How long does it take a plane that is traveling at 350 km/h to travel 1750 km?

What happens to the free energy released as electrons are passed from photosystem II to photosystem I through a series of electr

lana [24]

<span>It is used to establish and maintain a proton gradient.</span>

6 0

3 years ago

Which of the following events may occur when IR radiation is absorbed?

kherson [118]

can you have all of them as possibilities ???

5 0

3 years ago

An object is observed for a time interval of 20 seconds. From time 6.7 s the object experiences a Force of 106 N that lasts unti

shusha [124]

Answer:

1654 kg m/s

Explanation:

The impulse experienced by an object is equal to the product between the force exerted on the object and the time during which the force lasts:

$I=F\Delta t$

where:

I is the impulse

F is the force exerted on the object

$\Delta t$ is the time during which the force is applied

For the object in this problem, we have

$F=106 N$ (force applied)

$\Delta t= 15.6 s$ (time interval)

Therefore, the impulse experienced by the object is:

$I=(106)(15.6)=1654 kg m/s$

3 0

3 years ago

The apparent depth of an object at the bottom of a tank filled with a liquid of refractive index 1.3 is 7.7 cm. What is the actu

Juliette [100K]

$\orange{\underline{\huge{\bold{\textit{\green{\bf{QUESTION}}}}}}}$

The apparent depth of an object at the bottom of a tank filled with a liquid of refractive index 1.3 is 7.7 cm. What is the actual depth of the liquid in the tank?

${\bold{\blue{GIVEN}}}$

REFRACTIVE INDEX = 1.3

APPARENT DEPTH = 7.7 cm

${ \bold{\green{To \: Find}}}$

REAL DEPTH OF THE OBJECT.

${\red{FORMULA \: \: USED }}$

$Reflective \: Index = \frac{Real \: Depth}{Apparent \: Depth }$

$\huge\mathbb{\red A \pink{N}\purple{S} \blue{W} \orange{ER}}$

Refractive Index = 1.3

Apparent Depth = 7.7 cm

Putting the values in the formula:-

$Reflective \: Index = \frac{Real \: Depth}{Apparent \: Depth } \\ \\ 1.3 = \frac{Real \: Depth}{7.7 \: cm} \\ \\ 1.3 \times 7.7 = Real \: Depth \\ \\ 10.01 \: \: cm = Real \: Depth$

3 0

2 years ago

if a 60 kg person was standing on a platform at the surface of saturn and they jumped, they would have to push with a force grea

lidiya [134]

Answer:

A 60 kg person standing on a platform at the surface of Saturn and they jumped, they would have to push with a force greater than 540 N

Explanation:

The gravitational attraction between an object on the surface of a planet and the planet is given by the weight of the object

Therefore the force needed to be applied for an object to lift off the surface of a planet = The weight of the object

The weight of the object on the surface of a planet = m × g

Where;

m = The mass of the object

g = The strength of gravity on the planet's surface in N/kg

The given parameters are;

The mass of the person standing on a platform at the surface of Saturn, m = 60 kg

The strength of gravity on the surface of Saturn = 9 N/kg

Therefore, we have;

The weight of the person = The force greater than which the person would have to push on the surface of Saturn so as to Jump = The weight of the person on the surface of Saturn = 60 kg × 9 N/kg = 540 N

Therefore, for a 60 kg person standing on a platform at the surface of Saturn and they jumped, they would have to push with a force greater than 540 N.

4 0

3 years ago