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1 year ago

I need help with problem C, finding the area of the accel. v time graph

Physics

1 answer:

erastova [34]1 year ago

7 0

c) To determine the velocity of the motion at the given time, consider that the velocity is also the area under the line of the graph a vs t.

Between t=0s and t = 2 s, the area under the line is:

A1 = 2 X 2 = 4

Then, until t = 2s the velocity is 4 m/s

The area between t = 2s and t = 4 s is:

A2 = 2 x 0 = 0

The area between t = 4s and t = 5s, the area is:

A3 = 2 x 1 = 2

Now, consider that the velocity is the sum of the previous areas, but taking areas on the negative y-axis as negative ones. Then, you have:

A = A1 + A2 - A3 = 4 + 0 - 2 = 2

Hence, the velocity is 2 m/s

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A 3.5 kg object moving in two dimensions initially has a velocity v1 = (12.0 i^ + 22.0 j^) m/s. A net force F then acts on the o

lys-0071 [83]

Answer:

The work done by the force is 820.745 joules.

Explanation:

Let suppose that changes in potential energy can be neglected. According to the Work-Energy Theorem, an external conservative force generates a change in the state of motion of the object, that is a change in kinetic energy. This phenomenon is describe by the following mathematical model:

$K_{1} + W_{F} = K_{2}$

Where:

$W_{F}$ - Work done by the external force, measured in joules.

$K_{1}$ , $K_{2}$ - Translational potential energy, measured in joules.

The work done by the external force is now cleared within:

$W_{F} = K_{2} - K_{1}$

After using the definition of translational kinetic energy, the previous expression is now expanded as a function of mass and initial and final speeds of the object:

$W_{F} = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{1}^{2})$

Where:

$m$ - Mass of the object, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speeds of the object, measured in meters per second.

Now, each speed is the magnitude of respective velocity vector:

Initial velocity

$v_{1} = \sqrt{v_{1,x}^{2}+v_{1,y}^{2}}$

$v_{1} = \sqrt{\left(12\,\frac{m}{s} \right)^{2}+\left(22\,\frac{m}{s} \right)^{2}}$

$v_{1} \approx 25.060\,\frac{m}{s}$

Final velocity

$v_{2} = \sqrt{v_{2,x}^{2}+v_{2,y}^{2}}$

$v_{2} = \sqrt{\left(16\,\frac{m}{s} \right)^{2}+\left(29\,\frac{m}{s} \right)^{2}}$

$v_{2} \approx 33.121\,\frac{m}{s}$

Finally, if $m = 3.5\,kg$ , $v_{1} \approx 25.060\,\frac{m}{s}$ and $v_{2} \approx 33.121\,\frac{m}{s}$ , then the work done by the force is:

$W_{F} = \frac{1}{2}\cdot (3.5\,kg)\cdot \left[\left(33.121\,\frac{m}{s} \right)^{2}-\left(25.060\,\frac{m}{s} \right)^{2}\right]$

$W_{F} = 820.745\,J$

The work done by the force is 820.745 joules.

6 0

2 years ago

Answer these questions please and you will get the brain list

Naddika [18.5K]

I am using the equation F=ma (force equals mass times acceleration) to solve these problems.

1. You are looking for force, and have mass and acceleration. You just plug in the values for mass and acceleration to get the force needed.

F=(15kg)(5m/s^2)

F=75N

2. Again, you are looking for force, and just need to plug in the values for mass and acceleration

F=(3kg)(2.4m/s^2)

F=7.2N

3. In this problem, you have force and mass, but need to find acceleration. To do this, you need to get acceleration alone on one side of the equation - divide each side by m. Your equation will now be F/m=a

a=(5N)/(3.7kg)

a=18.5m/s^2

I did not use significant figures. Let me know if you need to do that and need any help on that. Hope this helps!

7 0

2 years ago

1L of gas at 101kPa is compressed to 0.473L. What is the final pressure of the gas?

koban [17]

Answer:

hijzjsmsmmsnsnwnwnsnnsnsnaj

6 0

3 years ago

How far from Earth must a space probe be along a line toward the Sun so that the Sun's gravitational pull on the probe balances

Tatiana [17]

Answer:

258774.9441 m

Explanation:

x = Distance of probe from Earth

y = Distance of probe from Sun

Distance between Earth and Sun = $x+y=149.6\times 10^6\ m$

G = Gravitational constant

$M_s$ = Mass of Sun = $1.989\times 10^{30}$

$M_e$ = Mass of Earth = $5.972\times 10^{24}\ kg$

According to the question

$\frac{GM_sm}{x^2}=\frac{GM_em}{y^2}\\\Rightarrow \frac{M_s}{x^2}=\frac{M_e}{y^2}\\\Rightarrow x=\sqrt{\frac{M_s\times y^2}{M_e}}\\\Rightarrow x=\sqrt{\frac{1.989\times 10^{30}\times y^2}{5.972\times 10^{24}}}\\\Rightarrow x=577.10852y$

$x+y=149.6\times 10^6\\\Rightarrow 577.10852y+y=149.6\times 10^6\\\Rightarrow 578.10852y=149.6\times 10^6\\\Rightarrow y=\frac{149.6\times 10^6}{578.10852}\\\Rightarrow y=258774.9441\ m$

The probe should be 258774.9441 m from Earth

7 0

2 years ago

What is the relationship between electric currents and magnetic fields?

Aleksandr [31]

The electric currents induces a magnetic field. They changing magnetic field can be induce an electric current.

7 0

3 years ago