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lisov135 [29]
3 years ago
5

The rate at which the earth's surface loses heat and keeps earth temperature at which living things can survive

Physics
1 answer:
agasfer [191]3 years ago
5 0
<span>I believe it's insulation.</span>
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A fire helicopter carries a 580-kg bucket of water at the end of a 20.0-m long cable. Flying back from a fire at a constant spee
inn [45]

Answer:

F = 41,954 N

Explanation:

given,

mass of bucket = 580 Kg

length of the cable = 20 m

velocity = 40 m/s

angle made = 38.0°

T cos 38° = m g..............(1)

T sin 38^0 = \dfrac{mv^2}{l} + F......(2)

dividing equation (2) by (1)

tan 38^0 = \dfrac{\dfrac{mv^2}{l} + F}{mg}

tan 38^0 = \dfrac{\dfrac{580\times 40^2}{20} + F}{580 \times 9.81}

4445.36 = \dfrac{580\times 40^2}{20} + F

F = -46400 + 4445.36

F = -41,954 N

hence, the force is acting in the opposite direction as assumed.

F = 41,954 N

4 0
3 years ago
If 100 J of electrical energy enter the bulb and 5 J of light energy leave the bulb, how many joules of heat energy leave the bu
Novay_Z [31]

As per energy conservation we know that

Energy enter into the bulb = Light energy + Thermal energy

so now we have

energy enter into the bulb = 100 J

Light energy = 5 J

now from above equation we have

100 = 5 + heat

Heat = (100 - 5) J

Heat = 95 J


6 0
3 years ago
It is warmer near the equator because that is where the Earth gets the most direct sunlight. T or f
Anni [7]
True this is true bc yes as you said you're at the most direct point of sunlight
4 0
3 years ago
Read 2 more answers
An airplane traveling 919 m above the ocean at 505 m/s is to drop a case of twinkies to the victims below. How much time before
Elena L [17]

Answer:

a. 13.7 s b. 6913.5 m

Explanation:

a. How much time before being directly overhead should the box be dropped?

Since the box falls under gravity we use the equation

y = ut - 1/2gt² where y = height of plane above ocean = 919 m, u = initial vertical velocity of airplane = 0 m/s, g = acceleration due to gravity = -9.8 m/s² and t = time it takes the airplane to be directly overhead.

So,

y = ut - 1/2gt²

y = 0 × t - 1/2gt²

y = 0 - 1/2gt²

y = - 1/2gt²

t² = -2y/g

t = √(-2y/g)

So, t = √(-2 × 919 m/-9.8 m/s²)

t = √(-1838 m/-9.8 m/s²)

t = √(187.551 m²/s²)

t = 13.69 s

t ≅ 13.7 s

So, the box should be dropped 13.69 s before being directly overhead.

b. What is the horizontal distance between the plane and the victims when the box is dropped?

The horizontal distance x between plane and victims, x = speed of plane × time it takes for box to drop = 505 m/s × 13.69 s = 6913.45 m ≅ 6913.5 m

6 0
3 years ago
Physics Kinematics question
just olya [345]
An interesting problem, and thanks to the precise heading you put for the question.

We will assume zero air resistance.
We further assume that the angle with vertical is t=53.13 degrees, corresponding to sin(t)=0.8, and therefore cos(t)=0.6.

Given:
angle with vertical, t = 53.13 degrees
sin(t)=0.8; cos(t)=0.6;
air-borne time, T = 20 seconds
initial height, y0 = 800 m

Assume g = -9.81 m/s^2

initial velocity, v m/s (to be determined)

Solution:

(i) Determine initial velocity, v.
initial vertical velocity, vy = vsin(t)=0.8v
Using kinematics equation, 
S(T)=800+(vy)T+(1/2)aT^2 ....(1)  
Where S is height measured from ground.

substitute values in (1):  S(20)=800+(0.8v)T+(-9.81)T^2  =>
v=((1/2)9.81(20^2)-800)/(0.8(20))=72.625 m/s  for T=20 s

(ii) maximum height attained by the bomb
Differentiate (1) with respect to T, and equate to zero to find maximum
dS/dt=(vy)+aT=0 =>
Tmax=-(vy)/a = -0.8*72.625/(-9.81)= 5.9225 s

Maximum height,
Smax
=S(5.9225)
=800+(0.8*122.625)*(5.9225)+(1/2)(-9.81)(5.9225^2)
= 972.0494 m

(iii) Horizontal distance travelled by the bomb while air-borne
Horizontal velocity = vx = vcos(t) = 0.6v = 43.575 m/s
Horizontal distace travelled, Sx = (vx)T = 43.575*20 = 871.5 m

(iv) Velocity of the bomb when it strikes ground
vertical velocity with respect to time
V(T) =vy+aT...................(2)
Substitute values, vy=58.1 m/s, a=-9.81 m/s^2
V(T) = 58.130 + (-9.81)T => 
V(20)=58.130-(9.81)(20) = -138.1 m/s (vertical velocity at strike)

vx = 43.575 m/s (horizontal at strike)
resultant velocity = sqrt(43.575^2+(-138.1)^2) = 144.812 m/s  (magnitude)
in direction theta = atan(43.575,138.1) 
= 17.5 degrees with the vertical, downward and forward. (direction)
4 0
3 years ago
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