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Natasha_Volkova [10]
3 years ago
7

Starting at the same time, an arrow and a ball are shot horizontally with a speeds of 50 m/s and 44 m/s respectively from the to

p of a building over flat ground. Which best describes what happens? A. The arrow and the ball land the same distance from the building at different times. B. The arrow and the ball land at different distances from the building at different times. C. The arrow and the ball land the same distance from the building at the same time. D. The arrow and the ball land at different distances from the building at the same time.
Physics
1 answer:
Yuri [45]3 years ago
5 0

Answer:

option D

Explanation:

given.

horizontal velocity of arrow and a ball given as  50 m/s and 44 m/s respectively from the top of a building over flat ground.

In vertical direction, they are both identical

In vertical direction the initial velocity of arrow and a ball  is 0  m/s

Their acceleration due to gravity  is same for both arrow and a ball  9.8 m/s²

they will react bottom at the same time

 time of flight is same for both

now,

In horizontal direction,

distance = speed × time

Since speed is more for arrow, it will travel more horizontal distance  at the same time.

the correct answer is option D

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Need answer ASAP!!!
Alina [70]

Hello!

Use <u>ohm law:</u>

I = V / R

Replacing:

I = 100 V / 20 O

I = 5 A

The current is <u>5 amperes.</u>

5 0
3 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
It takes 9 sec for a 10 newton force to move an object 4 meters to the right. What is direction &amp; magnitude of the force
lana66690 [7]

Answer:

the question is wrong

Explanation:

  1. M is not given
  2. after 9 second the acceleration multiply by the time divided by two then multiplied by the time is equal to 4 meter
  3. ((10/Mm/s *9s-)
3 0
4 years ago
If it takes a planet 2.8 × 108 s to orbit a star with a mass of 6.2 × 1030 kg, what is the average distance between the planet
Shtirlitz [24]

The average distance between the planet and the star is:

R=9.36*10^11 m

Orbital velocity  v=√{(G*M)/R},

G = gravitational constant =6.67*10^-11 m³ kg⁻¹ s⁻²,

M = mass of the star

R =distance from the planet to the star.

v=ωR, with ω as the angular velocity and R the radius

ωR=√{(G*M)/R},

ω=2π/T,

T = orbital period of the planet

To get R we write the formula by making R the subject of the equation

(2π/T)*R=√{(G*M)/R}

{(2π/T)*R}²=[√{(G*M)/R}]²,

(4π²/T²)*R²=(G*M)/R,

(4π²/T²)*R³=G*M,

R³=(G*M*T²)/4π²,

R=∛{(G*M*T²)/4π²},

Substitute values

R=9.36*10^11 m

As was already said, Earth is located roughly 150 million kilometres (93 million miles) from the Sun on average. It is 1 AU. Mars is on our fictitious football field's three-yard line. On average, the distance between the Sun and the red planet is around 142 million miles (228 million kilometres).

Learn more about average distance:

brainly.com/question/18366547

#SPJ4

The complete question is ''If it takes a planet 2.8 × 108 s to orbit a star with a mass of 6.2 × 10^30 kg, what is the average distance between the planet and the star? 1.43 × 10^9 m 9.36 × 10^11 m 5.42 × 10^13 m 9.06 × 10^17 m''.

4 0
1 year ago
A 2,000-kg test car, traveling 60 m/s hits a brick wall. Using motion pictures, the time involved is determined to be 0.050 s. W
Naily [24]

change in momentum of car is given as

\Delta P = P_f - P_i

here we have

\Delta P = m(v_f - v_i)

given that

m = 2000 kg

v_f = 0

v_i = 60 m/s

now we have

\Delta P = 2000(0 - 60)

\Delta P = - 1.2 \times 10^5 kg m/s

8 0
3 years ago
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