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2 years ago

Calculate the surface area of a cube with edge of 14.5cm

1 answer:

4 0

To find surface area of a cube, there's a formula that can be used: 6*s^2. You multiply 6 to a side or edge squared.

We know that an edge is 14.5 cm so we square it or multiply the edge by itself.

14.5*14.5 = 210.25 cm^2.
Now, we just multiply that number by 6.

210.25*6 = 1261.5 cm^2.
Hope this helps!

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A sample of an unknown volatile liquid was injected into a Dumas flask (mflask = 27.0928 g, Vflask = 0.1040 L) and heated until

NNADVOKAT [17]

Answer:

The gas was Hexane

Explanation:

taking the diference between the mass of the flask and the final mass qe can calculate the mass of liquid injected (assuming none escaped the flask):

$m_{l} = 27.4593g - 27.0928g = 0.3665g$

with the volume of the flask we can get the density of the gas at the indicated pressure and temperature:

$d_{g} = \frac{0.3665 g}{0.1040L} = 3.524 g/L$

From the ideal gases law we have that the density can be calculated as:

$d_{g} = \frac{P*M}{R*T}$

Where R is the ideal gases constant = , and M the molecular weight of the fluid. Solving for M:

$M=\frac{d_{g}*R*T}{P}=\frac{3.524g/L*0,082atmL/molK*291K}{0.976atm}$

$M=86.16 g/mol$

Note that the temperature is computed in Kelvin T= 18+273=291K

The gas with the closer molar mass is Hexane

4 0

2 years ago

Small frogs that are good jumpers are capable of remarkable accelerations. One species reaches a takeoff speed of3.7 m/s in60 ms

MAVERICK [17]

We know that acceleration is change in velocity by time taken for that change.

In this case velocity change is 3.7 m/s

Time taken for this change = 60 ms = $6 *10^{-3} seconds$

So acceleration of frog = $\frac{3.7}{60*10^{-3}}$

= 61.66 m/ $s^2$

So acceleration of frog is 61.66 m/ $s^2$

o it is evident that frog is capable of remarkable accelerations.

8 0

2 years ago

Read 2 more answers

A shuttle on Earth has a mass of 4.5 E 5 kg. Compare its weight on Earth to its weight while in orbit at a height of 6.3 E 5 met

faltersainse [42]

Answer:

83%

Explanation:

On the surface, the weight is:

W = GMm / R²

where G is the gravitational constant, M is the mass of the Earth, m is the mass of the shuttle, and R is the radius of the Earth.

In orbit, the weight is:

w = GMm / (R+h)²

where h is the height of the shuttle above the surface of the Earth.

The ratio is:

w/W = R² / (R+h)²

w/W = (R / (R+h))²

Given that R = 6.4×10⁶ m and h = 6.3×10⁵ m:

w/W = (6.4×10⁶ / 7.03×10⁶)²

w/W = 0.83

The shuttle in orbit retains 83% of its weight on Earth.

4 0

2 years ago

Before the student releases the cart by cutting the tinsel string, what forces are acting on the cart?

Naya [18.7K]

Answer: TENSION and WEIGHT

Explanation:

Force experienced by the spring is called TENSION while the WEIGHT is the gravitational pull on the body towards the earth surface. Therefore the forces acting on the cart are TENSION and WEIGHT(weight acts downwards (along negative y-axis) while the TENSION upward(along positive y-axis).

5 0

2 years ago

Which of the following are inertial reference frames? A. A car driving at steady speed on a straight and level road. B. A car dr

Aloiza [94]

Answers:

A. A car driving at steady speed on a straight and level road.

B. A car driving at steady speed up a 10∘ incline.

Explanation:

An object is said to be in an inertial reference frame if the net force acting on the object is zero. According to Newton's second law, this also means that the acceleration of the object is also zero:

$F=ma$

Since F=0, a=0 as well.

Let's now analyze each case.

A. A car driving at steady speed on a straight and level road. --> YES: this is an inertial reference frame, because the car is keeping a constant speed and a constant direction, so its velocity is not changing, and its acceleration is zero.

B. A car driving at steady speed up a 10∘ incline. --> YES: this is an inertial reference frame, because the car is keeping a constant speed and a constant direction, so its velocity is not changing, and its acceleration is zero.

C. A car speeding up after leaving a stop sign. --> NO: this is not an intertial reference frame, because the car is speeding up, so it is accelerating.

D. A car driving at steady speed around a curve. --> NO: this is not an inertial reference frame, because the car is changing direction, therefore its velocity is changing and so the car is accelerating.

So the only two choices which are correct are A and B.

8 0

2 years ago

Read 2 more answers