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2 years ago

Is my answer to part A correct ? And can someone please answer part 2 and if I am wrong correct part A ? !!!!!!

Physics

1 answer:

levacccp [35]2 years ago

5 0

In A, the left side is exactly what you should write, to lead you to the answer ... correct units and everything. But when you tried to walk across the 'equals' sign, you slipped, fell, and got demolished on the rocks below. "19.6" is not the product of the things on the left side. Do the arithmetic again please.

B. When the height doubles, so does the potential energy.

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Please help answer question

nika2105 [10]

Answer:

C = 1.01

Explanation:

Given that,

Mass, m = 75 kg

The terminal velocity of the mass, $v_t=60\ m/s$

Area of cross section, $A=0.33\ m^2$

We need to find the drag coefficient. At terminal velocity, the weight is balanced by the drag on the object. So,

R = W

$\dfrac{1}{2}\rho CAv_t^2=mg$

Where

$\rho$ is the density of air = 1.225 kg/m³

C is drag coefficient

So,

$C=\dfrac{2mg}{\rho Av_t^2}\\\\C=\dfrac{2\times 75\times 9.8}{1.225\times 0.33\times (60)^2}\\\\C=1.01$

So, the drag coefficient is 1.01.

4 0

2 years ago

An object is dropped from a vertical distance of 25.5 m above the ground, and it takes 2.28 sec to fall that distance. A second

DENIUS [597]

Answer:

The second object takes 2.28 s to fall the 25.5 m.

Explanation:

In this case, both objects take the same time to fall, since no vertical velocity is added to any of them.

You can also confirm this by sepparating the second's object movement into its two directions: in the horizontal one, we have linear uniform motion, and in the vertical one, we have free fall, with exactly the same characteristics as for the first object.

4 0

3 years ago

In his motorboat, Bill Ruhberg travels upstream at top speed to his favorite fishing spot, a distance of 120120 mi, in 33 hr.

photoshop1234 [79]

Answer:

The rate of the boat in still water is 44 mph and the rate of the current is 4 mph

Explanation:

x = the rate of the boat in still water

y = the rate of the current.

Distance travelled = 120 mi

Time taken upstream = 3 hr

Time taken downstream = 2.5 hr

Speed = Distance / Time

Speed upstream

$\frac{120}{3}=x-y\\\Rightarrow 40=x-y$

Speed downstream

$\frac{120}{2.5}=x+y\\\Rightarrow 48=x+y$

Adding both the equations

$48+40=x-y+x+y\\\Rightarrow 88=2x\\\Rightarrow 44=x$

$40=44-y\\\Rightarrow 40-44=-y\\\Rightarrow y=4$

The rate of the boat in still water is 44 mph and the rate of the current is 4 mph

8 0

3 years ago

If the distance between two objects is decreased to 1 10 of the original distance, how will it change the force of attraction be

aleksandr82 [10.1K]

(A) It will 100 times larger than the original force.

6 0

3 years ago

What was your electric potential relative to a metal pipe if a spark jumped 1.1 cm through dry air from your finger to the pipe?

romanna [79]

The electric field is given by volts/distance: $E= \frac{V}{d}$ . The breakdown voltage of dry air is about 3x10^6V/m. So solving for V we get
$V=Ed$
or $V=(3e6V/m)(0.011m)=33,000V$

6 0

3 years ago

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