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1 year ago

A ball is thrown upward from the top of a building at an angle of 30.0° to the horizontal and with

Physics

1 answer:

Ray Of Light [21]1 year ago

8 0

Answer:

See below

Explanation:

Vertical position = 45 + 20 sin (30) t - 4.9 t^2

when it hits ground this = 0

0 = -4.9t^2 + 20 sin (30 ) t + 45

0 = -4.9t^2 + 10 t +45 = 0 solve for t =4.22 sec

max height is at t= - b/2a = 10/9.8 =1.02

use this value of 't' in the equation to calculate max height = 50.1 m

it has 4.22 - 1.02 to free fall = 3.2 seconds free fall

v = at = 9.81 * 3.2 = 31.39 m/s VERTICAL

it will also still have horizontal velocity = 20 cos 30 = 17.32 m/s

total velocity will be sqrt ( 31.39^2 + 17.32^2) = 35.85 m/s

Horizontal range = 20 cos 30 * t = 20 * cos 30 * 4.22 = 73.1 m

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Two resistors, R1=3.85 Ω and R2=6.47 Ω , are connected in series to a battery with an EMF of 24.0 V and negligible internal resi

Bond [772]

Answer:

(a) 2.33 A

(b) 15.075 V

Explanation:

From the question,

The total resistance (Rt) = R1+R2 = 3.85+6.47

R(t) = 10.32 ohms.

Applying ohm's law,

V = IR(t)..........equation 1

Where V = Emf of the battery, I = current flowing through the circuit, R(t) = combined resistance of both resistors.

Note: Since both resistors are connected in series, the current flowing through them is the same.

Therefore,

I = V/R(t)............. Equation 2

Given: V = 24 V, R(t) = 10.32 ohms

Substitute these values into equation 2

I = 24/10.32

I = 2.33 A.

Hence the current through R1 = 2.33 A.

V2 = IR2.............. Equation 3

V2 = 2.33(6.47)

V2 = 15.075 V

7 0

3 years ago

A car moves at 20 m s' for 15 minutes. Calculate the distance travelled

Firdavs [7]

Explanation:

Calculating this according to the m/s rate

Now solving the question

If the car goes 20 m/s mathematically we can generate it as

15 × 60 = 600 secs

If the car goes 20 meters per every second it means

The car will go 20 meters for 900 secs

Which is 900 ×20 = 18000m/s

=18000÷1000 = 1km

Therefore the answer is 18km

6 0

2 years ago

A particle's trajectory is described by x = (0.5t^3-2t^2) meters and y = (0.5t^2-2t), where time is in seconds. What is the part

mestny [16]

Differentiate the components of position to get the corresponding components of velocity :

$v_x = \dfrac{\mathrm dx}{\mathrm dt} = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) t^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)t$

$v_y = \dfrac{\mathrm dy}{\mathrm dt} = \left(1\dfrac{\rm m}{\mathrm s^2}\right)t-2\dfrac{\rm m}{\rm s}$

At t = 5.0 s, the particle has velocity

$v_x = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) (5.0\,\mathrm s)^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s) = 17.5\dfrac{\rm m}{\rm s}$

$v_y = \left(1\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)-2\dfrac{\rm m}{\rm s} = 3.0\dfrac{\rm m}{\rm s}$

The speed at this time is the magnitude of the velocity :

$\sqrt{{v_x}^2 + {v_y}^2} \approx \boxed{17.8\dfrac{\rm m}{\rm s}}$

The direction of motion at this time is the angle $\theta$ that the velocity vector makes with the positive x-axis, such that

$\tan(\theta) = \dfrac{3.0\frac{\rm m}{\rm s}}{17.5\frac{\rm m}{\rm s}} \implies \theta = \tan^{-1}\left(\dfrac{3.0}{17.5}\right) \approx \boxed{9.73^\circ}$