answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
galben [10]
2 years ago
6

Isabella drops a pen off her balcony by accident while celebrating the successful completion of a physics problem. assuming air

resistance is negligible, how many seconds does it take the pen to reach a speed of 19.62 \,\dfrac{\text {m}}{\text s}19.62 âs â âm ââ 19, point, 62, space, start fraction, m, divided by, s, end fraction
Physics
1 answer:
expeople1 [14]2 years ago
8 0
U = 0, initial vertical velocity

Neglect air resistance, and g = 9.8 m/s².

The time, t, required for the pen to attain a vertical velocity of 19.62 m/s is given by
19.62 m/s = 0 + (9.8 m/s²)*(t s)
t = 19.62/9.8 = 2.00 s

Answer:  2.0 s
You might be interested in
A silver wire 2.6 mm in diameter transfers a charge of 420 Cin 80 min. Silver contains 5.8 x 10^{28} free electrons per cubic me
never [62]

1) Current in the wire: 0.0875 A

The current in the wire is given by:

I=\frac{Q}{t}

where

Q is the charge passing a given point in the conductor

t is the time elapsed

In this problem, we have

Q = 420 C is the total charge passing through a given point in a time of

t = 80 min = 4800 s

So, the current is

I=\frac{420 C}{4800 s}=0.0875 A

2) Drift velocity of the electrons: 1.78\cdot 10^{-6} m/s

The drift velocity of the electrons in the wire is given by:

u = \frac{I}{nAq}

where

I = 0.0875 A is the current

n=5.8\cdot 10^{28} is the number of free electrons per cubic meter

A is the cross-sectional area

q=1.6\cdot 10^{-19} C is the charge of one electron

The radius of the wire is

r=\frac{d}{2}=\frac{2.6 mm}{2}=1.3 mm=0.0013 m

So the cross-sectional area is

A=\pi r^2=\pi (0.0013 m)^2=5.31\cdot 10^{-6} m^2

So, the drift velocity is

u = \frac{(0.0875 A)}{(5.8\cdot 10^{28})(5.31\cdot 10^{-6})(1.6\cdot 10^{-19}C)}=1.78\cdot 10^{-6} m/s

4 0
2 years ago
Lexy used the formula shown to calculate the force of gravity on a space shuttle. Fg = G What does 3 × 105 kg represent? the dif
STALIN [3.7K]
<h2>Answer:</h2>

<u>This term shows the </u><u>mass of the space shuttle</u>

<h2>Explanation:</h2>

We know that the mass of the Earth is 5.972 × 10^24 kg. Similarly the sum of  mass of earth and the mass of shuttle must be a greater number as compared to the number given. It simply means that the mass of earth is itself 5.972 × 10^24 kg and the value given is 3 × 105 kg so it is obvious that if was the sum then it must be greater than the mass of earth. Therefore we can say that this not the mass of earth, neither the sum of mass of earth and shuttle, but this is only the mass of space shuttle which is the last multiple choice.

5 0
2 years ago
An object at rest is suddenly broken apart into two fragments by an explosion one fragment acquires twice the kinetic energy of
satela [25.4K]
<span>First, we use the kinetic energy equation to create a formula: Ka = 2Kb 1/2(ma*Va^2) = 2(1/2(mb*Vb^2)) The 1/2 of the right gets cancelled by the 2 left of the bracket so: 1/2(ma*Va^2) = mb*Vb^2 (1) By the definiton of momentum we can say: ma*Va = mb*Vb And with some algebra: Vb = (ma*Va)/mb (2) Substituting (2) into (1), we have: 1/2(ma*Va^2) = mb*((ma*Va)/mb)^2 Then: 1/2(ma*Va^2) = mb*(ma^2*Va^2)/mb^2 We cancel the Va^2 in both sides and cancel the mb at the numerator, leving the denominator of the right side with exponent 1: 1/2(ma) = (ma^2)/mb Cancel the ma of the left, leaving the right one with exponent 1: 1/2 = ma/mb And finally we have that: mb/2 = ma mb = 2ma</span>
8 0
2 years ago
A 50-kg load is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. By what distance will the wire stretch? Young'
lbvjy [14]

Answer:

3.5 cm

Explanation:

mass, m = 50 kg

diameter = 1 mm

radius, r = half of diameter = 0.5 mm = 0.5 x 10^-3 m

L = 11.2 m

Y = 2 x 10^11 Pa

Area of crossection of wire = π r² = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3  

                                              = 7.85 x 10^-7 m^2

Let the wire is stretch by ΔL.

The formula for Young's modulus is given by

Y =\frac{mgL}{A\Delta L}

\Delta L =\frac{mgL}{A\times Y}

ΔL = 0.035 m = 3.5 cm

Thus, the length of the wire stretch by 3.5 cm.

5 0
2 years ago
It requires 49 J of work to stretch an ideal very light spring from a length of 1.4 m to a length of 2.9 m. What is the value of
nadya68 [22]

Answer:

44 N/m

Explanation:

The extension, e, of the spring = 2.9 m - 1.4 m = 1.5 m

The work needed to stretch a spring by <em>e</em> is given by

W = \frac{1}{2} ke^2

where <em>k</em> is spring constant.

k = \dfrac{2W}{e^2}

Using the appropriate values,

k = \dfrac{2\times 49\text{ J}}{1.5^2\text{ m}^2} = 43.55\ldots\text{ N/m} \approx 44\text{ N/m}

3 0
2 years ago
Read 2 more answers
Other questions:
  • Which of the following are linear defects?. . An edge dislocation. . A Frenkel defect. . A screw dislocation. . A Schottky defec
    6·1 answer
  • Ddt is _____-soluble so it accumulates in _____.
    11·1 answer
  • A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. f(t) = 0.01t4 − 0.0
    8·1 answer
  • A glass jug can be used to play different pitched sounds by blowing air over the opening of the jug and vibrating the air molecu
    7·1 answer
  • 61. A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A
    5·1 answer
  • In the Bohr model of the hydrogen atom, the electron moves in a circular orbit of radius 5.3×10−11m with a speed of 2.2×106m/s.
    15·1 answer
  • A rigid, uniform bar with mass mmm and length bbb rotates about the axis passing through the midpoint of the bar perpendicular t
    10·1 answer
  • Lucy is cruising through space in her new spaceship. As she coasts along, a tiny spacebug drifts into her path and bounces off t
    11·1 answer
  • Kayla and her friends are setting up chairs for a school play each row will contain the same number of chairs Kayla knows that t
    11·1 answer
  • High-speed stroboscopic photographs show that the head of a 200 g golf club is traveling at 43.7 m/s just before it strikes a 45
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!