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Anna [14]
3 years ago
8

Xavier throws a tennis ball to his friend Olaf at a 20° angle relative to the ground with an initial velocity of 18 m/s. The ten

nis ball travels 30 m until Olaf catches itThe time the tennis ball was in the air, rounded to the nearest hundredth, is s.
Physics
1 answer:
Svetllana [295]3 years ago
3 0

Ball is thrown at an angle of 20 degree with velocity 18 m/s

so first we will find the two components of the velocity

v_x = 18 cos20

v_x = 16.91 m/s

similarly for other component

v_y = 18 sin20

v_y = 6.16 m/s

now the ball will cover horizontal distance of 30 m till it reach to other player

so here we can use the formula of time

x = v_x * t

30 = 16.91 * t

t = \frac{30}{16.91}

t = 1.77 s

so it will take 1.77 s to reach the other player

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Derive an expression for the block's centripetal acceleration ac in terms of m , θ , and physical constants, as appropriate
maxonik [38]

The expression for the block's centripetal acceleration is derived as ω²r or v²/r.

<h3>What is centripetal acceleration?</h3>

The centripetal acceleration of an object is the inward or radial acceleration of an object moving in a circular path.

The expression for the block's centripetal acceleration is derived as follows;

ω = dθ/dt

where;

  • ω is the angular speed
  • θ is the angular displacement
  • t is the time of motion

ac = ω²r

where;

  • r is the radius of the circular path

Also, ω = v/r

ac = (v/r)²r

ac = v²/r

Thus, the expression for the block's centripetal acceleration is derived as ω²r or v²/r.

Learn more about centripetal acceleration here: brainly.com/question/79801

7 0
2 years ago
Se deja caer una pelota inicialmente en reposo desde una altura de 50m sobre el nivel del suelo. ¿cuanto tiempo requiere para ll
umka2103 [35]

Answer:

a) t = 3.2 s

b) v_{f} = -32 m/s

Explanation:

a) El tiempo requerido para llegar al suelo se puede calcular usando la siguiente fórmula:

t = \sqrt{\frac{2y_{0}}{g}}

En donde:

y_{0}: es la altura inicial = 50 m

g: es la gravedad = 10 m/s²

t = \sqrt{\frac{2y_{0}}{g}} = \sqrt{\frac{2*50 m}{10 m/s^{2}}} = 3.2 s

Entonces, el tiempo requerido para llegar al suelo es 3.2 s.

b) La rapidez de la pelota justo antes del choque es el siguiente:

v_{f} = v_{0} - gt

En donde:

v_{0}: es la velocidad inicial = 0 (dado que se deja caer en resposo)

v_{f} = v_{0} - gt = 0 - 10 m/s^{2}*3.2 s = -32 m/s

Por lo tanto, la rapidez de la pelota justo en el momento anterior del choque es -32 m/s (el signo negativo es porque la pelota está cayendo).

Espero que te sea de utilidad!

6 0
3 years ago
What has more energy a 2 kg mass at 10 m/s or a 2 kg mass 5 m above the ground
Alexxandr [17]

Answer:

Both have same energy

Explanation:

Given:

Sample 1

Mass m = 2 kg

Velocity v = 10 m/s

Sample 2

Mass m = 2 kg

Height h = 5 m

Assume g = 10 m/s²

Find:

What has more energy

Computation:

In sample 1

Ke = 1/2(m)(v)²

Ke = 1/2(2)(10)²

Ke = 100 joule

In sample 2

Pe = mgh

Pe = (2)(10)(5)

Pe = 100  joules

Both have same energy

7 0
3 years ago
A car parked on level pavement exerts a force of 10,000 newtons on the ground. What force does the pavement exert back on the ca
natka813 [3]

Answer:

Normal force of 10,000N

Explanation:

From the question, the weight the car exerts on the pavement is 10,000N.

The pavement exerts upward and perpendicular contact force called normal force on the car to support its weight. Also, the normal force is equal and opposite to the weigh of the car.

Hence the pavement exerts normal force of 10,000N back on the car to prevent it from passing through it.

7 0
4 years ago
Topic: Chapter 10: Projectory or trajectile?
Charra [1.4K]

Answer:

The other angle is 75⁰

Explanation:

Given;

velocity of the projectile, v = 10 m/s

range of the projectile, R = 5.1 m

angle of projection, 15⁰

The range of a projectile is given as;

R = \frac{u^2sin(2\theta)}{g}

To find another angle of projection to give the same range;

5.1 = \frac{10^2 sin(2\theta)}{9.81} \\\\100sin(2\theta) = 50\\\\sin(2\theta) = 0.5\\\\2\theta = sin^{-1}(0.5)\\\\2\theta = 30^0\\\\\theta = 15^0\\\\since \ the \ angle \ occurs \ in \ \ the \ first \ quadrant,\  the \ equivalent \ angle \\ is \ calculated \ as;\\\\90- \theta = 15^0\\\\\theta = 90 - 15^0\\\\\theta = 75^0

<u>Check: </u>

sin(2θ) = sin(2 x 75) = sin(150) = 0.5

sin(2θ) = sin(2 x 15) = sin(30) = 0.5

5 0
3 years ago
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