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2 years ago

15

which has a greater kinetic energy, a bowling ball that has a mass of 5kg travelling at 6m/s, or a ship that has a mass of 12000

0kg and is moving at 0.02m/s?

1 answer:

AVprozaik [17]2 years ago

6 0

Answer:bowling ball has greater kinetic energy

Explanation:

Kinetic energy of bowling ball:

mass=m=5kg

Velocity=v=6m/s

Kinetic energy =ke

Ke=0.5 x m x v x v

Ke=0.5 x 5 x 6 x 6

Ke=90J

Kinetic energy of ship:

mass=m=120000kg

velocity=v=0.02m/s

Ke=0.5 x m x v x v

Ke=0.5 x 120000 x 0.02 x 0.02

Ke=24J

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A sample of a gas occupies a volume of 90 mL at 298 K and a pressure of 702 mm Hg. What is the correct expression for calculatin

aleksandr82 [10.1K]

Answer:

Explanation:

Given

volume of sample $V_1=90\ mL$

Temperature $T_1=298\ K$

Pressure $P_1=702\ mm\ of\ Hg$

for different conditions

Temperature $T_2=273\ K$

Pressure $P_2=760\ mm\ of\ Hg$

suppose $V_2$ is the volume of sample

Using ideal gas equation

$PV=nRT$

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$\frac{702\times 90}{298}=\frac{760\times V_2}{273}$

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Calculate the amount of hcn that gives the lethal dose in a small laboratory room measuring 14 × 15 × 8.0ft. the density of air

vova2212 [387]

Since we are given the density and volume, then perhaps we can determine the amount in terms of the mass. All we have to do is find the volume in terms of cm³ so that it will cancel out with the cm³ in the density. The conversion is 1 ft = 30.48 cm. The solution is as follows:

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A 0.305 kg book rests at an angle against one side of a bookshelf. The magnitude and direction of the total force exerted on the

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Answer

given,

$F_L= 1.52\ N$

$\theta_L= 31^0$

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so, from the diagram attached below

$F_L cos {\theta_L} + F_b sin {\theta_b} = m g$

$1.52 times cos {31^0} + F_b sin {\theta_b} = 0.305 \times 9.8$

$F_b sin {\theta_b} = 2.989 -1.303$

$F_b sin {\theta_b} = 1.686$

computing horizontal component

$F_b cos {\theta_b} = F_L sin {\theta_L}$

$cos {\theta_b} = \dfrac{F_L sin {\theta_L}}{F_b}$

$cos {\theta_b} = \dfrac{1.52 \times sin {31^0}}{1.686}$

$cos {\theta_b} = 0.464$

θ = 62.35°

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2 years ago

A guitar string has a linear density of 8.30 ✕ 10−4 kg/m and a length of 0.660 m. the tension in the string is 56.7 n. when the

Sedbober [7]

Ans: Beat Frequency = 1.97Hz

Explanation:
The fundamental frequency on a vibrating string is

$f = \sqrt{ \frac{T}{4mL} }$ <span> -- (A)</span>

<span>here, T=Tension in the string=56.7N,
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Plug in the values in Equation (A)

<span>so </span> $f = \sqrt{ \frac{56.7}{4*5.48*10^{-4}*0.66} }$ <span> = 197.97Hz </span>

<span>the beat frequency is the difference between these two frequencies, therefore:
Beat frequency = 197.97 - 196.0 = 1.97Hz
-i</span>

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2 years ago

Read 2 more answers

A 5.00-g bullet is shot through a 1.00-kg wood block suspended on a string 2.00 m long. The center of mass of the block rises a

o-na [289]

Answer:395.6 m/s

Explanation:

Given

mass of bullet $m=5 gm$

mass of wood block $M=1 kg$

Length of string $L=2 m$

Center of mass rises to an height of $0.38 cm$

initial velocity of bullet $u=450 m/s$

let $v_1$ and $v_2$ be the velocity of bullet and block after collision

Conserving momentum

$mu=mv_1+Mv_2$ -------------1

Now after the collision block rises to an height of 0.38 cm

Conserving Energy for block

kinetic energy of block at bottom=Gain in Potential Energy

$\frac{Mv_2^2}{2}=Mgh_{cm}$

$v_2=\sqrt{2gh_{cm}}$

$v_2=\sqrt{2\times 9.8\times 0.38}$

$v_2=0.272 m/s$

substitute the value of $v_2$ in equation 1

$5\times 450=5\times v_1+1000\times 0.272$

$v_1=395.6 m/s$

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2 years ago