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A dart is thrown horizontally with an initial speed of 10 m/s toward point P, the bull's-eye on a dart board. It hits at point Q

Kryger [21]

D=s(t) so it would be d=10(.19) d=.19 FOR BITH SNDWERS

3 0

4 years ago

The gage pressure in a liquid at a depth of 3 m is read to be 39 kPa. Determine the gage pressure in the same liquid at a depth

ioda

Answer: 117 kPa

Explanation:

For the liquid at depth 3 m, the gauge pressure is equal to = P₁=39 kPa

For the liquid at depth 9m, the gauge pressure is equal to= P₂

Now we are given the condition that the liquid is same. That must imply that the density must be same throughout the depth.

So, For finding gauge pressure we have formula P= ρ * g * h

Also gravity also remains same for both liquids

So taking ratio of their respective pressures we have

$\frac{P_{1} }{\\P_2}$ = $\frac{density * g * h_1}{density * g * h_2}$

So $\frac{39}{P_2}$ = $\frac{3}{9}$

Or P₂= 39 * 3 = 117 kPa

5 0

3 years ago

Name the fundamental units involved in pascal

gtnhenbr [62]

Explanation:

Hey there!

Here,

Pascal is a unit of pressure.

$pressure = \frac{f}{a}$

Now, As per the formula the units are:

kg, m and s^2.

Hope it helps...

5 0

4 years ago

Two cyclists start on a race between points A and D on two different routes. Cyclist X takes the route passing through the equid

ollegr [7]

The displacement is the shortest distance between two points, which is 546.41. The displacement for both is 546.41 meters

Average velocity of X = (200 + 200 + 200) / 30
Average velocity of X = 20 m/s

Average velocity of Y = 546.41 / 30 = 18.2 m/s

8 0

4 years ago

Read 2 more answers

(a) The Sun orbits the Milky Way galaxy once each 2.60 x 108y , with a roughly circular orbit averaging 3.00 x 104 light years i

PilotLPTM [1.2K]

Answer:

Part a)

$a_c = 1.67 \times 10^{-10} m/s^2$

Part b)

$v = 2.18 \times 10^5 m/s$

Explanation:

Time period of sun is given as

$T = 2.60 \times 10^8 years$

$T = 2.60 \times 10^8 (365 \times 24 \times 3600) s$

$T = 8.2 \times 10^{15} s$

Now the radius of the orbit of sun is given as

$R = 3.00 \times 10^4 Ly$

$R = 3.00 \times 10^4 (3\times 10^8)(365 \times 24 \times 3600)m$

$R = 2.84 \times 10^20 m$

Part a)

centripetal acceleration is given as

$a_c = \omega^2 R$

$a_c = \frac{4\pi^2}{T^2} R$

$a_c = \frac{4\pi^2}{(8.2\times 10^{15})^2}(2.84 \times 10^{20})$

$a_c = 1.67 \times 10^{-10} m/s^2$

Part b)

orbital speed is given as

$v = \frac{2\pi R}{T}$

$v = \frac{2\pi (2.84 \times 10^{20})}{8.2 \times 10^{15}}$

$v = 2.18 \times 10^5 m/s$

5 0

4 years ago