All categories

4 years ago

A horizontal force of 400 N is exerted on a 2.0-kg ball as it rotates (at

Physics

2 answers:

ycow [4]4 years ago

7 0

Answer:

The speed of the ball is 10 meters per second.

Explanation:

The fact that ball rotates uniformly means that speed of the ball is constant and resulting acceleration is centripetal, which means that external horizontal force exerted on the ball goes towards the center of rotation. We can estimate the speed of the ball by this equation of equilibrium:

$\Sigma F_{r} = F = m\cdot \frac{v^{2}}{R}$ (Eq. 1)

Where:

$F$ - External force exerted on the ball, measured in newtons.

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

$v$ - Linear rotation speed, measured in meters per second.

$R$ - Radius of rotation, measured in meters.

We clear the linear speed in (Eq. 1):

$v^{2} = \frac{F\cdot R}{m}$

$v = \sqrt{\frac{F\cdot R}{m} }$

If we know that $F = 400\,N$ , $m = 2\,kg$ and $R = 0.50\,m$ , the speed of the ball is:

$v =\sqrt{\frac{(400\,N)\cdot (0.50\,m)}{2\,kg} }$

$v = 10\,\frac{m}{s}$

The speed of the ball is 10 meters per second.

frutty [35]4 years ago

6 0

Answer:

the speed of the ball is 10 m/s

Explanation:

Given;

magnitude of exerted force, F = 400 N

mass of the ball, m = 2 kg

radius of the circle, r = 0.5

The speed of the ball is calculated by applying centripetal force formula;

$F = \frac{mv^2}{r} \\\\v^2 = \frac{Fr}{m}\\\\v = \sqrt{\frac{Fr}{m}}\\\\ v = \sqrt{\frac{400*0.5}{2}}\\\\v = 10 \ m/s$

Therefore, the speed of the ball is 10 m/s

You might be interested in

A uniform plank of mass 10kg and length 10m rests on two supports, A and B as shown. A boy of weight 500N stands at a distance o

kifflom [539]

Answer:

U² = 142.86 N

U¹ = 357.14 N

Explanation:

Taking summation of the moment about point A, we get the following equilibrium equation: (taking clockwise direction as positive)

$W(2\ m) - U^2(7\ m) = 0$

where,

W = weight of boy = 500 N

U² = reaction ay B = ?

Therefore,

$(500\ N)(2\ m)-(U^2)(7\ m)=0\\U^2=\frac{1000\ Nm}{7\ m}\\$

U² = 142.86 N

Now, taking summation of forces on the plank. Taking upward direction as positive, for equilibrium position:

$W-U^1-U^2=0\\500\ N - 142.86\ N = U^1\\$

U¹ = 357.14 N

3 0

3 years ago

Only the smallest particles of soil can be displaced by suspension because

mote1985 [20]

The choices can be found elsewhere and as follows:

a. they are so small that they stay close to the ground due to the attractive properties of charged soil particles.

b. they are easily carried by the wind.

c. they easily dissolve in liquid droplets.

d. it is easier for then to roll along the small crevices in the ground.

I think the correct answer from the choices listed above is option B. Only the smallest particles of soil can be displaced by suspension because they are so small that they are easily carried by the wind. Hope this answers the question. Have a nice day. Feel free to ask more questions.

7 0

4 years ago

Read 2 more answers

A satellite completes one revolution of a planet in almost exactly one hour. At the end of one hour, the satellite has traveled

ad-work [718]

Answer:

shown below

Explanation:

2 x 10⁷ as a number is 20,000,000

20,000,000 - 10 = 19,999,990

It went 19,999,990 m/h

in km/h:

19,999,990 / 1000 = 19,999.99 km/h

in km/s

19,999,990 / 3,600,000 = ~5.56 km/s

in m/s

19,999,990 / 3600 = ~5555.56 m/s

7 0

2 years ago

Arthur and Betty start walking toward each other when they are 100 m apart. Arthur has a

Gnoma [55]

The distance covered by Arthur is
3t
The distance covered by Betty from where Arthur started is
100 - 2t
They meet each other when this distance is equal
3t = 100 - 2t
t = 20 s

Since the dog has a constant speed of 5m/s, the distance that Spot has run is
5 m/s (20s) = 100 m

6 0

3 years ago

Read 2 more answers

A particle with an initial linear momentum of 2.00 kg-m/s directed along the positive x-axis collides with a second particle, wh

ladessa [460]

Answer:

a) p₂ = 1.88 kg*m/s

θ = 273.4 º

b) Kf = 37% of Ko

Explanation:

Assuming no external forces acting during the collision, total momentum must be conserved.
Since momentum is a vector, their components (projected along two axes perpendicular each other, x- and y- in this case) must be conserved too.
The initial momenta of both particles are directed one along the x-axis, and the other one along the y-axis.
So for the particle moving along the positive x-axis, we can write the following equations for its initial momentum:

$p_{o1x} = 2.00 kg*m/s (1)$

$p_{o1y} = 0 (2)$

We can do the same for the particle moving along the positive y-axis:

$p_{o2x} = 0 (3)$

$p_{o2y} = 4.00 kg*m/s (4)$

Now, we know the value of magnitude of the final momentum p1, and the angle that makes with the positive x-axis.
Applying the definition of cosine and sine of an angle, we can find the x- and y- components of the final momentum of the first particle, as follows:

$p_{f1x} = 3.00 kg*m/s * cos 45 = 2.12 kg*m/s (5)$

$p_{f1y} = 3.00 kg*m/s sin 45 = 2.12 kg*m/s (6)$

Now, the total initial momentum, along these directions, must be equal to the total final momentum.
We can write the equation for the x- axis as follows:

$p_{o1x} + p_{o2x} = p_{f1x} + p_{f2x} (7)$

We know from (3) that p₀₂ₓ = 0, and we have the values of p₀1ₓ from (1) and pf₁ₓ from (5) so we can solve (7) for pf₂ₓ, as follows:

$p_{f2x} = p_{o1x} - p_{f1x} = 2.00kg*m*/s - 2.12 kg*m/s = -0.12 kg*m/s (8)$

Now, we can repeat exactly the same process for the y- axis, as follows:

$p_{o1y} + p_{o2y} = p_{f1y} + p_{f2y} (9)$

We know from (2) that p₀1y = 0, and we have the values of p₀₂y from (4) and pf₁y from (6) so we can solve (9) for pf₂y, as follows:

$p_{f2y} = p_{o1y} - p_{f1y} = 4.00kg*m*/s - 2.12 kg*m/s = 1.88 kg*m/s (10)$

Since we have the x- and y- components of the final momentum of the second particle, we can find its magnitude applying the Pythagorean Theorem, as follows:

$p_{f2} = \sqrt{p_{f2x} ^{2} + p_{f2y} ^{2} } = \sqrt{(-0.12m/s)^{2} +(1.88m/s)^{2}} = 1.88 kg*m/s (11)$

We can find the angle that this vector makes with the positive x- axis, applying the definition of tangent of an angle, as follows:

$tg \theta = \frac{p_{2fy} }{p_{2fx} } = \frac{1.88m/s}{(-0.12m/s} = -15.7 (12)$

The angle that we are looking for is just the arc tg of (12) which measured in a counter-clockwise direction from the positive x- axis, is just 273.4º.

Assuming that both masses are equal each other, we find that the momenta are proportional to the speeds, so we find that the relationship from the final kinetic energy and the initial one can be expressed as follows:

$\frac{K_{f}}{K_{o} } = \frac{v_{f1}^{2} + v_{f2} ^{2}}{v_{o1}^{2} + v_{o2} ^{2} } = \frac{12.5}{20} = 0.63 (13)$

So, the final kinetic energy has lost a 37% of the initial one.

6 0

3 years ago