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3 years ago

A ball is dropped from rest on a cliff what is the speed of the ball 5 seconds later

Physics

1 answer:

diamong [38]3 years ago

4 0

Answer:

The velocity of the ball after 5 seconds will be 49 m/s

Explanation:

v = final velocity

u = initial velocity

g = acceleration due to gravity

t = time

Initial velocity of the ball = 0 (As the ball is dropped from rest )

Acceleration due to gravity = 9.8 m/s

Time taken = 5 sec

As the acceleration due to gravity is constant in both the cases we can use the equations of motion in order to solve this question

Part I :- As we already know the values of u,g,ant t we can use the first equation of motion in order to find v

Part II :- As we know the values of u, t , g we can use the second equation of motion in order to find s.

Velocity of the ball after 5 seconds

Distance covered by the ball in 5 sec

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If you swim with the current in a river, your speed is increased by the speed of the water; if you swim against the current, you

Blababa [14]

Answer:

11.23%

Explanation:

Lets take

Speed of man in still water =u= 1.73 m/s

Speed of flow of water = v=0.52 m/s

When swims in downward direction then speed of man = u + v

When swims in upward direction then speed of man = u - v

Lets time taken by man when he swims in downward direction is $t_1$ and when he swims in downward direction is $t_2$

Lets distance is d and it will be remain constant in both the case

$d=(u+v)t_1$

$d=(u-v)t_2$

$(1.73+0.52)t_1=(1.73-0.52)t_2$

$t_2=1.85t_1$

Time taken in still water

2 d= t x 1.73

t=1.15 x d sec

$t_1=0.44d\ sec$

$t_2=0.82d\ sec$

total time in current = 0.82 +0.44 d=1.26 d sec

So the percentage time

$percentage\ time =\dfrac{1.28-1.15}{1.15}$

Percentage time =11.32%

So it will take 11.32% more time as compare to still current.

5 0

3 years ago

The coefficent of static friction between the floor of a truck and a box resting on it is 0.38. The truck is traveling at 87.9 k

nekit [7.7K]

Answer:

$d=79.9m$

Explanation:

From the question we are told that:

coefficient of static friction $\mu=0.38$

Velocity $v=87.9=>24.41667m/s$

Generally the equation for Conservation of energy is mathematically given by

$\mu*mgd = 0.5 m v^2$

$d=\frac{0.5*24.42^2}{0.38*9.8}$

$d=79.9m$

3 0

3 years ago

A 4.1-kg ball is thrown from the top of a 4-m tall building (point A) with a speed of 1 m/s at an angle 2 degrees above the hori

lubasha [3.4K]

Answer:

$K.E = 163 J$

Explanation:

From the question we are told that:

Mass $m=4.1kg$

Height $h=4m$

Speed $V=1m/s$

Angle $\theta= 2 \textdegree$

Generally the equation for K.E is mathematically given by

Since

The potential energy and kinetic energy is equal to the kinetic energy as it hits the ground.

Therefore

$K.E=mgH + 0.5mV^2$

$K.E = (4.1)(9.81)(4) + (4.1)(1)^2*0.5$

$K.E = 163 J$

6 0

3 years ago

A listener triples his distance from a source that emits sound uniformly in all directions. by how many decibels does the sound

Novay_Z [31]

Sound intensity goes inversely as the square of the distance from the source. This is easy to understand if you think of the total sound energy spread out over the surface of a sphere with the source at the center. The surface area of the sphere goes as r^2, so the power density goes inversely as such.

The decibel change is simply

Change = 10 log10( (1/3)^2 ) = -9.54 dB

8 0

3 years ago

Please help me on this 1 assignment im giving 25 points GOOD LUCK!

kati45 [8]

Answer:

Explanation:

im sorry if this doesn't help but i think its A

3 0

3 years ago