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Andre45 [30]
3 years ago
5

A soccer ball is kicked from the ground with an initial speed of 20.1 m/s at an upward angle of 47.8˚. A player 53.0 m away in t

he direction of the kick starts running to meet the ball at that instant. What must be his average speed if he is to meet the ball just before it hits the ground? Neglect air resistance.
Physics
1 answer:
e-lub [12.9K]3 years ago
5 0

Answer:3.95 m/s

Explanation:

Given

Initial velocity(u)=20.1 m/s

launch angle=47.8^{\circ}

Player is 53 m away

Range of projectile

R=\frac{u^2sin2\theta }{g}

R=\frac{20.1^2\times sin95.6}{9.8}

R=41.02 m

so he need to run 53-41.02=11.98 m

Time of flight of projectile =\frac{2usin\theta }{g}

T=3.03 s

thus average speed of boy s_{avg}=\frac{11.98}{3.03}=3.95 m/s

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A heavy hall with a mass of 3.5 kg is observed to accelerate
LenKa [72]

Answer:

<h2>21 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 3.5 × 6

We have the final answer as

<h3>21 N</h3>

Hope this helps you

6 0
3 years ago
An astronaut has a mass of 75 kg and is floating in space 500 m from his 125,000 kg spacecraft. What will be the force of gravit
nikdorinn [45]

Answer:

1. 2.5×10¯⁹ N

2. 3.33×10¯¹¹ m/s²

Explanation:

1. Determination of the force of attraction.

Mass of astronaut (M₁) = 75 Kg

Mass of spacecraft (M₂) = 125000 Kg

Distance apart (r) = 500 m

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Force of attraction (F) =?

The force of attraction between the astronaut and his spacecraft can be obtained as follow:

F = GM₁M₂ /r²

F = 6.67×10¯¹¹ × 75 × 125000 / 500²

F = 2.5×10¯⁹ N

Thus, the force of attraction between the astronaut and his spacecraft is 2.5×10¯⁹ N

2. Determination of the acceleration of the astronaut.

Mass of astronaut (m) = 75 Kg

Force (F) = 2.5×10¯⁹ N

Acceleration (a) of astronaut =?

The acceleration of the astronaut can be obtained as follow:

F = ma

2.5×10¯⁹ N = 75 × a

Divide both side by 75

a = 2.5×10¯⁹ / 75

a = 3.33×10¯¹¹ m/s²

Thus, the acceleration the astronaut is 3.33×10¯¹¹ m/s²

5 0
2 years ago
Write the device modeling equation for the first and second law of thermodynamics
anyanavicka [17]
1st <span>the total </span>energy<span> of an </span>isolated system<span> is constant; energy can be transformed from one form to another, but can be neither created nor destroyed. ▲U=Q-W
</span><span> 
2nd the total </span>entropy<span> can never decrease over time for an </span>isolated system, that is, a system in which neither energy nor matter can enter nor leave.
DS (Greater than or equal to) 0
8 0
3 years ago
Investigators are most likely to use the case history method when they study
user100 [1]
Investigators are most likely to use the case history method when they study <span>a rare behavior or an unusual person.
They do this to obtain some sort of basis that they could use as a pointer to make their decision regarding the similar case (after figuring out the difference in situation between each period)</span>
4 0
3 years ago
A small, 300 g cart is moving at 1.20 m/s on an air track when it collides with a larger, 2.00 kg cart at rest?
stiv31 [10]

Answer:

The speed of the large cart after collision is 0.301 m/s.

Explanation:

Given that,

Mass of the cart, m_1 = 300\ g = 0.3\ kg

Initial speed of the cart, u_1=1.2\ m/s

Mass of the larger cart, m_2 = 2\ kg

Initial speed of the larger cart, u_2=0

After the collision,

Final speed of the smaller cart, v_1=-0.81\ m/s (as its recolis)

To find,

The speed of the large cart after collision.

Solution,

Let v_2 is the speed of the large cart after collision. It can be calculated using conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

m_1u_1+m_2u_2-m_1v_1=m_2v_2

v_2=\dfrac{m_1u_1+m_2u_2-m_1v_1}{m_2}

v_2=\dfrac{0.3\times 1.2+0-0.3\times (-0.81)}{2}

v_2=0.301\ m/s

So, the speed of the large cart after collision is 0.301 m/s.

4 0
3 years ago
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