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Morgarella [4.7K]
3 years ago
14

From the same height and at the same time one ball is dropped and another ball is fired horizontally which ball has the greater

magnitude
Physics
1 answer:
Pani-rosa [81]3 years ago
7 0
The ball thrown horizontally
The weight of the ball dropped down is a factor in magnitude with gravity playing an important role as well.
On the other hand, the ball thrown horizontally has speed which allows magnitude and gravity playing against it, eventually the ball will hit the ground harder than the one dropped straight down
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A cat dozes on a stationary merry-go-round, at a radius of 5.7 m from the center of the ride. Then the operator turns on the rid
tatyana61 [14]

Answer:

0.76

Explanation:

we are given:

radius (r) =5.7 m

speed (s) = 1 revolution in 5.5 seconds

acceleration due to gravity (g) = 9.8 m/s^{2}

coefficient of friction (Uk) = ?

 we can get the minimum coefficient of friction from the equation below

centrifugal force = frictional force

m x r x ω^{2} = Uk x m x g

r x ω^{2} = Uk x g

Uk = \frac{ r x ω^{2} }{g}

where ω (angular velocity) = \frac{2π}{time}

= \frac{2π }{5.5} = 1.14

Uk = \frac{ 5.7 x 1.14^{2} }{9.8} = 0.76

6 0
3 years ago
A 0.2 kg block sliding on a horizontal table slows down from 25 m/s to 20 m/s. How much energy does the block lose due to fricti
Papessa [141]

Answer:

the kinetic energy lost due to friction is 22.5 J

Explanation:

Given;

mass of the block, m = 0.2 kg

initial velocity of the block, u = 25 m/s

final velocity of the block, v = 20 m/s

The kinetic energy lost due to friction is calculated as;

\Delta K.E= K.E_f - K.E_i\\\\\Delta K.E= \frac{1}{2}mv^2 -  \frac{1}{2}mu^2\\\\\Delta K.E= \frac{1}{2}m(v^2 -u^2)\\\\\Delta K.E= \frac{1}{2} \times 0.2 (20^2 - 25^2)\\\\\Delta K.E= -22.5 \ J

Therefore, the kinetic energy lost due to friction is 22.5 J

7 0
3 years ago
Read 2 more answers
A policeman is chasing a criminal across a rooftop at 10 m/s. He decides to jump to the next building which is 2 meters across f
valina [46]

Answer:

They will meet at a distance of 7.57 m

Given:

Initial velocity of policeman in the x- direction, u_{x} = 10 m/s

The distance between the buildings, d_{x} = 2.0 m

The building is lower by a height, h = 2.5 m

Solution:

Now,

When the policeman jumps from a height of 2.5 m, then his initial velocity, u was 0.

Thus

From the second eqn of motion, we can write:

h = ut + \frac{1}{2}gt^{2}

h = \frac{1}{2}gt^{2}

2.5 = \frac{1}{2}\times 10\times t^{2}

t = 0.707 s

Now,

When the policeman was chasing across:

d_{x} = u_{x}t + \frac{1}{2}gt^{2}

d_{x} = 10\times 0.707 + \frac{1}{2}\times 10\times 0.5 = 9.57 m

The distance they will meet at:

9.57 - 2.0 = 7.57 m

   

8 0
3 years ago
How long does it take a car travelling at 60 m.p.h to cover 5 miles?
k0ka [10]

Answer:

Time = 5 minutes

6 0
3 years ago
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Which state of matter is the least compressible
kramer
Incompressible. Compressibility is determine by the amount of space between particles in each state.  
7 0
3 years ago
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