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3 years ago

9

Mr. Stephenson drives his car from Dallas to Town X, a distance of 437 km. The trip takes 7.27 hours. What is the average speed

of this trip in meters per second (m/s)?

2 answers:

Rom4ik [11]3 years ago

7 0

I believe the answer is <span>216396.148556 meters per second</span>

PtichkaEL [24]3 years ago

6 0

the answer is 60.1 m/s

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The answer is d (head for cover)

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2 years ago

What is the minimum force required to increase the energy of a car by 69 J over a distance of 42 m? Assume the force is constant

sineoko [7]

ANSWER: 1.6N (forth option)

EXPLANATION:
E= Force x displacement
69=F(42)
69/42 = F
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Answer : 1.6N

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2 years ago

A 50-kg copper block initially at 140Â°c is dropped into an insulated tank that contains 90 l of water at 10Â°c. Determine the f

xxMikexx [17]

Answer:

$T_f=24.71$

Explanation:

From the question we are told that:

Mass of block $m=50$

Temperature of block $T_b =140 \textdegree C$

Volume of water $V= 90L$

Temperature of water $T_w=10 \textdegree C$

Density of water $\rho=1000kg/m^3$

Specific heat of water $C_w=4.18KJ/kg-k$

Specific heat of copper $C_p=0.96KJ/kg-k$

Generally the equation for equilibrium stage is mathematically given by

$mC_p(T_b-T_f)=\rho*VV*c(T_f-T_w)$

$50*0.96(140-T_f)=1000*90*10^-3*c_w(T_f-10)$

$48(140-T_f)=376.2(T_f-10)$

$140-T_f=7.8375(T_f-10)$

$140-T_f=7.8375T_f-78.375$

$-8.8375T_f=-218.375$

$T_f=\frac{-218.375}{-8.8375}$

$T_f=\frac{-218.375}{-8.8375}$

$T_f=24.71$

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2 years ago

A student places his hand in front of a plane mirror as shown in the diagram. Which terms correctly describe the image that the

ANTONII [103]

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2 years ago

If a basball is project upwards from the ground level with an initial velovaity of 32 feet per second, then it's height is a fun

inessss [21]

Answer:

Maximum height reached by the ball is 32 meters.

Explanation:

It is given that,

If a baseball is project upwards from the ground level with an initial velocity of 32 feet per second, then it's height is a function of time. The equation is given as :

$s=-8t^2+32t$ ...........(1)

t is the time taken

s is the height attained as a function of time.

Maximum height achieved can be calculated as :

$\dfrac{ds}{dt}=0$

$\dfrac{d(-8t^2+32t)}{dt}=0$

-16 t + 32 = 0

t = 2 seconds

Put the value of t in equation (1) as :

$s=-8(2)^2+32(2)$

s = 32 meters

So, the maximum height reached by the ball is 32 meters. Hence, this is the required solution.

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3 years ago