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3 years ago

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Physics

2 answers:

blagie [28]3 years ago

5 0

Answer:

The last option.

Explanation:

Since you are going down, the gravitational potential energy would go down too. Thus, the gravitational potential energy decreases.

Since the gravitational potential energy is converted to kinetic energy when you move down, there is an increase in kinetic energy.

Goryan [66]3 years ago

4 0

Answer:

The fourth option

Explanation:

You might be interested in

A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 2.5 s

gulaghasi [49]

(a) 0.96 m/s

The period of the wave corresponds to the time taken for one complete oscillation of the boat, from the highest point to the highest point again. Since the time between the highest point and the lowest point is 2.5 s, the period is twice this time:

$T=2\cdot 2.5 s=5.0 s$

The frequency of the waves is the reciprocal of the period:

$f=\frac{1}{T}=\frac{1}{5.0 s}=0.20 Hz$

The wavelength instead is just the distance between two consecutive crests, so

$\lambda=4.8 m$

And the wave speed is given by:

$v=\lambda f=(4.8 m)(0.20 Hz)=0.96 m/s$

(b) 0.265 m

The total distance between the highest point of the wave and its lowest point is

d = 0.53 m

The amplitude is just the maximum displacement of the wave from the equilibrium position, so it is equal to half of this distance. So, the amplitude is

$A=\frac{d}{2}=\frac{0.53 m}{2}=0.265 m$

(c) Amplitude: 0.15 m, wave speed: same as before

In this case, the amplitude of the wave would be lower. In fact,

d = 0.30 m

So the amplitude would be

$A=\frac{d}{2}=\frac{0.30 m}{2}=0.15 m$

Instead, the wave speed would not change, since neither the frequency nor the wavelength of the wave have changed.

8 0

3 years ago

The law of conservation of momentum states that...

Brums [2.3K]

Answer:

The momentum before is equal to the momentum after

Explanation:

It is equal and should level out in an equation.

4 0

3 years ago

A point charge (–5.0 µC) is placed on the x axis at x = 4.0 cm, and a second charge (+5.0 µC) is placed on the x axis at x = –4.

AveGali [126]

Answer:

The magnitude of electric force is $7.2\times10^{-3} N$

Explanation:

Coulomb's Law:

The force of attraction or repletion is

directly proportional to the products of charges i.e $F\propto q_1q_2$

inversely proportional to the square of distance i.e $F\propto \frac{1}{r^2}$

$\therefore F\propto \frac{q_1q_2}{r^2}$

$\Rightarrow F=k \frac{q_1q_2}{r^2}$ [ k is proportional constant=9×10⁹N m²/C²]

There are two types of force applied on Q=+2.5 μC=2.5×10⁻⁶ C

Let F₁ force be applied on Q =+2.5 μC by q₁= -5.0 μC = - 5.0×10⁻⁶ C

and F₂ force be applied on Q=+2.5 μC by q₂= 5.0 μC= 5.0×10⁻⁶ C

Since the magnitude of F₁ and F₂ are same. Therefore their y component cancel.

If we draw a line from q₁ to Q .

The it forms a triangle whose base = 4.0 cm and altitude =3.0 cm.

Let hypotenuse = r

Therefore, $r=\sqrt{altitude^2+base^2} =\sqrt{3^2+4^2} =5$

we know,

$cos \theta = \frac{base }{hypotenuse}$

$\Rightarrow cos \theta = \frac{4 }{r}$

Total force $F_Q = 2.F_1 cos\theta \hat{i}$

$=2k\frac{Qq_1}{r^2} cos\theta \hat i$

$=2\ \frac{9\times1 0^9\times2.5 \times 5\times 10^{-12}}{r^2} \frac{4}{r} \hat i$

$=8\ \frac{9\times10^9\times2.5 \times 5\times 10^{-12}}{5^3} \hat i$ [ r=5]

$=7.2\times10^{-3}\hat i$ N

The magnitude of electric force is $7.2\times10^{-3} N$

3 0

3 years ago

NASA is designing a Mars-lander that will enter the Martian atmosphere at high speed. To land safely it must slow to a constant

Viktor [21]

Answer:

a) maximum mass of the Mars lander to ensure it can land safely is 200 kg

b) area of the parachute required is 480 m² which is larger than 400 m²

c) area of the parachute should be 12.68 m²

Explanation:

Given the data in the question;

V = 20 m/s

A = 200 m²

drag co-efficient CD = 1.855

g = 3.71 m/s²

density of the atmospheric pressure β = 0.01 kg/m³

a. Calculate the maximum mass of the Mars lander to ensure it can land safely?

Drag force FD = 1/2 × CD × β × A × V²

we substitute

FD = 1/2 × 1.855 × 0.01 kg/m × 200 m² × ( 20 m/s )²

FD = 742 N

we know that;

FD = Fg

Fg = gravity force

Fg = mg

FD = mg

m = FD/g

we substitute

m = 742 N / 3.71 m/s²

m = 200 kg

Therefore, the maximum mass of the Mars lander to ensure it can land safely is 200 kg

b. The mission designers consider a larger lander with a mass of 480 kg. Show that the parachute required would be larger than 400 m²;

Given that;

M = 480 kg

Show that the parachute required would be larger than 400 m²

we know that;

FD = Fg = Mg = 480 kg × 3.71 m/s²

FD = 1780.8 N

Now, FD = 1/2 × CD × β × A × V², we solve for A

A = FD / 0.5 × CD × β × V²

we substitute

A = 1780.8 / 0.5 × 1.855 × 0.1 × (20)²

A = 1780.8 / 3.71

A = 480 m²

Therefore, area of the parachute required 480 m² which is larger than 400 m²

c. To test the lander before launching it to Mars, it is tested on Earth where g = 9.8 m/s^2 and the atmospheric density is 1.0 kg m-3. How big should the parachute be for the terminal speed to be 20 m/s, if the mass of the lander is 480 kg?

Given that;

g = 9.8 m/s²,

β" = 1 kg/m³

v" = 20 m/s

M" = 480 kg

we know that;

FD = Fg = M"g

FD = 480 kg × 9.8 m/s² = 4704 N

from the expression; FD = 1/2 × CD × β × A × V²

A = FD / 0.5 × CD × β" × V"²

we substitute

A = 4704 / 0.5 × 1.855 × 1 × (20)²

A = 4704 / 371

A = 12.68 m²

Therefore area of the parachute should be 12.68 m²

3 0

3 years ago

A ray of light strikes a plane mirror at an angle of incidence 35°. If the mirror is rotated through 10°.

xxTIMURxx [149]

Answer:

1 35°

2 20°

3 50°

..............

7 0

3 years ago