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harina [27]
3 years ago
7

A shotputter throws the shot with an initial speed of 15.6 m/s at a 30.0 degree angle to the horizontal. Calculate the horizonta

l distance traveled by the shot if it leaves the athlete's hand at a height of 2.20m above the ground.
Physics
1 answer:
vfiekz [6]3 years ago
8 0
To calculate the horizontal distance traveled by the shot if it leaves the athlete's hand at a height of 2.20 above the ground we can get the root of the quad equation for time are t=-0.24 or t =1.84 taking the t = 1.84, so the equation will be:
x = 15.6cos(30) * 1.86, x = 24.79m
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1. (30 pts) Let x(t) = cos(πt/2) be a continuous-time signal,
posledela

Given that the function of the wave is f(x) = cos(π•t/2), we have;

a. The graph of the function is attached

b. 4 units of time

c. Even

d. 4.935 J/kg

e. 1.234 W/kg

<h3>How can the factors of the wave be found?</h3>

a. Please find attached the graph of the signal created with GeoGebra

b. The period of the signal, T = 2•π/(π/2) = <u>4</u>

c. The signal is <u>even</u>, given that it is symmetrical about the y-axis

d. The energy of the signal is given by the formula;

\frac{1}{2}  \cdot  \mu^{2} \cdot \omega ^{2}  \cdot \:  {a}^{2}  \times  \lambda

Which gives;

E = 0.5 × 1.571² × 1² × 4 = <u>4.935 J/kg</u>

e. The power of the wave is given by the formula;

E = 0.5 × 1.571² × 1² × 4 × 0.25 = <u>1.234 W/</u><u>kg</u>

Learn more about waves here:

brainly.com/question/14015797

8 0
2 years ago
A rod 16.0 cm long is uniformly charged and has a total charge of -25.0 µC. Determine the magnitude and direction of the electri
Vlada [557]

Answer:

-1.4x10^6N/C

Explanation:

Pls see attached file

8 0
3 years ago
A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.35 kg to a friend standing in front of him
nataly862011 [7]

Answer:

a) u_c=0\ m.s^{-1}       &        m_c.v_c=m_b.v_b\times \cos\theta

b) v_c=0.0566\ m.s^{-1}

c) p_e=2.9218\ kg.m.s^{-1}

Explanation:

Given:

mass of the book, m_b=1.35\ kg

combined mass of the student and the skateboard, m_c=97\ kg

initial velocity of the book, v_b=4.61\ m.s^{-1}

angle of projection of the book from the horizontal, \theta=28^{\circ}

a)

velocity of the student before throwing the book:

Since the student is initially at rest and no net force acts on the student so it remains in rest according to the Newton's first law of motion.

u_c=0\ m.s^{-1}

where:

u_c= initial velocity of the student

velocity of the student after throwing the book:

Since the student applies a force on the book while throwing it and the student standing on the skate will an elastic collision like situation on throwing the book.

m_c.v_c=m_b.v_b\times \cos\theta

where:

v_c= final velcotiy of the student after throwing the book

b)

m_c.v_c=m_b.v_b\times \cos\theta

97\times v_c=1.35\times 4.61\cos28

v_c=0.0566\ m.s^{-1}

c)

Since there is no movement of the student in the vertical direction, so the total momentum transfer to the earth will be equal to the momentum of the book in vertical direction.

p_e=m_b.v_b\sin\theta

p_e=1.35\times 4.61\times \sin28^{\circ}

p_e=2.9218\ kg.m.s^{-1}

6 0
3 years ago
a golfer hits a 0.05 kg golf ball with an impulse of 3 N-s what is the change in velocity of the ball
Rainbow [258]

so your saying the start is 0 N and when he/she hits the ball its inertia is 3 N. if that is so m*v=

.05*3=<u>.15</u>

7 0
3 years ago
Which element is found in period 3, group 2?
kicyunya [14]

Answer:

Mg (Magnesium)

Explanation:

7 0
3 years ago
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