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Salsk061 [2.6K]

2 years ago

9

a bus travelling on a straight road at 25m/s accelerates uniformly at 5m/s squared for 2 seconds. find its speed in kilometres p

er hour

1 answer:

notka56 [123]2 years ago

5 0

Answer:

126 kmh⁻¹

Explanation:

We can simply solve this by applying motion equations

where

v - final velocity

u - initial velocity

a-acceleration

t - time

v = u + at

<em>= 25 + 5×2 = 35 ms⁻¹</em>

<em> = (35/1000)×3600 = 126 kmh⁻¹</em>

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A car’s bumper is designed to withstand a 4.0-km/h (1.1-m/s) collision with an immovable object without damage to the body of th

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Answer:

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Explanation:

For this problem we can use Newton's second law, to calculate the average force and acceleration we can find it by kinematics.

vf² = v₀² - 2 ax

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3 years ago

When air resistance is ignored, _____ of the projectile affect(s) the range and maximum height of the projectile.

FromTheMoon [43]

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The height from the ground at the top most position of projectile is referred to as maximum height.

When air resistance is ignored, initial velocity of the projectile affect the range and maximum height of the projectile.

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8 0

1 year ago

A vessel with an unknown volume is filled with 10 kg of water at 90oC. Inspection of the vessel at equilibrium shows that 8 kg o

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Hey there!

In this case, it is possible to solve this problem by using the widely-known steam tables which show that at 90 °C, the pressure that produces a vapor-liquid mixture at equilibrium is about 70.183 kPa (Cengel, Thermodynamics 5th edition).

Moreover, for the calculation of the volume, it is necessary to calculate the volume of the vapor-liquid mixture, given the quality (x) it has:

$x=\frac{m_{steam}}{m_{total}}$

Thus, since 8 kg correspond to liquid water, 2 kg must correspond to steam, so that the quality turns out:

$x=\frac{2kg}{10kg} =0.20$

Now, at this temperature and pressure, the volume of a saturated vapor is 2.3593 m³/kg whereas that of the saturated liquid is 0.001036 m³/kg and therefore, the volume of the mixture is:

$v=0.001036m^3/kg+0.2(2.3593-0.001036 )m^3/kg=0.4727m^3/kg$

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8 0

2 years ago

A particle moves along the x axis. It is initially at the position 0.180 m, moving with velocity 0.060 m/s and acceleration -0.3

shusha [124]

Answer:

a

$x_2 = -2.3356$

b

$v = -1.384 \ m/s$

Explanation:

From the question we are told that

The initial position of the particle is $x_1 = 0.180 \ m$

The initial velocity of the particle is $u = 0.060 \ m/s$

The acceleration is $a = -0.380 \ m/s^2$

The time duration is $t = 3.80 \ s$

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=> $v = 0.060 + (-0.380 * 3.80)$

=> $v = -1.384 \ m/s$

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Here s is the distance covered by the particle, so

$(-1.384)^2 = (0.060)^2 + 2(-0.380)* s$

=> $s = -2.5156 \ m$

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$x_2 = x_1 + s$

=> $x_2 = 0.180 + (-2.5156)$

=> $x_2 = -2.3356$

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2 years ago