strategy. " is True

This is further explained below.

<h3>What is DeWayne's strategy?</h3>

Generally, While DeWayne's strategy is eating at a restaurant, he meets a potential employer who provides him with his phone number.

Because he does not have anything with which to write, he must recite the telephone number to himself many times in order to commit it to memory.

In conclusion,  It is quite improbable that he will remember this phone number later on in the evening if you use this strategy.

Read more about  DeWayne's strategy

brainly.com/question/15263096

#SPJ1

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A boy sleds down a hill and onto a frictionless ice- covered lake at 10.0 m/s. In the middle of the lake is a 1000-kg boulder. W
mina [271]

Answer:

The speed of the sled is 9.2 m/s

The speed of the boulder is 0.82 m/s

Solution:

As per the question:

Mass of the boulder, m_{B} = 1000\ kg

Mass of the sled, m_{S} = 2.50\ kg

Mass of the boy, m_{b} = 40\ kg

Initial Velocity, v = 10.0 m/s

Now,

To calculate the speed of both the sled and the boulder after the occurrence of the collision:

m = m_{b} + m_{S} = 40 + 2.50 = 42.50\ kg

Initial velocity of the boulder, v_{B} = 0\ m/s

Since, the collision is elastic, both the energy and momentum rem,ain conserved.

Now,

Using the conservation of momentum:

mv + m_{B}v_{B} = mv' + m_{B}v'_{B}

where

v' = final velocity of the the system of boy and sled

v'_{B} = final velocity of the boulder

42.50\times 10 + m_{B}.0 = 42.50v' + 1000v'_{B}

42.50v' + 1000v'_{B} = 425            (1)

Now,

Using conservation of energy:

\frac{1}{2}mv^{2} + \frac{1}{2}m_{B}v_{B}^{2} = \frac{1}{2}mv'^{2} + \frac{1}{2}m_{B}v'_{B}^{2}

42.50\times 10^{2} + m_{B}.0 = 42.50v'^{2} + 1000v'_{B}^{2}

42.50v'^{2} + 1000v'_{B}^{2} = 4250         (2)

Now, from  eqn (1) and (2):

v' = \frac{m - m_{B}}{m + m_{B}}\times v

v' = \frac{42.50 - 1000}{42.5 + 1000}\times 10 = - 9.2\ m/s

Now,

v'_{B} = \frac{2m}{m + m_{B}}\times v

v'_{B} = \frac{2\times 42.50}{42.5 + 1000}\times 10 = 0.82\ m/s

5 0
3 years ago
Read 2 more answers
1. A 12 kg barrel is pulled up by a rope. The barrel accelerates at 1.2 m/s2. Find the force exerted by the rope. Show all your
allochka39001 [22]
F-mg=ma
F-(12kg)(9.8m/s^2)=(12kg)(1.2m/s^2)
F-117.6N=14.4N
F=132 Newtons 
5 0
4 years ago
Assuming the starting height is 0.0 m, calculate the potential energy of the cart after it has been elevated to a height of 0.5
Bogdan [553]
The potential energy is most often referred to as the "energy at rest" and is dependent on the elevation of an object. This can be calculated through the equation,

     E = mgh

where E is the potential energy, m is the mass, g is the acceleration due to gravity, and h is the height. In this item, we are not given with the mass of the cart so we assume it to be m. The force is therefore,

   E = m(9.8 m/s²)(0.5 m) = 4.9m

Hence, the potential energy is equal to 4.9m.
8 0
4 years ago
Peter and Fred are skateboarding in a large semicircular halfpipe. Peter starts out from rest at a height h and collides with Fr
jolli1 [7]

Answer:

The answer is h/4

Explanation:

When Peter collides with Fred, the collision is inelastic & they both proceed with a velocity of V/2.

let m represent the masses for Peter & Fred

v represent the initial velocity of Peter

V represent final velocity of both of them

mv + 3m × 0 = (m+m)V

V = v/2

Using the expression; H = v² / 2g .............Eqn 1

Upon substitution of V/2 into Eqn 1 above,

H = (V/2)² / 2g

H = (V²/4) ÷ 2g

Therefore height will be h/4

6 0
3 years ago
(4A) The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km.
faltersainse [42]

Answer:

x₁ = 345100 km

Explanation:

The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and  (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

Let  call "x₁"  distance between center of the earth and the object, and

"x₂" the distance between center of the moon and the object, Mt mass of the earth, Ml mass of the moon, m₀ mass of the object

we can express:

F₁  ( force between earth and the object )

F₁ = K *  Mt * m₀/ ( x₁)²        K is a gravitational constant

F₂  (force between mn and the object)

F₂ = K * Ml * m₀ / (x₂)²

Then:

F₁ = F₂               K*Mt*m₀ / x₁²   =  K*Ml*m₀ /x₂²

Or  simplifying the expression

Mt/ x₁²  =  Ml/ x₂²

We know that   x₁   +  x₂  = 384000 Km then

x₁ =  384000 - x₂

Mt/( 384000 - x₂)²  =  Ml / x₂²

Mt *  x₂²  =  Ml *( 384000 - x₂)²

We need to solve for x₂

Mt *  x₂²  =  Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

                                7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂²  = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂²  = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂²  + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂  =  -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  =  -56.43*10∧5 + √3184*10∧10 + 254880*10∧10  / 1160

x₂ ₁  = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁  =  -56.43*10∧5 + 508* 10∧5  / 1160

x₂ ₁  =  451.27*10∧5/1160

x₂ ₁  =  4512.7*10∧4 /1160

x₂ ₁  = 3.89*10∧4  km (distance between the moon  and the object)

x₂ ₁  = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁  = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

5 0
3 years ago
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