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mafiozo [28]
2 years ago
12

a child riding a bicycle at 15 meters per second accelerates at -3,0 meters per second? for 4.0 seconds. What is the child's spe

ed A at the end of this 4.0-second interval? (with work, all steps))
Physics
1 answer:
marin [14]2 years ago
3 0

Answer:

<h2> 27m/s</h2>

Explanation:

Given data

initital velocity u=15m/s

deceleration a=3m/s^2

time t= 4 seconds

final velocity v= ?

Applying the expression

v=u+at------1

substituting our data into the expression we have

v=15+3*4

v=15+12

v=27m/s

The velocity after 4 seconds is 27m/s

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Two blocks, with masses m and 3m, are attached to the ends of a string with negligible mass that passes over a pulley, as shown
olasank [31]

Answer:

a)  v = √ g x , b)  W = 2 m g d , c)    a = ½ g

Explanation:

a) For this exercise we use Newton's second law, suppose that the block of mass m moves up

            T-W₁ = m a

            W₃ - T = M a

            w₃ - w₁ = (m + M) a

            a = (3m - m) / (m + 3m) g

            a = 2/4 g

            a = ½ g

the speed of the blocks is

          v² = v₀² + 2 ½ g x

          v = √ g x

b) Work is a scalar, therefore an additive quantity

light block s

           W₁ = -W d = - mg d

3m heavy block

             

            W₂ = W d = 3m g d

the total work is

             W = W₁ + W₂

             W = 2 m g d

c) in the center of mass all external forces are applied, they relate it is

                      a = ½ g

8 0
2 years ago
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Roberto creates an animation titled “Flying through an Atom.” The animation makes it seem as if the viewer passes all the way th
vladimir2022 [97]
<span>The viewer travels through the electron cloud, then the nucleus, and then the electron cloud again.</span>
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2 years ago
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thin films of soap sometime display an array of colours. this display is the result of: a) reflection, diffraction, and interfer
kumpel [21]
The array of colors observed when viewing films of soap are caused by the interference, and from the reflections of light from two nearby surfaces (outer surface and inner surface). Among the choices, the closest answer would be B. reflection, refraction, and interference.
7 0
2 years ago
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A certain resistor dissipates 0.5 W when connected to a 3 V potential difference. When connected to a 1 V potential difference,
Stels [109]

Answer:

<h2>0.056 W</h2>

Explanation:

Power = IV

From ohms law we know that

V= IR\\\\I= \frac{V}{R} \\\\Power= \frac{V}{R}*V\\\\Power= \frac{V^2}{R}

Given data

P1 = 0.5 Watt

P2 = ?

V1= 3 Volts

V2= 1 Volt

Thus we can solve for the power dissipated as follows

P1= \frac{V1^2}{R1}\\\\P2= \frac{V2^2}{R2}

\frac{P1}{P2} = \frac{V1^2}{V2^2}\\\\ P2=\frac{ V2^2}{ V1^2} *P1\\\\ P2=\frac{ 1^2}{ 3^2} *0.5= 0.055= 0.056 W

<em>The  resistor will dissipate 0.056 Watt</em>

7 0
2 years ago
If a 20.0 g object at a temperature of 35.0∘C has a specific heat of 2.89Jg∘C, and it releases 450. J into the atmosphere, what
nataly862011 [7]

Answer:

The final temperature of the object will be 42.785 °C

Explanation:

When the heat added or removed from a substance causes a change in temperature in it, this heat is called sensible heat.

In other words, sensible heat is the amount of heat that a body absorbs or releases without any changes in its physical state (phase change), so that the temperature varies.

The equation for calculating the heat exchanges in this case is:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature.

In this case:

  • Q= 450 J
  • c= 2.89 \frac{J}{g*C}
  • m= 20 g
  • ΔT= Tfinal - Tinitial= Tfinal - 35 °C

Replacing:

450 J= 2.89 \frac{J}{g*C} *20 g* (Tfinal - 35°C)

Solving for Tfinal:

\frac{450 J}{2.89\frac{J}{g*C}*20g} =Tfinal -35C

7.785 °C=Tfinal - 35°C

7.785 °C + 35°C= Tfinal

42.785 °C=Tfinal

<u><em>The final temperature of the object will be 42.785 °C</em></u>

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2 years ago
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