All categories

3 years ago

What are some types of landforms on Earth’s surface? PLS ANSWER QUICK 11 POINTS

Physics

2 answers:

Brrunno [24]3 years ago

5 0

Answer:

plateau, mountains, hills, plains

DochEvi [55]3 years ago

4 0

Plateau, butte, canyon, valley, mountain, hill

You might be interested in

Select all that apply. which of the following astronomers supported the sun-centered system? tycho brahe johannes kepler coperni

Alex73 [517]

<h3 />

-Tycho Brahe and

-Ptolemy

<h3 />

The sun is a hot ball of glowing gases which is a star whose gravity holds the solar system together and also keeping all the planet and smallest particles of debris in its orbit.

3 0

3 years ago

Read 2 more answers

A 50.0 kg person is walking horizontally with constant acceleration of 0.25 m/s² inside an elevator. The elevator is also accele

Alexxandr [17]

Answer:

The acceleration is in 2 D as in between east and south.

Explanation:

mass, m = 50 kg

acceleration, a = 0.25 m/s^2 horizontal

acceleration of elevator, a' = 1 m/s^2 downwards

When a person on the ground the resultant acceleration of the person with respect to the ground is between east and south direction so the path os parabolic in nature. It graph is shown below:

6 0

2 years ago

2.5 g of helium at an initial temperature of 300 K interacts thermally with 9.0 g of oxygen at an initial temperature of 620 K .

muminat

Answer:

Explanation:

2.5 g of He = 2.5 / 4 mole

= .625 moles

9 g of oxygen = 9/32

= .28 mole of oxygen

C_p of He = 3/2 R

C_p of O₂ = 5/2 R

A ) Initial thermal energy of He = 3/2 n R T

= 1.5 x .625 x 8.32 x 300

= 2340 J

Initial thermal energy of O₂ = 5/2 n R T

= 2.5 x .28 x 8.32 x 620

= 3610.88 J

B ) If T be the equilibrium temperature after mixing

gain of heat by helium

= n C_p Δ T

= .625 x 3/2 R x ( T - 300 )

Loss of heat by oxygen

n C_p Δ T

= .28 x 5/2 R x ( 620 - T )

Loss of heat = gain of heat

.625 x 3/2 R x ( T - 300 ) = .28 x 5/2 R x ( 620 - T )

1.875 T- 562.5 = 868- 1.4 T

3.275 T = 1430,5

T = 436.8 K

Thermal energy of He

= 1.5 x .625 x 8.32 x 436.8

= 3407 J

thermal energy of O₂

= 2.5 x .28 x 8.32 x 436.8

= 2543.92 J

C )

Heat energy transferred

= .28 x 5/2 R x ( 620 - T )

= .28 x 5/2 x 8.32 x ( 620 - 436.8 )

1066.95 J

Heat will flow from O₂ to He

Final temperature is 436.8 K

7 0

3 years ago

A football punter wants to kick the ball so that it is in the air for 4.5 s and lands 50 m from where it was kicked. Assume that

irakobra [83]

Answer:

(a) The angle of projection is 63 degree.

(b) The velocity of projection is 24.5 m/s.

Explanation:

Height, h = 1 m

horizontal distance, d = 50 m

time, t = 4.5 s

Let the initial velocity is u and the angle is A.

(a) Horizontal distance = horizontal velocity x time

50 = u cos A x 4.5

u cos A = 11.1 .....(1)

Use second equation of motion in vertical direction

$h = u t + 0.5 gt^2\\\\- 1 = u sin A \times 4.5 - 0.5 \times 9.8\times 4.5^2\\\\u sin A = 21.8 ..... (2)$

Divide (2) by (1)

tan A = 1.97

A = 63 degree

(b) Substitute the value of A in equation (2)

u x sin 63 = 21.8

u = 24.5 m/s

7 0

3 years ago

What is the matching nitrogen base sequence for the gene below?

grigory [225]

Answer:

Explanation:

T-G-T is the answer .

the matching Nitrogen base sequence for gene ATC is A

6 0

3 years ago