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galben [10]
2 years ago
6

Isabella drops a pen off her balcony by accident while celebrating the successful completion of a physics problem. assuming air

resistance is negligible, how many seconds does it take the pen to reach a speed of 19.62 \,\dfrac{\text {m}}{\text s}19.62 âs â âm ââ 19, point, 62, space, start fraction, m, divided by, s, end fraction
Physics
1 answer:
expeople1 [14]2 years ago
8 0
U = 0, initial vertical velocity

Neglect air resistance, and g = 9.8 m/s².

The time, t, required for the pen to attain a vertical velocity of 19.62 m/s is given by
19.62 m/s = 0 + (9.8 m/s²)*(t s)
t = 19.62/9.8 = 2.00 s

Answer:  2.0 s
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A chair of mass 12.0 kg is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force F
Tanya [424]

Answer: Normal force, N = 141.64 Newton

Explanation:

All the forces acting on the system and described in free body diagram  are:

1) gravitational pull in downward direction  

2) Normal force in upward direction

3) External force of 40 N acting at an angle of 37° with the horizontal can be resolved in two rectangular components:

  i) F Cos 37° along the horizontal plane  in forward direction and

  ii) F Sin 37° along the vertical plane in downward direction

Applying the Newton's second law, net forces in the vertical plane are:

Net force, f = N - (mg + F Sin 37°)

As there is no acceleration in the vertical plane hence, net force f = 0.

So,

N - (mg + F Sin 37°) = 0

Adding (mg + F Sin 37°) both the sides in above equation, we get

N = mg + F Sin 37°

N = 12 \times 9.8 + 40 \times 0.601      because (Sin 37° = 0.601)

N = 117.6 + 24.04

N = 141.64 Newton

3 0
1 year ago
The absolute pressure, in kilopascals, a depth 10m below sea level is most nearly?
saul85 [17]

Answer:

option A

Explanation:

given,

depth of the sea level = 10 m

g = 10 m/s²

Pressure underwater = ?

we know,

P = ρ g h

where ρ is the density of water which is equal to 1000 kg/m³

h is the depth of sea level

P = ρ g h

P = 1000 x 10 x 10

P = 100000 Pa

P = 100 kPa

Hence, the correct answer is option A

8 0
2 years ago
Terminal velocity. A rider on a bike with the combined mass of 100kg attains a terminal speed of 15m/s on a 12% slope. Assuming
Firlakuza [10]

Answer:

0.9378

Explanation:

Weight (W) of the rider = 100 kg;

since 1 kg = 9.8067 N

100 kg will be = 980.67 N

W = 980.67 N

At the slope of 12%, the angle θ is calculated as:

tan \ \theta = \dfrac{12}{100} \\ \\  tan \ \theta = 0.12 \\ \\  \theta = tan^{-1}(0.12) \\\\ \theta = 6.84^0

The drag force D = Wsinθ

\dfrac{1}{2}C_v \rho AV^2 = W sin \theta

where;

\rho = 1.23 \ kg/m^3

A = 0.9 m²

V = 15 m/s

∴

Drag coefficient C_D = \dfrac{2 *W*sin \theta}{\rho *A *V^2}

C_D =\dfrac{2 *980.67*sin 6.84}{1.23 *0.9 *15^2}

C_D =0.9378

8 0
1 year ago
The difference in heights of the liquid in the two sides of the manometer is 43.4 cm when the atmospheric pressure is 755 mm hg.
Schach [20]

The atmospheric P is greater than the P in the flask, since the Hg level is lacking down lower on the side open to the atmosphere. 

43.4 cm x (10 mm / 1 cm) = 435 mm 

the density of Hg is 13.6 / 0.791 = 17.2 times better than the liquid in the manometer. This means that 1 mmHg = 17.2 mm of manometer liquid. 

435 mm manometer liquid x (1 mm Hg / 17.2 mm manometer liquid) = 25.3 mm Hg 

The pressure in the flask is 755 - 25.3 = 729.7 mmHg. 

729.7 mmHg x (1 atm / 760 mmHg ) = 0.960 atm.

4 0
2 years ago
Recent findings in astrophysics suggest that the observable universe can be modeled as a sphere of radius R = 13.7 × 109 light-y
-BARSIC- [3]

Answer:

3.7\times 10^{51}) kg

Explanation:

R = radius of the sphere modeled as universe = 13\times 10^{25} m

Volume of sphere is given as

V = \frac{4\pi R^{3}}{3}

V = \frac{4(3.14) (13\times 10^{25})^{3}}{3}

V = 9.2\times 10^{78} m³

\rho = average total mass density of universe = 1\times 10^{-26} kg/m³

m = Total mass of the universe = ?

We know that mass is the product of volume and density, hence

m = \rho V

m = (1\times 10^{-26}) (9.2\times 10^{78})

m = 9.2\times 10^{52} kg

M = mass of "ordinary" matter  = ?

mass of "ordinary" matter is only about 4% of total mass, hence

M = (0.04) m

M = (0.04)(9.2\times 10^{52})

M = 3.7\times 10^{51} kg

6 0
1 year ago
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