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3 years ago

At which location could you place the north pole of a bar magnet so that it would be pushed away from the magnet shown?

Physics

2 answers:

Gala2k [10]3 years ago

7 0

It’s a because b makes more since shown as the magnet

skelet666 [1.2K]3 years ago

4 0

Answer:

The answer is C (A)

Explanation:

I just did it on Apex

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The following statement is false. An atom can be broken into smaller atoms. Reword the statement so it is true.

Softa [21]

Answer:

an atom can not be broken into smaller pieces

Explanation:

4 0

4 years ago

A 910 kg car is approaching a loop-the-loop. The loop has a diameter of 50 m. Determine the minimum speed the car must have at t

scoundrel [369]

Answer:

The minimum speed the car must have at the top of the loop to not fall = 35 m/s

Explanation:

Anywhere else on the loop, the speed needed to keep the car in the loop is obtained from the force that keeps the body in circular motion around the loop which has to just match the force of gravity on the car. (Given that frictional force = 0)

mv²/r = mg

v² = gr = 9.8 × 25 = 245

v = 15.65 m/s

But at the top, the change in kinetic energy of the car must match the potential energy at the very top of the loop-the-loop

Change in kinetic energy = potential energy at the top

Change in kinetic energy = (mv₂² - mv₁²)/2

v₁ = velocity required to stay in the loop anywhere else = 15.65 m/s

v₂ = minimum velocity the car must have at the top of the loop to not fall

And potential energy at the top of the loop = mgh (where h = the diameter of the loop)

(mv₂² - mv₁²)/2 = mgh

(v₂² - v₁²) = 2gh

(v₂² - (15.65)²) = 2×9.8×50

v₂² - 245 = 980

v₂² = 1225

v₂ = 35 m/s

Hence, the minimum speed the car must have at the top of the loop to not fall = 35 m/s

4 0

3 years ago

A cross-country skier can complete a 7. 5 km race in 45 min. What is the skier’s average speed? km/h.

never [62]

The average speed is a scalar quantity. The average speed of the skier is 10 km/hrs.

<h3>What is the average speed?</h3>

It is calculated by dividing the total distance by the total time elapsed.

The

$\overline{v}={\frac{\Delta x}{\Delta t}}$

Where,

$\overline{v}$ = average velocity

${\Delta x}$ = displacement = 7.5 km

${\Delta t}$ = change in time = 45 min = 0.75 hours

Put the values in the formula,

$\overline{v}={\frac{7.5 }{0.75 }}\\\\\overline{v}= 10 \rm \ km/hrs$

Therefore, the average speed of the skier is 10 km/hrs.

Learn more about average speed:

https://brainly.in/question/32459982

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3 years ago

Students had two batteries and two different resistors. During four trials, they build four different circuits and measure the c

Darina [25.2K]

Answer:

bhi jo bhi of gp oh oh gi IG 7u to uff do if goo td to yd do FP ae rt 7g hi pic vo icon

Explanation:

bh hi h bhi vc di oh x At jb jo iv hp of di of dr hi o hc x gh ki vc hi jo

5 0

3 years ago

A baseball player hits a homerun, and the ball lands in the left field seats, which is 103m away from the point at which the bal

Sati [7]

(a) The ball has a final velocity vector

$\mathbf v_f=v_{x,f}\,\mathbf i+v_{y,f}\,\mathbf j$

with horizontal and vertical components, respectively,

$v_{x,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\cos(-38^\circ)\approx16.2\dfrac{\rm m}{\rm s}$

$v_{y,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\sin(-38^\circ)\approx-12.6\dfrac{\rm m}{\rm s}$

The horizontal component of the ball's velocity is constant throughout its trajectory, so $v_{x,i}=v_{x,f}$ , and the horizontal distance x that it covers after time t is

$x=v_{x,i}t=v_{x,f}t$

It lands 103 m away from where it's hit, so we can determine the time it it spends in the air:

$103\,\mathrm m=\left(16.2\dfrac{\rm m}{\rm s}\right)t\implies t\approx6.38\,\mathrm s$

The vertical component of the ball's velocity at time t is

$v_{y,f}=v_{y,i}-gt$

where g = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve for the vertical component of the initial velocity:

$-12.6\dfrac{\rm m}{\rm s}=v_{y,i}-\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)\implies v_{y,i}\approx49.9\dfrac{\rm m}{\rm s}$

So, the initial velocity vector is

$\mathbf v_i=v_{x,i}\,\mathbf i+v_{y,i}\,\mathbf j=\left(16.2\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(49.9\dfrac{\rm m}{\rm s}\right)\,\mathbf j$

which carries an initial speed of

$\|\mathbf v_i\|=\sqrt{{v_{x,i}}^2+{v_{y,i}}^2}\approx\boxed{52.4\dfrac{\rm m}{\rm s}}$

and direction θ such that

$\tan\theta=\dfrac{v_{y,i}}{v_{x,i}}\implies\theta\approx\boxed{72.0^\circ}$

(b) I assume you're supposed to find the height of the ball when it lands in the seats. The ball's height y at time t is

$y=v_{y,i}t-\dfrac12gt^2$

so that when it lands in the seats at t ≈ 6.38 s, it has a height of

$y=\left(49.9\dfrac{\rm m}{\rm s}\right)(6.38\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)^2\approx\boxed{119\,\mathrm m}$

6 0

4 years ago