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3 years ago

What is 9/15 as a percentage

1 answer:

4 0

The cheat was is to punch it in a calculator and multiply the answer by 100. But let's actually figure it out.

Get the denominator to a factor of 100.
The most convenient one for this is 5, since that's also a factor of 15.
Reduce the fraction with 3/3 (divide both the numerator and denominator by 3, in case that's unclear), and you get 3/5.
Multiply both by 20, and you get 60/100.
60%

You might be interested in

In which position would an object have a greater amount of potential energy

valentinak56 [21]

Answer:

At the maximum height

Explanation:

The gravitational potential energy of an object is the energy possessed by the object due to its position in the gravitational field.

Mathematically, it is given by:

$PE=mgh$

where

m is the mass of the object

g is the acceleration due to gravity

h is the height of the object relative to the ground

From the formula, we see that the potential energy is directly proportional to the height of the object above the ground: therefore, the larger the value of h, the larger the potential energy of the object.

Therefore, the object has the maximum potential energy when it is at the highest position above the ground.

7 0

3 years ago

Read 2 more answers

A 4.25-g bullet traveling horizontally with a velocity of magnitude 375 m/s is fired into a wooden block with mass 1.12 kg, init

Tems11 [23]

Answer:

0.9904 m/s

Explanation:

To solve this problem we need to use the conservation of momentum:

m1v1 + m2v2 = m1'v1' + m2'v2'

Using this equation, we have:

m_bullet*v_bullet = m_bullet_after*v_bullet_after + m_block*v_block

0.00425 * 375 = 0.00425 * 114 + 1.12 * v_block

1.12 * v_block = 1.5938 - 0.4845

1.12 * v_block = 1.1093

v_block = 1.1093 / 1.12

v_block = 0.9904 m/s

3 0

4 years ago

Read 2 more answers

A subway train starts from rest at a station and accelerates at a rate of 1.68 m/s2 for 14.2 s. It runs at constant speed for 68

liubo4ka [24]

Answer:

total distance = 1868.478 m

Explanation:

given data

accelerate = 1.68 m/s²

time = 14.2 s

constant time = 68 s

speed = 3.70 m/s²

to find out

total distance

solution

we know train start at rest so final velocity will be after 14 .2 s is

velocity final = acceleration × time ..............1

final velocity = 1.68 × 14.2

final velocity = 23.856 m/s²

and for stop train we need time that is

final velocity = u + at

23.856 = 0 + 3.70(t)

t = 6.44 s

and

distance = ut + 1/2 × at² ...........2

here u is initial velocity and t is time for 14.2 sec

distance 1 = 0 + 1/2 × 1.68 (14.2)²

distance 1 = 169.37 m

and

distance for 68 sec

distance 2= final velocity × time

distance 2= 23.856 × 68

distance 2 = 1622.208 m

and

distance for 6.44 sec

distance 3 = ut + 1/2 × at²

distance 3 = 23.856(6.44) - 0.5 (3.70) (6.44)²

distance 3 = 76.90 m

total distance = distance 1 + distance 2 + distance 3

total distance = 169.37 + 1622.208 + 76.90

total distance = 1868.478 m

7 0

3 years ago

Read 2 more answers

A 3.0-kg block is on a horizontal surface. The block is at rest when, at t = 0, a force (magnitude P = 12 N) acting parallel to

GarryVolchara [31]

Explanation:

It is given that,

Mass of the block, m = 3 kg

Initially, the block is at rest, u = 0

Force acting on the block, P = 12 N

The coefficient of kinetic friction between the block and the surface is, $\mu_k=0.2$

We need to find the rate is the force P doing work on the block at t = 2.0 s. The rate at which work is done is called the power. Let is equal to P'

Frictional force acting on the block, $f=\mu_k mg$

$f=0.2\times 3\ kg\times 9.8\ m/s^2=5.88\ N$

So, the net force acting on the block, F = P - f

$F=12-5.88=6.12\ N$

Let a is the acceleration of the block, $a=\dfrac{F}{m}$

$a=\dfrac{6.12}{3}=2.04\ m/s^2$

Let v is the velocity of the block after 2 seconds. So,

$v=u+at$

$v=0+2.04\times 2$

v = 4.08 m/s

Power, $P'=\dfrac{W}{t}=\dfrac{F.d}{t}=F.v$

$P'=12\ N\times 4.08\ m/s=48.96\ Watts$

So, the force P is doing work on the block at the rate of 48.96 watts.

3 0

3 years ago

Why is a car racetrack built with a steep inward slope?

Step2247 [10]

<span>So we want to know why is a car racetrack built with a steep inward slope. When a car is moving on a circular path, it experiences a centrifugal force. This force is bigger as the car speed rises. So to counteract that force the steep inward slope is creating a force component towards the center of the circle that comes from the car's own weight so the car doesn't slide off the road. </span>

5 0

4 years ago