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3 years ago

what is the kinetic energy of a 1 kilogram ball is thrown into the air with an initial velocity of 3 m/sec

1 answer:

3 0

Data:
m (mass) = 1 Kg
s (speed) = 3 m/s
Kinetic energy = ? (Joule)

Formula (Kinetic energy)
$E_{k} = \frac{m*s^2}{2}$

Solving:
$E_{k} = \frac{m*s^2}{2}$
$E_{k} = \frac{1*3^2}{2}$
$E_{k} = \frac{1*9}{2}$
$E_{k} = \frac{9}{2}$
$\boxed{\boxed{E_{k} = 4.5\:J}}\end{array}}\qquad\quad\checkmark$

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10 points!! If you help!!!

marta [7]

For Mass

K.E = (1/2*mv^2)

Explanation:

Kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity squared. For example, if a an object with a mass of 10 kg (m = 10 kg) is moving at a velocity of 5 meters per second (v = 5 m/s), the kinetic energy is equal to 125 Joules, or (1/2 * 10 kg) * 5 m/s2.

4 0

2 years ago

the train accelerates from 30 km/h to 45 km/h in 15 secs. a. find its acceleration. b. distance it travels during this time ...

Sonbull [250]

Acceleration = v-u/t
= (45-30)/15
= 1 km/h

Distance = ut + 1/2 at^2
s = [30 x 15] + 1/2 x 1 x 15^2
= 562.5 km

Hope this helps

3 0

3 years ago

Read 2 more answers

A 5.30kg block hangs from a spring with a spring constant 1700 N/m. The block is pulled down 4.50cm from the equilibrium positio

Andru [333]

To solve this problem it is necessary to apply the concepts related to the frequency in a spring, the conservation of energy and the total mechanical energy in the body (kinetic or potential as the case may be)

PART A) By definition the frequency in a spring is given by the equation

$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$

Where,

m = mass

k = spring constant

Our values are,

k=1700N/m

m=5.3 kg

Replacing,

$f = \frac{1}{2\pi} \sqrt{\frac{1700}{5.3}}$

$f=2.85 Hz$

PART B) To solve this section it is necessary to apply the concepts related to the conservation of energy both potential (simple harmonic) and kinetic in the spring.

$\frac{1}{2}kA^2 = \frac{1}{2}mv^2 + \frac{1}{2} kY^2$

Where,

k = Spring constant

m = mass

y = Vertical compression

v = Velocity

This expression is equivalent to,

$kA^2 =mV^2 +ky^2$

Our values are given as,

k=1700 N/m

V=1.70 m/s

y=0.045m

m=5.3 kg

Replacing we have,

$1700*A^2=5.3*1.7^2 +1700*(0.045)^2$

Solving for A,

$A^2 = \frac{5.3*1.7^2 +1700*(0.045)^2}{1700}$

$A ^2 = 0.011035$

$A=0.105 m \approx 10.5 cm$

PART C) Finally, the total mechanical energy is given by the equation

$E = \frac{1}{2}kA^2$

$E=\frac{1}{2}1700*(0.105)^2$

$E= 9.3712 J$

3 0

3 years ago

Please answer these diagrammatic questions ASAP and please no spam answers

SVEN [57.7K]

Answer:

i. The pressure of due to the water, P, is given according to the following equation;

P = ρ·g·h

Where;

ρ = The density of the water (a constant) = 997 kg/m³

g = The acceleration due to gravity = 9.81 m/s²

h = The height of the water (minimum h = h₁, maximum h = h₂)

The pressure is directly proportional to the water height, and we have;

The pressure, P, will be maximum when the water height, h, is maximum or h = h₂, which is the level DC

ii. The thrust = The force acting on the body = Pressure × Area

The maximum areas exposed to the water are on side AB and DC

However, the pressure at level DC, which is the location of the maximum pressure, is larger than the pressure at level AB, therefore, the maximum thrust will be at the level DC

Explanation:

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3 years ago

A group of hikers hear an echo 2.6 s after they shout. The temperature is 20 ◦C. How far away is the mountain that reflected the

Vadim26 [7]

Answer:

$x = 445.9\,m$