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NeX [460]
2 years ago
13

A car with a mass of 500 kg is moving at a speed of 12 m/s. How much kinetic energy does it have?

Physics
1 answer:
shtirl [24]2 years ago
5 0

Answer:

36000

Explanation:

1/2mv²=0.5×500×12²=36000j

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Water flows with constant speed through a garden hose that goes up to 27.5 cm high. if the water pressure is 132kpa at the botto
sergejj [24]

Answer:

The pressure at the top of the step is 129.303 kilopascals.

Explanation:

From Hydrostatics we find that the pressure difference between extremes of the water column is defined by the following formula, which is a particular case of the Bernoulli's Principle (v_{bottom}\approx v_{top}):

p_{bottom}-p_{top} = \rho\cdot g\cdot \Delta h (1)

p_{bottom}, p_{top} - Total pressures at the bottom and at the top, measured in pascals.

\rho - Density of the water, measured in kilograms per cubic meter.

\Delta h - Height difference of the step, measured in meters.

If we know that p_{bottom} = 132000\,Pa, \rho = 1000\,\frac{kg}{m^{3}}, g = 9.807\,\frac{m}{s^{2}} and \Delta h = 0.275\,m, then the pressure at the top of the step is:

p_{top} = p_{bottom}-\rho\cdot g\cdot \Delta h

p_{top} = 132000\,Pa-\left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.275\,m)

p_{top} = 129303.075\,Pa

p_{top} = 129.303\,kPa

The pressure at the top of the step is 129.303 kilopascals.

6 0
2 years ago
The faster am igneous rock cools the_ the crystal size
HACTEHA [7]
Bigger is the correct answer. the faster an igneous rock cools the bigger the bigger the crystal size will be
7 0
2 years ago
What is the average velocity of atoms in 1.00 mol of neon (a monatomic gas) at 465 K? For m, use 0.0202 kg.
TiliK225 [7]

Answer: 757m/s

Explanation:

Given the following :

Mole of neon gas = 1.00 mol

Temperature = 465k

Mass = 0.0202kg

Using the ideal gas equation. For calculating the average kinetic energy molecule :

0.5(mv^2) = 3/2 nRt

Where ;

M = mass, V = volume. R = gas constant(8.31 jK-1 mol-1, t = temperature in Kelvin, n = number of moles

Plugging our values

0.5(0.0202 × v^2) = 3/2 (1 × 8.31 × 465)

0.0101 v^2 = 5796.225

v^2 = 5796.225 / 0.0101

v^2 = 573883.66

v = √573883.66

v = 757.55109m/s

v = 757m/s

5 0
3 years ago
66. Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, and (d) 2.00 s for a ball thrown st
kozerog [31]

Answer:

a) t=1s

y = 10.1m

v=5.2m/s

b) t=1.5s

y =11.475 m

v=0.3m/s

c) t=2s

y =10.4 m

v=-4.6m/s  (The minus sign (-) indicates that the ball is already going down)

Explanation:

Conceptual analysis

We apply the free fall formula for position (y) and speed (v) at any time (t).

As gravity opposes movement the sign in the equations is negative.:  

y = vi*t - ½ g*t2 Equation 1

v=vit-g*t  Equation 2

y: The vertical distance the ball moves at time t  

vi: Initial speed

g= acceleration due to gravity

v= Speed the ball moves at time t  

Known information

We know the following data:

Vi=15 m / s

g =9.8 \frac{m}{s^{2} }

t=1s ,1.5s,2s

Development of problem

We replace t in the equations (1) and (2)  

a) t=1s

y = 15*1 - ½ 9.8*1^{2}=15-4.9=10.1m

v=15-9.8*1 =15-9.8 =5.2m/s

b) t=1.5s

y = 15*1.5 - ½ 9.8*1.5^{2}=22.5-11.025=11.475 m

v=15-9.8*1.5 =15-14.7=0.3m/s

c) t=2s

y = 15*2 - ½ 9.8*2^{2}= 30-19.6=10.4 m

v=15-9.8*2 =15-19.6=-4.6m/s  (The minus sign (-) indicates that the ball is already going down)

3 0
3 years ago
Two liquids, A and B, have equal masses and equal initial temperatures. Each is heated for the same length of time over identica
DochEvi [55]

Answer:

So the specific heat of the liquid B is greater than that of A.

Explanation:

Liquid A is hotter than the liquid B after both the liquids are heated identically for the same duration of time from the same initial temperature then according to heat equation,

Q=m.c.\Delta T

where:

m = mass of the body

c = specific heat of the body

\Delta T= change in temperature of the body

The identical heat source supplies the heat for the same amount of time then the quantity of heat supplied is also equal.

So for constant heat, constant mass the temperature change is inversely proportional to the specific of heat of the liquid.

\Delta T=\frac{Q}{m} \times \frac{1}{c}

\Delta T\propto\frac{1}{c}

So the specific heat of the liquid B is greater than that of A.

5 0
3 years ago
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