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3 years ago

If hot air rises then why is it cold in space please explain in detail thank you

2 answers:

7 0

Hot air doesn’t rise, its displaced by colder air because it is less dense. Also, space doesn’t exactly have a temperature. The gases are cold, especially the farther you are away from stars, planets and other objects.

Helen [10]3 years ago

5 0

Hot air has a lower density than the surrounding atmosphere, therefore,it will rise.. The reason it is "cold" in the upper atmosphere is actually because of a lower air density.

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A mass hanging from a spring is set in motion and its ensuing velocity is given by v (t )equals 2 pi cosine pi t for tgreater th

lianna [129]

Answer:

2(maximum), -2(minimum), -2(maximum).

Explanation:

V(t)= 2πcos πt--------------------------------------------------------------------------------(1).

Therefore, there is a need to integrate v(t) to get S(t).

S(t)= 2×sinπt + C ------------------------------------------------------------------------------(2).

Applying the condition given, we have s(0)= 0.

S(0)= 2sin ×π(0) + C.

Which means that; 0+C= 0. That is; C=0.

S(t)= 2 sin πt.

The mass moves to its highest positions at time,t=half(1/2=.5) and time,t=2.5.

Take note that; sin(π/2) = sin(5π/2) = 1 .

Also, the mass moves to its lowest position at time,t=(3/2); also, sin(3π/2) = -1.

Therefore, we have that 2 maximum; -2 minimum and -2 maximum.

7 0

3 years ago

Which statements correctly describe the properties of air? Select three options. Air has mass but no weight. Air pressure is cre

jolli1 [7]

Answer:

2,4,5

Air pressure is created by the weight of the atmosphere pushing on Earth’s surface.

Denser air is heavier than less dense air.

Air is less dense at higher altitudes.

4 0

3 years ago

Read 2 more answers

A cart loaded with bricks has a total mass of 9.13 kg and is pulled at constant speed by a rope. The rope is inclined at 24.7 ◦

blagie [28]

Answer:

W = 0.63 KJ

Explanation:

Work (W) is defined as the point product of force (F) by the distance (d)the body travels due to this force.

W= F*d *cosα Formula (1)

F : force (N)

d : displacement (m)

α : angle between force and displacement

Newton's second law:

∑F = m*a Formula (2)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration (m/s²)

We define the x-axis in the direction parallel to the movement of the cart on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the cart

W: Weight of the cart : In vertical direction

FN : Normal force : perpendicular to the floor

f : Friction force: parallel to the floor

T : tension Force, inclined at θ=24.7° above the horizontal

Calculated of the W

W= m*g

W= 9.13 kg* 9.8 m/s² = 89.47 N

x-y components o the tension force (T)

Tx = Tcosθ = T*cos 24.7° (N)

Ty = Tsin θ = T*sin 24.7° (N)

Calculated of the FN

We apply the formula (2)

∑Fy = m*ay ay = 0

FN +Ty- W = 0

FN = W-Ty

FN = 89.47-T*sin 24.7°

Calculated of the friction force (f)

f = μk*FN

f =(0.597)*( 89.47-T*sin 24.7° )

f= 53.41-0.249T

Calculated of the tension force of the rope (f)

We apply the formula (2) :

∑Fx = m*ax , ax= 0 ,because the speed of the cart is constant

Tx - f = 0

T*cos 24.7°-( 53.41 - 0.249T )= 0

T*cos 24.7° + 0.249T = 53.41

(1.1575)T = 53.41

T= (53.41) / (1.1575)

T= 46.14 N

Work done on the cart by the rope

We apply the formula (1)

W=T*d *cosα

W= (46.14 N)*(15.1 m) *(cos24.7)

W = 632.97 (N*m) = 632.97 (J)

W = 0.63 KJ

6 0

4 years ago

A model train engine was moving at a constant speed on a straight horizontal track. As the engine moved along, a marble was fir

bagirrra123 [75]

Answer:

The marble was moving in a projectile and the speed of the engine was 2.716 m/s

Explanation:

The vertical component of the marble's flight path relative to the train

is given by the equation y(t) = v*t - (4.9)*t^2,

where, v is the initial upward velocity of the marble relative to the train.

So with y(1) = v - 4.9 = 0 we have

v = 4.9 m/s.

The marble will reach maximum height after 0.5 seconds, at which the

height will be y(0.5) = (4.9)*(0.5) - (4.9)*(0.5)^2 = (4.9)*(0.25) = 1.225 m.

Now, the marble has a vertical velocity component of 4.9 m/s and a horizontal velocity component

of V m/s such that tan(61) = 4.9 / V

V = 4.9 / tan(61) = 2.716 m/s

This horizontal velocity component of the marble is the same as the

speed of the train i.e. 2.716 m/s.

3 0

3 years ago

The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou

puteri [66]

Answer:

$t=5.3687\ s$ is the time taken by the car to accelerate the desired range of the speed from zero at full power.

Explanation:

Given:

Range of speed during which constant power is supplied to the wheels by the car is $0\ mph\ to\ 70\ mph$ .

Initial velocity of the car, $v_i=0\ mph$

final velocity of the car during the test, $v_f=32\ mph=14.3052\ m.s^{-1}$

Time taken to accelerate form zero to 32 mph at full power, $t=1.2\ s$

initial velocity of the car, $u_i=0\ mph$

final desired velocity of the car, $u_f=64\ mph=28.6105\ m.s^{-1}$

Now the acceleration of the car:

$a=\frac{v_f-v_i}{t}$

$a=\frac{14.3052-0}{1.2}$

$a=11.921\ m.s^{-1}$

Now using the equation of motion:

$u_f=u_i+a.t$

$64=0+11.921\times t$

$t=5.3687\ s$ is the time taken by the car to accelerate the desired range of the speed from zero at full power.

8 0

3 years ago