Ask question

Ask question

All categories

2 years ago

14

A man on the moon throws a ball vertically upwards and it is noticed that the ball travels 3.0m less in the fifth second of its

upward motion than it does in the third second. What is the acceleration due to gravity on the moon?

1 answer:

sdas [7]2 years ago

8 0

<h2>Acceleration due to gravity in moon is 1.5 m/s²</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

Here the ball travels 3 m less distance in fifth second compared to third second.

That is

s₃ = s₅ + 3

Now we have

Distance traveled in third second, s₃ = u x 3 - 0.5 x g x 3² - u x 2 - 0.5 x g x 2²

s₃ = u - 2.5 g

Also

Distance traveled in fifth second, s₅ = u x 5 - 0.5 x g x 5² - u x 4 - 0.5 x g x 4²

s₅ = u - 4.5 g

That is

u - 2.5 g = u - 4.5 g + 3

2 g = 3

g = 1.5 m/s²

Acceleration due to gravity in moon = 1.5 m/s²

You might be interested in

Parallel light rays with a wavelength of 563 nm fall on a single slit. On a screen 3.30 m away, the distance between the first d

MrMuchimi

Answer:

The width of the slit is 0.4 mm (0.00040 m).

Explanation:

From the Young's interference expression, we have;

(λ ÷ d) = (Δy ÷ D)

where λ is the wavelength of the light, D is the distance of the slit to the screen, d is the width of slit and Δy is the fringe separation.

Thus,

d = (Dλ) ÷ Δy

D = 3.30 m, Δy = 4.7 mm (0.0047 m) and λ = 563 nm (563 × $10^{-9}$ m)

d = (3.30 × 563 × $10^{-9}$ ) ÷ (0.0047)

= 1.8579 × $10^{-6}$ ÷ 0.0047

= 0.0003951 m

d = 0.00040 m

The width of the slit is 0.4 mm (0.00040 m).

3 0

3 years ago

Ferdinand the frog is hopping from lily pad to lily pad in search of a good fly

loris [4]

Answer: $36.86\°$

Explanation:

According to the described situation we have the following data:

Horizontal distance between lily pads: $d=2.4 m$

Ferdinand's initial velocity: $V_{o}=5 m/s$

Time it takes a jump: $t=0.6 s$

We need to find the angle $\theta$ at which Ferdinand jumps.

In order to do this, we first have to find the <u>horizontal component (or x-component)</u> of this initial velocity. Since we are dealing with parabolic movement, where velocity has x-component and y-component, and in this case we will choose the x-component to find the angle:

$V_{x}=\frac{d}{t}$ (1)

$V_{x}=\frac{2.4 m}{0.6 s}$ (2)

$V_{x}=4 m/s$ (3)

On the other hand, the x-component of the velocity is expressed as:

$V_{x}=V_{o}cos\theta$ (4)

Substituting (3) in (4):

$4 m/s=5 m/s cos\theta$ (5)

Clearing $\theta$ :

$\theta=cos^{-1} (\frac{4 m/s}{5 m/s})$

$\theta=36.86\°$ This is the angle at which Ferdinand the frog jumps between lily pads

4 0

3 years ago

Under the influence of its drive force, a snowmobile is moving at a constant velocity along a horizontal patch of snow. When the

balandron [24]

Answer:

a) Δx = 11.6 m

b) t = 3.9 s

Explanation:

a)

Since the snowmobile is moving at constant speed, and the drive force is 195 N, this means that thereis another force equal and opposite acting on it, according to Newton's 2nd Law, due to there is no acceleration present in the horizontal direction .
This force is just the force of kinetic friction, and is equal to -195 N (assuming the positive direction as the direction of the movement).
Once the drive force is shut off, the only force acting on the snowmobile remains the friction force.
According Newton's 2nd Law, this force is causing a negative acceleration (actually slowing down the snowmobile) that can be found as follows:

$a = \frac{F_{fr} }{m} = \frac{-195N}{128kg} = -1.5 m/s2 (1)$

Assuming the friction force keeps constant, we can use the following kinematic equation in order to find the distance traveled under this acceleration before coming to an stop, as follows:

$v_{f} ^{2} -v_{o} ^{2} = 2* a* \Delta x (2)$

Taking into account that vf=0, replacing by the given (v₀) and a from (1), we can solve for Δx, as follows:

$\Delta x =- \frac{v_{o}^{2}}{2*a} =- \frac{(5.90m/s)^{2}}{2*(-1.5m/s2)} = 11.6 m (3)$

b)

We can find the time needed to come to an stop, applying the definition of acceleration, as follows:

$v_{f} = v_{o} + a*\Delta t (4)$

Since we have already said that the snowmobile comes to an stop, this means that vf = 0.
Replacing a and v₀ as we did in (3), we can solve for Δt as follows:

$\Delta t = \frac{-v_{o} }{a} = \frac{-5.9m/s}{-1.5m/s2} = 3.9 s (5)$

7 0

2 years ago

A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac

Y_Kistochka [10]

Answer:

See Explanation

Explanation:

a) We know that;

v = λf

Where;

λ = wavelength of the wave

f = frequency of the wave

v = velocity of the wave

So;

T = 2 * 2.10 s = 4.2 s

Hence f = 1/4.2 s

f = 0.24 Hz

The wavelength = 6.5 m

Hence;

v = 6.5 m * 0.24 Hz

v = 1.56 m/s

b)The amplitude of the wave is;

A = 0.600 m/2 = 0.300 m

c) Since the wave speed does not depend on the amplitude of the wave then the answer in (a) above remains the same

Where d = 0.30 m

A = 0.30 m/2 = 0.15 m

6 0

2 years ago

Ten high-technology batteries are tested for 200 hours each. One failed at 20 hours; another failed at 140 hours; all others com

Bas_tet [7]

Answer:

Failure rate = 20%

MTBF = 880 hours

Explanation:

given data

batteries = 10

tested = 200 hours

one failed = 20 hours

another fail at = 140 hours

solution

we know that Mean Time between Failures is express as = (Total up time) ÷ (number of breakdowns) ....................1

so here Total up time will be

Total up time = 200 × 10

Total up time = 2000

and here

Number of breakdown = 1 at 20 hour and another at 140 hour = 2

so it will be = (Total up time) ÷ (number of breakdowns) .......2

= $\frac{2000}{2}$ = 1000

so here gap between occurrences is

gap between occurrences= 140 - 20

gap between occurrences = 120 hour

and

MTBF will be

MTBF = 1000 - 120

MTBF = 880 hours

and

Failure rate (FR) will be

Failure rate (FR) = 1 ÷ MTBF ................3

Failure rate (FR) = R÷T ......................4

as here R is the number of failures and T is total time

so Failure rate (FR) = 20%

4 0

3 years ago