Physics, 09.10.2019 21:00 jjjones2233

From the window of a building, a ball is tossed from a height y0 above the ground with an initial velocity of 8.10 m/s and angle of 18.0° below the horizontal. it strikes the ground 4.00 s later. (a) if the base of the building is taken to be the origin of the coordinates, with upward the positive y-direction, what are the initial coordinates of the ball? (use the following as necessary: y0.) xi = yi = (b) with the positive x-direction chosen to be out the window, find the x- and y-components of the initial velocity. vi, x = m/s vi, y = m/s (c) find the equations for the x- and y- components of the position as functions of time. (use the following as necessary: y0 and t. let the variable t be measured in seconds.) x = m y = m (d) how far horizontally from the base of the building does the ball strike the ground? m (e) find the height from which the ball was thrown. m (f) how long does it take the ball to reach a point 10.0 m below the level of launching? s

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Answer from: nothingworksoutforme

Part a)

y = 88.5 m

Part b)

v_x = 7.7 m/s

v_y = 2.5 m/s

Part c)

equation of position in x direction is given as

x = v_x t

equation of position in y direction is given as

$y = v_y t + \frac{1}{2}gt^2$

Part d)

x = 30.8 m

Part e)

H = 88.5 m

Part f)

t = 1.2 s

Explanation:

As we know that ball is projected with speed

v = 8.10 m/s at an angle 18 degree below the horizontal

so we will have

v_x = 8.10 cos18

v_x = 7.7 m/s

v_y = 8.10 sin18

v_y = 2.5 m/s

Part a)

Since it took t =4 s to reach the ground

so its initial y coordinate is given as

$y = v_y t + \frac{1}{2}a_y t^2$

$y = 2.5(4) + \frac{1}{2}(9.81)(4^2)$

y = 88.5 m

Part b)

components of the velocity is given as

v_x = 8.10 cos18

v_x = 7.7 m/s

v_y = 8.10 sin18

v_y = 2.5 m/s

Part c)

equation of position in x direction is given as

x = v_x t

equation of position in y direction is given as

$y = v_y t + \frac{1}{2}gt^2$

Part d)

distance where it will strike the floor is given as

x = v_x t

$x = 7.7 \times 4$

x = 30.8 m

Part e)

Height from which it is thrown is same as initial y coordinate of the ball

so it is given as

H = 88.5 m

Part f)

time taken by ball to reach 10 m below is given as

$y = v_y t + \frac{1}{2}gt^2$

$10 = 2.5t + \frac{1}{2}(9.81) t^2$

t = 1.2 s

Answer from: Quest

a solenoid is created by wrapping a l = 90 m long wire around a hollow tube of diameter d = 4.5 cm. the wire diameter is d = 0.9 mm. the solenoid wire is then connected to a power supply so that a current of i = 9 a flows through the wire.

randomized variablesl = 90 m

d = 4.5 cm

d = 0.9 mm

i = 9 a

write an expression for the number of turns, n, in the solenoid. you do not need to take into account the diameter of the wire in this calculation.

calculate the number of turns, n, in the solenoid.

write an expression for the length of the solenoid (l2) in terms of the diameter of the hollow tube d.

calculate the length of the solenoid (l2) in meters.

calculate the magnitude of the magnetic field at the center of the solenoid in teslas.

explanation:

Answer from: Quest

answer would be letter choice (a), a ball resting on a shelf. would be another example of potential energy > stored energy > not moving.

a ball spinning on a finger > kinetic energy > moving energy

a ball rolling on the ground > kinetic energy > moving energy

a ball flying through the air > kinetic energy > moving energy

hope that : )

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