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3 years ago

A man is sitting on a chair with wheels. He grabs a 2.1 kg book from the desk and throws

Physics

1 answer:

geniusboy [140]3 years ago

8 0

The speed of the man and the chair after the book is thrown is 0.2 m/s.

The given parameters:

mass of the book, m₁ = 2.1 kg
speed of the book, u₁ = 7.2 m/s
mass of the man, M = 70 kg
mass of the book, m = 9.2 kg

The total mass of the man and the book is calculated as follows;

m₂ = 70 kg + 9.2 kg

m₂ = 79.2 kg

The speed of the man and the chair after the book is thrown is determined by applying the principle of conservation of linear momentum;

$m_1 u_1 = m_2u_2\\\\u_2 = \frac{m_1u_1}{m_2} \\\\u_2 = \frac{2.1 \times 7.2}{79.2} \\\\u_2 = 0.2 \ m/s$

Thus, the speed of the man and the chair after the book is thrown is 0.2 m/s.

Learn more about conservation of linear momentum here: brainly.com/question/7538238

You might be interested in

A seagull flies at a velocity of 9.00 m/s straight into the wind.

RideAnS [48]

a)If it takes the bird 18.0 minutes to fly 6 km away from the earth, the wind's speed will be 4 m/s.

b) The bird would need 7 minutes and 42 seconds to fly back 6 kilometers if he turned around and flew with the wind.

c)Compared to the 133.33 seconds it would take without the wind, the overall round-trip time is affected by the wind.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

A seagull flies at a velocity, $\rm v_{SA} = 9 \ m/sec$

The time the bird takes,t=18.0 min

The distance traveled relative to the earth = 6.00 km

The seagull's relative velocity with reference to the ground as;

$\rm v_{sg} = \frac{6.00 \times 10^3 \ m }{(20 min) \times \frac{60 s }{1 \ min}} \\\\ v_{sg}= 5.00 \ m/sec$

Air velocity with reference to the ground is;

$\rm v_{AG}= v_{SG}-v_{SA} \\\\ v_{AG} = 5.00 \ m/sec - 9.00 \ m/sec \\\\ v_{AG} = -4.00 \ m/sec$

If the bird turns around and flies with the wind, The time will he take to return 6.00 km is;

$\rm v_{SG}=v_{SA}+v_{AG} \\\\ v_{SG}=-900 \ m/sec +(-4.00 \ m/sec) \\\\ v_{SG}= -13.00 \ m/sec$

The time the bird takes;

$\rm t = \frac{x_{SG}}{v_{SG}} \\\\ t = \frac{6.00 \times 10^3 \ m }{13.00 \ m/sec } \\\\ t = 462 m/sec \\\\ t = 7 \ min \ and \ 42 \ sec$

c)\

The total round-trip time compared to what it would be with no wind. is;

$\rm t = 20 \ min( \frac{60 \ sec }{1 \ min} )+ 462 \ sec \\\\ t = 1200 \ sec +6 462 \ ec \\\\ t= 1662 \ sec$

The time for the round trip is;

$\rm t = \frac{12 \times 10^ 3 }{ 9 \ m/sec } \\\\ t = 1333.33 \ sec$

Hence the wind's speed, the time bird would need to fly back the total round-trip time will be 4 m/s, 7 minutes and 42 seconds and 1333.33 sec.

To learn more about the velocity, refer to the link: brainly.com/question/862972.

#SPJ1

4 0

2 years ago

A device that converts high voltages to lower voltages is a(n) ________.

AysviL [449]

Answer: A) Transformer

Explanation: A Transformer is an apparatus, device or a component in a system that is used to convert high voltage into low voltage. It is used to either increase or decrease the voltage of an alternating current.

A transformer uses the basic principle of electro magnetic induction, having two or more coils, the voltage is changed from one coil to another but with thesame frequency as alternating current energy passes through them.

7 0

3 years ago

What happens to the force between two charges when each charge is doubled and the distance between them is 1/4 its original

DIA [1.3K]

Answer:

F' = 64 F

Explanation:

The electric force between charges is given by :

$F=\dfrac{kq_1q_2}{r^2}$

Where

q₁ and q₂ are charges

r is the distance between charges

When each charge is doubled and the distance between them is 1/4 its original magnitude such that,

q₁' = 2q₁, q₂' = 2q₂ and r' = (r/4)

New force,

$F'=\dfrac{kq_1'q_2'}{r'^2}$

Apply new values,

$F'=\dfrac{k\times 2q_1\times 2q_2}{(\dfrac{r}{4})^2}\\\\=\dfrac{k\times 4q_1q_2}{\dfrac{r^2}{16}}\\\\=64\times \dfrac{kq_1q_2}{r^2}\\\\=64F$

So, the new force becomes 64 times the initial force.

7 0

3 years ago

On level ground a shell is fired with an initial velocity of 49.0 m/s at 70.0 โ above the horizontal and feels no appreciable ai

xenn [34]

Answer:

Horizontal component = 16.8 m/s

Vertical component = 46.0 m/s

Explanation:

If we denote the initial velocity by v and the angle above the horizontal by θ,

the horizontal component of this initial velocity is given by

$v_x = v\cos \theta$

$v_x = (49.0\text{ m/s})\cos\,(70.0^\circ) = 16.8\text{ m/s}$

The vertical component is given by

$v_y = v\sin\theta$

$v_x = (49.0\text{ m/s})\sin\,(70.0^\circ) = 46.0\text{ m/s}$

3 0

3 years ago

Describe two experiments to determine the speed of propagation of a transverse wave on a rope. You have the following tools to u

AnnZ [28]

Answer:

#See solution for details.

Explanation:

Tools: $stopwatch, \ meter \ stick, \ mass \ measuring \ scale , \ force \ measuring \ device$ .

$Experiment \ 1$ :Calculate the speed of the wave using the time, $t$ it takes to travel along the rope. Rope's length, $L$ is measured using the meter stick.

-Attach one end of rope to a wall or post, shake from the unfixed end to generate a pulse. Measure the the time it takes for the pulse to reach the wall once it starts traveling using the stopwatch.

-Speed of the pulse can then be obtained as:

$v=\frac{L}{t}$

$Experiment \ 2$ : Apply force of known value to the rope then use the following relation equation to find the speed of a pulse that travels on the rope.

$v=\sqrt{\frac{F}{\mu}}\ ,\mu=\frac{m}{L}$

-Use the measuring stick and measuring scale to determine $L,m$ values of the rope then obtain $\mu$ .

-Use the force measuring constant to determine $F$ . These values can the be substituted in $Experiment \ 1$ to obtain $v.$

4 0

3 years ago