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valkas [14]
2 years ago
11

A ball is dropped from the rest from a height of 6. 0 meters above the ground. The ball falls freely and reaches the ground at t

he 1.1 seconds later. What is the average speed of the ball
Physics
2 answers:
igor_vitrenko [27]2 years ago
8 0
What is the velocity of the ball<span> when it </span>reaches<span> its highest point? ... Then double it for the "hang time"-the time one's feet are off the </span>ground<span>. ... Calculate the </span>speed<span> of a bowling </span>ball<span> that travels 5.0 </span>meters<span> in 2.4</span>seconds<span>. ... An object </span>falls freely<span> from </span>rest<span> on a planet where the acceleration due to gravity is twice as much as it is ...</span>6';"'A ball is dropped<span> from </span>rest from a height<span> 6.0 </span>meters above<span> t e </span>ground<span>. The hall </span><span>falls freely and reaches the ground 1.1 seconds later.

</span>
makkiz [27]2 years ago
8 0
The speed is =6m/1.1s

Therefore the speed is 5.454545... meter per second
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The blade of metal cutter is shorter than that scissors of tailors<br><br>​
Naily [24]

Explanation:it is beause they are sharper and also have less surface area and therefore more pressure

8 0
2 years ago
1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n
sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

(1).\: x =v_0t

(2).\: y= 15m-\dfrac{1}{2}gt^2

where v_0 is the initial velocity.

(a).

When the projectile hits the 50m mark, y=0; therefore,

0=15-\dfrac{1}{2}gt^2

solving for t we get:

t= 1.75s.

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

50m = v_0(1.75s)

which gives

\boxed{v_0 = 28.58m/s.}

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

v_x = 28.58m/s.

the vertical component of the velocity is

v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.

which gives a speed v of

v = \sqrt{v_x^2+v_y^2}

\boxed{v =33.3m/s.}

4 0
1 year ago
A rectangular loop of wire of width 10 cm and length 20 cm has a current of 2.5 A flowing through it. Two sides of the loop are
Dahasolnce [82]

Answer:

(a) 0.05 Am^2

(b) 1.85 x 10^-3 Nm

Explanation:

width, w = 10 cm = 0.1 m

length, l = 20 cm = 0.2 m

Current, i = 2.5 A

Magnetic field, B = 0.037 T

(A) Magnetic moment, M = i x A

Where, A be the area of loop

M = 2.5 x 0.1 x 0.2 = 0.05 Am^2

(B) Torque, τ = M x B x Sin 90

τ = 0.05 x 0.037 x 1

τ = 1.85 x 10^-3 Nm

4 0
2 years ago
A 0.0140 kg bullet traveling at 205 m/s east hits a motionless 1.80 kg block and bounces off it, retracing its original path wit
makvit [3.9K]

Answer:

Final velocity of the block = 2.40 m/s east.

Explanation:

Here momentum is conserved.

Initial momentum = Final momentum

Mass of bullet = 0.0140 kg

Consider east as positive.

Initial velocity of bullet = 205 m/s

Mass of Block = 1.8 kg

Initial velocity of block = 0 m/s

Initial momentum = 0.014 x 205 + 1.8 x 0 = 2.87 kg m/s

Final velocity of bullet = -103 m/s

We need to find final velocity of the block( u )

Final momentum = 0.014 x -103+ 1.8 x u = -1.442 + 1.8 u

We have

            2.87 = -1.442 + 1.8 u

               u = 2.40 m/s

Final velocity of the block = 2.40 m/s east.

7 0
2 years ago
A car is traveling in a race.The car went from initial velocity of 35m/s to the final velocity of 65m/s in 5 seconds what was th
I am Lyosha [343]
Acceleration is the change in velocity divided by time. The change in velocity is -30m/s and time is 5s. If you divide -30m/s by 5s, you get -6m/s<span>².</span>
8 0
2 years ago
Read 2 more answers
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