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taurus [48]
3 years ago
13

A boy throws a baseball onto a roof and it rolls back down and off the roof with a speed of 3.05 m/s. If the roof is pitched at

40.0° below the horizon and the roof edge is 4.50 m above the ground, find the time the baseball spends in the air and the horizontal distance from the roof edge to the point where the baseball lands on the ground.
Physics
1 answer:
vekshin13 years ago
7 0

1) Time in the air: 0.78 s

The motion of the ball is a projectile motion, which consists of two independent motions:

- A horizontal motion with constant horizontal velocity

- A vertical motion with constant downward acceleration of

g=-9.8 m/s^2 (acceleration of gravity)

The initial vertical velocity is

u_y = u sin \theta = (3.05)(sin(-40^{\circ}))=-1.96 m/s

where the negative sign means the direction is downward.

The vertical position of the ball is given by

y(t) = h + u_y t + \frac{1}{2}gt^2

where

h = 4.50 m is the initial heigth of the ball when it starts falling down

The ball reaches the ground when y = 0, so we have:

0 = 4.50 -1.96t-4.9t^2

This is a second-order equation; solving for t, we get

t = -1.18 s

t = 0.78 s

We discard the negative solution since it has no physical meaning, so we can say that the ball spent 0.78 s in the air.

2) Horizontal distance: 1.83 m

For this second part of the problem, we just have to consider the horizontal motion of the ball.

As we said previously, the motion of the ball along the horizontal direction is a uniform motion with constant velocity, which is given by

v_x = u cos \theta = (3.05)(cos (-40.0^{\circ}))=2.34 m/s

where u = 3.05 m/s is the initial speed and \theta the angle of projection.

For a uniform motion, we can use the following relationship between distance covered and velocity:

d=v_x t

and substituting t = 0.78 s, we find the total distance travelled along the horizontal direction by the ball before reaching the ground:

d=(2.34)(0.78)=1.83 m

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hram777 [196]

Answer:

8 signals received by twin A during the trip.

Explanation:

Given that,

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Speed = 0.6 c

Time = 1 year

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Using formula of time

T=t\sqrt{\dfrac{1+\dfrac{v}{c}}{1-\dfrac{v}{c}}}

Put the value into the formula

T=1\sqrt{\dfrac{1+0.6}{1-0.6}}

T=2\ years

Similarly,

The expression for distance cover by A

D=d\sqrt{1-\dfrac{v^2}{c^2}}

D=12\sqrt{1-(0.6)^2}

D=9.6\ ly

We need to calculate the time

Using formula of time

t=\dfrac{D}{v}

t=\dfrac{9.6}{0.6}

t=16\ years

We need to calculate the signals received by twin A

Using formula for number of signals

n=\dfrac{t}{T}

Put the value into the formula

n=\dfrac{16}{2}

n=8\ signals

Hence, 8 signals received by twin A during the trip.

7 0
3 years ago
The potential energy of a particle as a function of position will be given as U(x) = A x2 + B x + C, where U will be in joules w
satela [25.4K]

Answer:

F = - 2 A x - B

Explanation:

The force and potential energy are related by the expression

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Where i ^, j ^, k ^ are the unit vectors on the x and z axis

The potential they give us is

     U (x) = A x² + B x + C

Let's calculate the derivatives

    dU / dx = A 2x + B + 0

The other derivatives are zero because the potential does not depend on these variables.

Let's calculate the strength

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3 0
3 years ago
A force of 960 newtons stretches a spring 4 meters. A mass of 60 kilograms is attached to the end of the spring and is initially
Drupady [299]

Answer:

x(t) = - 6 cos 2t

Explanation:

Force of spring = - kx

k= spring constant

x= distance traveled by compressing

But force = mass × acceleration

==> Force = m × d²x/dt²

===> md²x/dt² = -kx

==> md²x/dt² + kx=0   ------------------------(1)

Now Again, by Hook's law

Force = -kx

==> 960=-k × 400

==> -k =960 /4 =240 N/m

ignoring -ve sign k= 240 N/m

Put given data in eq (1)

We get

60d²x/dt² + 240x=0

==> d²x/dt² + 4x=0

General solution for this differential eq is;

x(t) = A cos 2t + B sin 2t   ------------------------(2)

Now initially

position of mass spring

at time = 0 sec

x (0) = 0 m

initial velocity v= = dx/dt=  6m/s

from (2) we have;

dx/dt= -2Asin 2t +2B cost 2t = v(t) --- (3)

put t =0 and dx/dt = v(0) = -6 we get;

-2A sin 2(0)+2Bcos(0) =-6

==> 2B = -6

B= -3

Putting B = 3 in eq (2) and ignoring first term (because it is not possible to find value of A with given initial conditions) - we get

x(t) = - 6 cos 2t

==>  

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Basile [38]

Answer:ball, ramp, and angle

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8 0
2 years ago
A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the c
puteri [66]

Correct question:

A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the center of the solenoid is 2.8 x 10⁻² T, what is the number of turns per meter for this solenoid?

Answer:

the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.

Explanation:

Given;

length of solenoid, L= 0.35 m

diameter of the solenoid, d = 0.04 m

current through the solenoid, I = 5.0 A

magnetic field in the center of the solenoid, 2.8 x 10⁻² T

The number of turns per meter for the solenoid is calculated as follows;

B =\mu_o  I(\frac{N}{L} )\\\\B =  \mu_o  I(n)\\\\n = \frac{B}{\mu_o I} \\\\n = \frac{2.8 \times 10^{-2}}{4 \pi \times 10^{-7} \times 5.0} \\\\n = 4.5 \times 10^3 \ turns/m

Therefore, the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.

3 0
2 years ago
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