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3 years ago

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The position equation for a particle is s of t equals the square root of the quantity 2 times t plus 1 where s is measured in fe

et and t is measured in seconds. find the acceleration of the particle at 4 seconds. 3 ft/sec2 one third ft/sec2 negative 1 over 27 ft/sec2 none of these

1 answer:

loris [4]3 years ago

8 0

Given that,

s = √(2t + 1)

Time, t = 4 s

Acceleration , a = ??

Since,

Acceleration = velocity / time

Velocity = distance/ time

Acceleration = distance/ time²

s/t² = √(2t+1)/t²

putting t = 4 sec, we have

a = √(2*4+1)/4²

a = √(5)/16

a= 0.139 ft/s²

Therefore, acceleration of the given particle will be 0.139 feet/ second².

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A physics major is working to pay her college tuition by performing in a traveling carnival. She rides a motorcycle inside a hol

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Answer:

v = 12.1 m/s

Explanation:

When at the top of the circle, there are two forces acting on the combined mass of the rider and the motorcycle.
These are the force of gravity (downward) and the normal force, which is directed from the surface away from it, perpendicular to the surface.
In this case, as the motorcycle runs in the interior of the circle, at the top point this force is completely vertical, and is also downward.
Since the motorcycle is moving in a vertical circle, there must be a force, keeping the object moving around a circle.
This force is the centripetal force, aims towards the center of the circle, and is just the net force aiming in this direction at any point.
At the top point, this force is just the sum of the normal force and the weight of the mass of the rider and the motorcycle combined, as follows (we take the direction towards the center as positive):

$F_{c} = N + m*g (1)$

Now, we know that the centripetal force is related with the tangential speed at this point and the radius of the circle as follows:

$F_{c} = m*\frac{v^{2}}{r} (2)$

Since the normal force takes any value as needed to make (1) equal to (2), if the speed diminishes, it will be needed less force to keep the equality valid.
In the limit, when the motorcyvle tires barely touch the surface, this normal force becomes zero.
In this condition, from (1) and (2), we can find the minimum possible value of the speed that still keeps the motorcycle touching the surface, as follows:
$v_{min} =\sqrt{r*g} =\sqrt{15.0m*9.8m/s2} = 12.1 m/s (3)$

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3 years ago

The change in a pitch of a sound as an object moves closer or farther away is a result of

MrMuchimi

The Doppler effect is a change in the frequency of sound waves that occurs when the source of the sound waves is moving relative to a stationary listener. As the source of sound waves approaches a listener, the sound waves get closer together, increasing their frequency and the pitch of the sound.

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A square coil (length of side = 24 cm) of wire consisting of two turns is placed in a uniform magnetic field that makes an angle

dalvyx [7]

Explanation:

It is given that,

Length of side of a square, l = 24 cm = 0.24 m

The uniform magnetic field makes an angle of 60° with the plane of the coil.

The magnetic field increases by 6.0 mT every 10 ms. We need to find the magnitude of the emf induced in the coil. The induced emf is given by :

$\epsilon=N\dfrac{d\phi}{dt}$

$\dfrac{d\phi}{dt}$ is the rate of change if magnetic flux.

$\phi=BA\ cos\theta$

$\theta$ is the angle between the magnetic field and the normal to area vector.

$\theta=90-60=30$

$\epsilon=NA\dfrac{dB}{dt}\times cos30$

$\epsilon=2\times (0.24\ m)^2\times \dfrac{6\ mT}{10\ mT}\times cos(30)$

$\epsilon=0.0598\ T$

$\epsilon=59.8\ mT$

or

EMF = 60 mT

So, the magnitude of emf induced in the coil is 60 mT. Hence, this is the required solution.

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3 years ago

What is the main reason why fewer nuclear power plants are being built today, compared to 40 years ago?

shusha [124]

<span>nuclear power plants have become to expensive these days

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3 years ago

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.19 s s

ziro4ka [17]

A) Initial velocity of ball 1: 9.53 m/s upward

B) Height of the building: 6.48 m

C) Maximum velocity: 11.7 m/s

D) Minimum velocity: 5.83 m/s

Explanation:

A)

The y-position of the 1st ball at time t is given by the equation for free fall motion:

$y_1 = h + v_0 t - \frac{1}{2}gt^2$ (1)

where

h = 21.0 m is the initial height of the ball, the height of the building

$v_0$ is the initial velocity of the ball, upward

$g=9.8 m/s^2$ is the acceleration of gravity

The y-position of the 2nd ball instead, dropped from the roof 1.19 s later, is given by

$y_2 = h-\frac{1}{2}g(t-1.19)^2$

where

h = 21.0 m is the initial height of the ball, the height of the building

t' = 1.19 s is the delay in time of the 2nd ball (we can verify that at t = 1.19 s, then $y_2=h$ , so the ball is still on the roof

The 2nd ball reaches the ground when $y_2=0$ , so:

$0=h-\frac{1}{2}g(t-1.19)^2\\0=(21.0)-4.9(t^2-2.38t+1.42)\\4.9t^2-11.66t-14.04=0$

Which has two solutions:

t = -0.88 s (negative, we discard it)

t = 3.26 s (this is our solution)

The 1st ball reaches the ground at the same time, so we can substitute t = 3.26 s into eq.(1) and $y_1=0$ , so we find the initial velocity:

$0=h+v_0 t -\frac{1}{2}gt^2\\v_0 = \frac{1}{2}gt-\frac{h}{t}=\frac{1}{2}(9.8)(3.26)-\frac{21.0}{3.26}=9.53 m/s$

B)

In this case, the height of the building h is unknown, while the initial velocity of ball 1 is known:

$v_0 = 8.70 m/s$

When the two balls reach the ground at the same time, there position is the same, so we can write:

$y_1=y_2\\h+v_0 t - \frac{1}{2}gt^2 = h-\frac{1}{2}g(t-1.19)^2$

Solving the equation, we find:

$v_0t=1.19gt-\frac{1}{2}g(1.19)^2\\t=\frac{0.5g(1.19)^2}{1.19g-v_0}=2.34 s$

This is the time at which both balls reache the ground; and substituting into the eq. of ball 2, we find the height of the building:

$0=h-\frac{1}{2}g(t-1.19)^2\\h=0.5g(t-1.19)^2=0.5(9.8)(2.34-1.19)^2=6.48 m$

C)

If $v_0$ is greater than some value $v_{max}$ , then there is no value of h such that the two balls hit the ground at the same time. This situation occurs when the demoninator of the formula found in part b:

$t=\frac{0.5g(1.19)^2}{1.19g-v_0}$

becomes negative: in that case, the time becomes negative, so no solution is possible.

The denominator becomes negative when

$1.19g-v_0 < 0$

Therefore when

$v_0>1.19g=(1.19)(9.8)=11.7 m/s$

So, if the initial velocity of ball 1 is greater than 11.7 m/s, the two balls cannot reach the ground at the same time.

D)

There is also another condition that must be true in order for the two balls to reach the ground at the same time: the time at which ball 1 reaches the ground must be larger than 1.19 s (because ball 2 starts its motion after 1.19 s). This means that the following condition must be true

$t=\frac{0.5g(1.19)^2}{1.19g-v_0}>1.19$

Solving the equation for $v_0$ , we find:

$0.5g(1.19)^2>1.19(1.19g-v_0)\\6.94>13.88-1.19v_0\\1.19v_0>6.94$

Which gives

$v_0>5.83 m/s$

Therefore, the minimum speed of ball 1 at the beginning must be 5.83 m/s.

Learn more about free fall motion:

brainly.com/question/1748290

brainly.com/question/11042118

brainly.com/question/2455974

brainly.com/question/2607086

#LearnwithBrainly

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3 years ago