Ask question

Ask question

All categories

2 years ago

15

[Assume g = 10m/s?] If a girl is running along a straight road with a uniform velocity 1.5 m/s, find her acceleration. (Ans: 0)

a.

1 answer:

DerKrebs [107]2 years ago

7 0

Answer:

Her acceleration is 0 m/s²

Explanation:

We note that the motion of the girl is on a straight road, therefore;

The vertical acceleration (e.g. due to gravity, <em>g)</em> on the horizontal motion = 0

The horizontal acceleration, a = (Change in velocity, Δv)/(Change in time, Δt)

For uniform velocity, the change in velocity, Δv = 0

Therefore, fore any change in time, Δt, we have;

a = Δv/Δt

Her acceleration, a = 0/Δt = 0

Her acceleration, a = 0 m/s²

You might be interested in

Questions 15 out of 20

Olin [163]

Answer:

D

Explanation:

7 0

3 years ago

HELPPPPPPPP PLSSSSSSSS A car moves with constant velocity along a straight road. Its position is x1 = 0 m at t1 = 0 s and is x2

yawa3891 [41]

Answer:

Explanation:

6.2 miles

6 0

3 years ago

Read 2 more answers

A fluid in an aquifer is 23.6 m above a reference datum, the fluid pressure (in gage pressure) is 4390 n/m2 and the flow velocit

Phoenix [80]

As per Bernuolli's Theorem total energy per unit mass is given as

$\frac{P}{\rho} + \frac{1}{2}v^2 + gH = E$

now from above equation

$P = 4390 N/m^2$

$\rho = 0.999 * 10^3$

$v = 7.22 * 10^{-4} m/s$

$H = 23.6 m$

now by above equation

$\frac{4390}{0.999*10^3} + \frac{1}{2}*(7.22*10^{-4})^2 + 9.8*23.6 = E$

$E = 235.7 J/kg$

Part B)

Now energy per unit weight

$U = \frac{E}{g}$

$U = \frac{235.7}{9.8}$

$U = 24 m$

7 0

3 years ago

Enter the expression 2cos2(θ)−1, where θ is the lowercase Greek letter theta.

vlabodo [156]

GAVNSTVVHUYCZ swffggff

4 0

3 years ago

About 65 million years ago an asteroid struck Earth in the area of the Yucatán Peninsula and wiped out the dinosaurs and many ot

9966 [12]

Answer:

$2.44156\times 10^{13}\ m^3$

29010.53917 m

Explanation:

$\rho$ = Density of asteroid = 2 g/cm³

V = Volume

d = Diameter = 10 km

r = Radius = $\dfrac{d}{2}=\dfrac{10}{2}=5\ km$

v = Velocity = 11 km/s

$H_v$ = Heat vaporization of water = $2.26\times 10^6\ J/kg$

$\Delta T$ = Change in temperature = 100-20

Mass is given by

$m=\rho V\\\Rightarrow m=\rho\dfrac{4}{3}\pi r^3\\\Rightarrow m=2000\dfrac{4}{3}\times \pi\times 5000^3\\\Rightarrow m=1.0472\times 10^{15}\ kg$

The kinetic energy is

$K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}1.0472\times 10^{15}\times 11000^2\\\Rightarrow K=6.33556\times 10^{22}\ J$

Heat is given by

$Q=mc\Delta T+mH_v\\\Rightarrow 6.33556\times 10^{22}=m\times (4186\times (100-20)+2.26\times 10^6)\\\Rightarrow m=\dfrac{ 6.33556\times 10^{22}}{4186\times (100-20)+2.26\times 10^6}\\\Rightarrow m=2.44156\times 10^{16}\ kg$

Mass of water is $2.44156\times 10^{16}\ kg$

Volume is $\dfrac{2.44156\times 10^{16}}{10^3}=2.44156\times 10^{13}\ m^3$

Amount of water is $2.44156\times 10^{13}\ m^3$

If it were a cube

$h=V^{\dfrac{1}{3}}\\\Rightarrow h=(2.44156\times 10^{13})^{\dfrac{1}{3}}\\\Rightarrow h=29010.53917\ m$

The height of the water would be 29010.53917 m

4 0

2 years ago