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1 year ago

An object on a number line moved from x = 15 cm to x = 165 cm and then

Physics

2 answers:

mel-nik [20]1 year ago

5 0

Answer:

The average velocity of the object is 0.1cm/s

Explanation:

Given that the object travels from point 15cm to 165cm and back to 25cm within 100 seconds

The average velocity is calculated as thus.

Average Velocity = ∆D/t

Where ∆D represent the displacement.

The displacement is calculated as follows.

∆D = End point - Start Point.

From the question, the end and start point are 25cm and 15cm respectively.

Hence,

∆D = 25cm - 15cm

∆D = 10cm.

t = 100 seconds

So, Average Velocity = 10cm/100s

Average Velocity = 0.1cm/s

Hence, the average velocity of the object is 0.1cm/s

olasank [31]1 year ago

3 0

Answer:

v_avg = 2.9 cm/s

Explanation:

The average velocity of the object is the sum of the distance of all its trajectories divided the time:

$v_{avg}=\frac{x_{all}}{t}$

x_all is the total distance traveled by the object. In this case you have that the object traveled in the first trajectory 165cm-15cm = 150cm, and in the second one, 165cm - 25cm = 140cm

Then, x_all = 150cm + 140cm = 290cm

The average velocity is, for t = 100s

$v_{avg}=\frac{290cm}{100s}=2.9\frac{cm}{s}$

hence, the average velocity of the object in the total trajectory traveled is 2.9 cm/s

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electromagnetic

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A single-turn current loop carrying a 4.00 A current, is in the shape of a right-angle triangle with sides of 50.0 cm, 120 cm, a

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Given that,

Current = 4 A

Sides of triangle = 50.0 cm, 120 cm and 130 cm

Magnetic field = 75.0 mT

Distance = 130 cm

We need to calculate the angle α

Using cosine law

$120^2=130^2+50^2-2\times130\times50\cos\alpha$

$\cos\alpha=\dfrac{120^2-130^2-50^2}{2\times130\times50}$

$\alpha=\cos^{-1}(0.3846)$

$\alpha=67.38^{\circ}$

We need to calculate the angle β

Using cosine law

$50^2=130^2+120^2-2\times130\times120\cos\beta$

$\cos\beta=\dfrac{50^2-130^2-120^2}{2\times130\times120}$

$\beta=\cos^{-1}(0.923)$

$\beta=22.63^{\circ}$

We need to calculate the force on 130 cm side

Using formula of force

$F_{130}=ILB\sin\theta$

$F_{130}=4\times130\times10^{-2}\times75\times10^{-3}\sin0$

$F_{130}=0$

We need to calculate the force on 120 cm side

Using formula of force

$F_{120}=ILB\sin\beta$

$F_{120}=4\times120\times10^{-2}\times75\times10^{-3}\sin22.63$

$F_{120}=0.1385\ N$

The direction of force is out of page.

We need to calculate the force on 50 cm side

Using formula of force

$F_{50}=ILB\sin\alpha$

$F_{50}=4\times50\times10^{-2}\times75\times10^{-3}\sin67.38$

$F_{50}=0.1385\ N$

The direction of force is into page.

Hence, The magnitude of the magnetic force on each of the three sides of the loop are 0 N, 0.1385 N and 0.1385 N.

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2 years ago

A car drives at a constant speed around a banked circular track with a diameter of 136 m . The motion of the car can be describe

galina1969 [7]

Answer:

speed = 44.9m/s

x = 35.5 m, y = 58.0m

Explanation:

A car on a circular track with constant angular velocity ω can be described by the equation of position r:

$\overrightarrow {r(t)} = Rsin(\omega t)\hat{i} + Rcos(\omega t)\hat{j}$

The velocity v is given by:

$\overrightarrow {v(t)} = \overrightarrow{\frac{dr}{dt}}= \omega Rcos(\omega t)\hat{i} - \omega Rsin(\omega t)\hat{j}$

The acceleration a:

$\overrightarrow {a(t)} = \overrightarrow{\frac{dv}{dt}}= -\omega^2 Rsin(\omega t)\hat{i} - \omega^2 Rcos(\omega t)\hat{j}$

From the given values we get two equations:

$-\omega^2 Rsin(\omega t)=-15.4\\-\omega^2 Rcos(\omega t)=-25.4$

We also know:

$\overrightarrow {a(t)} = -\omega^2 Rsin(\omega t)\hat{i} - \omega^2 Rcos(\omega t)\hat{j}=-\omega^2\overrightarrow{r(t)}$

The magnitude of the acceleration a is:

$a=\sqrt{(-15.4)^2+(-25.4)^2}=29.7$

The magnitude of position r is:

$r=R=68m$

Plugging in to the equation for a(t):

$\overrightarrow {a(t)} =-\omega^2\overrightarrow{r(t)}$

and solving for ω:

$|\omega|=0.66$

Now solve for time t:

$\frac{sin(0.66t)}{cos(0.66t)}=tan(0.66t)=\frac{15.4}{25.4}\\t=0.83$

Using the calculated values to compute v(t):

$\overrightarrow {v(t)}= \omega Rcos(\omega t)\hat{i} - \omega Rsin(\omega t)\hat{j}\\\overrightarrow {v(t)}=44.88cos(0.55)\hai{i}-44.88sin(0.55)\hat{j}\\\overrightarrow {v(t)}=38.3\hat{i}-23.5\hat{j}$

The speed of the car is:

$\sqrt{38.3^2 + (-23.5)^2} = 44.9$

The position r:

$\overrightarrow {r(t)} = Rsin(\omega t)\hat{i} + Rcos(\omega t)\hat{j}\\\overrightarrow {r(t)} = 68sin(0.55)\hat{i} + 68cos(0.55)\hat{j}\\\overrightarrow {r(t)} = 35.5{i} + 58.0\hat{j}$

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Answer:

$8.0\cdot 10^5 N/C$

Explanation:

Outside the sphere's surface, the electric field has the same expression of that produced by a single point charge located at the centre of the sphere.

Therefore, the magnitude of the electric field ar r = 5.0 cm from the sphere is:

$E=k\frac{q}{(R+r)^2}$

where

$k=8.99\cdot 10^9 N m^2C^{-2}$ is the Coulomb's constant

$q=2.0 \mu C=2.0 \cdot 10^{-6}C$ is the charge on the sphere

$R=10 cm = 0.10 m$ is the radius of the sphere

$r=5.0 cm = 0.05 m$ is the distance from the surface of the sphere

Substituting, we find

$E=(8.99\cdot 10^9 Nm^2 C^{-2})\frac{2.0\cdot 10^{-6} C}{(0.10 m+0.05 m)^2}=8.0\cdot 10^5 N/C$

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2 years ago

A balloon drifts 140m toward the west in 45s ; then the wind suddenly changes and the balloon flies 90m toward the east in the n

Bogdan [553]

Answer: 140 m

Explanation:

Let's begin by stating clear that motiont is the change of position of a body at a certain time. So, during this motion, the balloon will have a trajectory and a displacement, being both different:

The trajectory is the path followed by the body, the distance it travelled (is a scalar quantity).

The displacement is the distance in a straight line between the initial and final position (is a vector quantity).

So, according to this, the distance the balloon traveled during the first 45 s (its trajectory) is 140 m.

But, if we talk about displacement, we have to draw a straight line between the initial position of the balloon (point 0) to its final position (point 90 m). Being its displacement 95 m.

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1 year ago