7) The acceleration due to gravity near the surface of Mars is about one-third of the value

Lady bird [3.3K]

I don't know!!!!I don't know!!!!I don't know!!!!plz give me brainliest

5 0

3 years ago

If 1.34 ✕ 1020 electrons move through a pocket calculator during a full day's operation, how many coulombs of charge moved throu

ioda

Given :

Number of operations move through a pocket calculator during a full day's operation , $n=1.34 \times 10^{20}$ .

To Find :

How many coulombs of charge moved through it .

Solution :

We know , charge in one electron is :

$e^-=-1.6\times 10^{-19}\ coulombs$

So , charge on n electron is :

$C=e^-\times n\\C=-1.6\times 10^{-19}\times 1.34\times 10^{20} \ C\\C=-21.44\ C$

Therefore , -21.44 coulombs of charge is moved through it .

Hence , this is the required solution .

3 0

3 years ago

Transverse thrusters are used to make large ships fully maneuverable at low speeds without tugboat assistance. A transverse thru

suter [353]

Answer:

Thrust developed = 212.3373 kN

Explanation:

Assuming the ship is stationary

Determine the Thrust developed

power supplied to the propeller ( Punit ) = 1900 KW

Duct distance ( diameter ; D ) = 2.6 m

first step : calculate the area of the duct

A = π/4 * D^2

= π/4 * ( 2.6)^2 = 5.3092 m^2

next : calculate the velocity of propeller

Punit = (A*v*β ) / 2 * V^2 ( assuming β = 999 kg/m^3 ) also given V1 = 0

∴V^3 = Punit * 2 / A*β

= ( 1900 * 10^3 * 2 ) / ( 5.3092 * 999 )

hence V2 = 8.9480 m/s

Finally determine the thrust developed

F = Punit / V2

= (1900 * 10^3) / ( 8.9480)

= 212.3373 kN

8 0

3 years ago

Carbon-14 is used to determine the age of ancient objects. If a sample today contains 0.060 g of carbon-14, how much carbon-14 m

MakcuM [25]

Answer: 86.47 g of carbon-14 must have been present in the sample 11,430 years ago.

Explanation:

Half-life of sample of carbon -14= 5,730 days

$\lambda=\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{5,730 days}=0.00012 day^{-1}$

Let the sample present 11,430 years(t) ago = $N_o$

Sample left till today ,N= 0.060 g

$N=N_o\times e^{-\lambda t}$

$ln[N]=ln[N]_o-\lambda t$

$\log[0.060 g]=\log[N_o]-2.303\times 0.00012 day^{-1}\times 11,430 days$

$\log[N_o]=1.9369$

$N_o=86.47 g$

86.47 g of carbon-14 must have been present in the sample 11,430 years ago.

5 0

3 years ago

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Ball X of mass 1.0 kg and ball Y of mass 0.5kg travel toward each other on a horizontal surface. Both balls travel with a consta

bazaltina [42]

Answer:

B. Ball Y will travel at a speed less than 5 m/s in the opposite direction of travel as before the collision.

Explanation:

Impulse created by ball Y on ball X = 40 x 1/6 Ns

Ball X will also create impulse 40 / 6 on ball Y .

impulse = change in momentum .

impulse in Y = change in momentum in Y .

Initial momentum of Y = .5 x 5 = 2.5

Let final velocity of Y after collision be v in opposite direction .

change in momentum of Y = v - (-2.5 )

so,

v + 2.5 = 40 / 6 = 6.67

v = 4.17 m / s .

Option B is correct .

B. Ball Y will travel at a speed less than 5 m/s in the opposite direction of travel as before the collision.

4 0

3 years ago

Read 2 more answers