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3 years ago

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1. A person is lifting a suitcase into the trunk of a car. Assuming the suitcase is already in the air, draw the FBD and write t

he sum of forces equation for the box. (Assume the person is lifting straight up).

1 answer:

Alik [6]3 years ago

7 0

Answer:

Explanation:Assume you are lifting an object with mass 20 kg from the ground to a height of 1.5 m. Assume that you are exerting a constant force in the upward direction and that you are moving the object upward with uniform velocity. The net force on the object is zero. The force you are exerting is equal in magnitude and opposite in direction to the force of gravity. As you are lifting the object you are doing work on the object.

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What's the frequency of a wave with a wavelength of 10 and velocity of 200m/s?

uranmaximum [27]

Answer:

$\boxed {\boxed {\sf 20 \ Hz}}$

Explanation:

The frequency of a wave can be found using the following formula.

$f=\frac{v}{\lambda}$

where <em>f</em> is the frequency, <em>v</em> is the velocity/wave speed, and λ is the wavelength.

The wavelength is 10 meters and the velocity is 200 meters per second.

1 m/s can also be written as 1 m*s^-1

Therefore:

$v= 200 \ m*s^{-1} \\\lambda = 10 \ m$

Substitute the values into the formula.

$f=\frac{200 \ m*s^{-1}}{10 \ m}$

Divide and note that the meters (m) will cancel each other out.

$f=\frac{200 \ s^{-1}}{10 \ }$

$f=20 \ s^{-1}$

1 s^-1 is equal to Hertz
Therefore, our answer of 20 s^-1 is equal to 20 Hz

$f= 20 \ Hz$

The frequency of the wave is <u>20 Hertz</u>

7 0

2 years ago

What happens when a penny is dropped from a height of 20 meters? A.it falls at a constant velocity B.its mass doubles for every

Travka [436]

Answer: Option (c) is the correct answer.

Explanation:

When a penny is dropped from a height of 20 meters then it will achieve an acceleration.

As acceleration is the rate of change in velocity of an object with respect to time. Therefore, the velocity does not remain constant.

Whereas mass of the penny will remain the same as it will not get affected when it falls. Also, there will be no change in direction of the penny as it is falling only in one direction.

The acceleration of penny is due to the force of gravity.

Thus, we can conclude that the force of gravity causes it to accelerate.

4 0

2 years ago

Read 2 more answers

For Part A, Sebastian decided to use the copper cylinder. How would the magnitude of his q and ∆H compare if he were to redo Par

Vitek1552 [10]

The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

The given parameters;

<em>initial temperature of metals, = </em> $T_m$ <em />
<em>initial temperature of water, = </em> $T_i$ <em> </em>
<em>specific heat capacity of copper, </em> $C_p$ <em> = 0.385 J/g.K</em>
<em>specific heat capacity of aluminum, </em> $C_A$ = 0.9 J/g.K
<em>both metals have equal mass = m</em>

The quantity of heat transferred by each metal is calculated as follows;

Q = mcΔt

<em>For</em><em> copper metal</em><em>, the quantity of heat transferred is calculated as</em>;

$Q_p = (m_wc_w + m_pc_p)(T_m - T_i)\\\\Q_p = (T_m - T_i)(m_wc_w ) + (T_m - T_i)(m_pc_p)\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m_p(T_m - T_i)\\\\m_p = m\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m(T_m - T_i)\\\\let \ (T_m - T_i)(m_wc_w ) = Q_i, \ \ \ and \ let \ (T_m- T_i) = \Delta t\\\\Q_p = Q_i + 0.385m\Delta t$

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>copper metal</em>;

$\Delta H = Q_p - Q_i\\\\\Delta H = (Q_i + 0.385m \Delta t) - Q_i\\\\\Delta H = 0.385 m \Delta t$

<em>For </em><em>aluminum metal</em><em>, the quantity of heat transferred is calculated as</em>;

$Q_A = (m_wc_w + m_Ac_A)(T_m - T_i)\\\\Q_A = (T_m -T_i)(m_wc_w) + (T_m -T_i) (m_Ac_A)\\\\let \ (T_m -T_i)(m_wc_w) = Q_i, \ and \ let (T_m - T_i) = \Delta t\\\\Q_A = Q_i \ + \ m_Ac_A\Delta t\\\\m_A = m\\\\Q_A = Q_i \ + \ 0.9m\Delta t$

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>aluminum metal </em><em>;</em>

$\Delta H = Q_A - Q_i\\\\\Delta H = (Q_i + 0.9m\Delta t) - Q_i\\\\\Delta H = 0.9m\Delta t$

Thus, we can conclude that the magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

Learn more here:brainly.com/question/15345295

6 0

2 years ago

Olin [163]

Answer:

Explanation:

THE FOURTEEN WEEKS' COURSES

IN

NATURAL SCIENCE,

BY

J. DORMAN STEELE, A.M., PHI.D.

Fourteen Weeks iq Natural Philosophy,

Fourteen Weeks iq Ctlenqistry,

Fourteen Weeks iq Descriptive Astroqonqy,

Fourteel Weeks iq Popular Geology,

Fourteeq Weeks iQ2 Human P1ysiology,

Fourteen Weeks iq Zoology,

Fourteeq Weeks iq Botany,

A Key, containing Answers to the Questions

and Problems in Steele's I4 Weeks' Courses,

4 1ISTORIC4L SERIES,

ON THE PLAN OF STEELE'S 14 WEEKS IN THE SCIENCES.

A Brief History of the Urlited States,

A Brief History of France,

The same publishers also offer the following standard scientific

vworks, being more extended or difficult treatises than those of

Prof. Steele, though still of Academic grade.

Peck's Ganot's Natural Philosophy,

Porter's Principles of Chemistry,

Jarvis' Physiology and Laws of Healtl,

Wood's Botanist and Florist,

Clanlbers' Elenments of Zoology,

lcIqtyre's Astroqomy and tle Globes,

Page's. Elen~ents of Geology,

Entered according to Act of Congress, in the year 1869, by

A. S. BARNES & CO.,

In the Clerk's Office of the District Court of the United States

for the Southern District of New York.

sTERLE'S KEY.

6 0

2 years ago

What is the force exerted on a charge of 2.5 µC moving perpendicular through a magnetic field of 3.0 × 102 T with a velocity of

OverLord2011 [107]

Given:

B = $3 \times 10^{2}$ T

V= $5 \times 10^{3} \frac{m}{s}$

q = 2.5 × $10^{-6}$ C

α = 90

To find:

Force = ?

Formula used:

Force on the moving charge is given by,

F = q V B sin α

Where F = force exerted on moving charge

V = velocity of charge

q = charge

α = angle between direction of V and B

Solution:

F = q V B sin α

Where F = force exerted on moving charge

V = velocity of charge

q = charge

α = angle between direction of V and B

F = $2.5 \times 10^{-6} \times 3 \times 10^{2} \times 5 \times 10^{3}$

F = 37.5 × $10^{-1}$

F = 3.75 Newton

Thus, the force acting on the moving charge is 3.75 Newton.

8 0

3 years ago

Read 2 more answers