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d1i1m1o1n [39]
4 years ago
5

he acceleration due to gravity on the surface of Mars is about one third the acceleration due to gravity on Earth’s surface. The

weight of a space probe on the surface of Mars is about _____.
Physics
1 answer:
Ahat [919]4 years ago
5 0

1/3 the weight than it is on earth

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At a given point on a horizontal streamline in flowing air, the static pressure is â2.0 psi (i.e., a vacuum) and the velocity is
Nastasia [14]
At a point on the streamline, Bernoulli's equation is
p/ρ + v²/(2g) = constant
where
p = pressure
v = velocity
ρ = density of air, 0.075 lb/ft³ (standard conditions)
g = 32 ft/s²

Point 1:
p₁ = 2.0 lb/in² = 2*144 = 288 lb/ft²
v₁ = 150 ft/s

Point 2 (stagnation):
At the stagnation point, the velocity is zero.

The density remains constant.
Let p₂ = pressure at the stagnation point.
Then,
p₂ = ρ(p₁/ρ + v₁²/(2g))
p₂ = (288 lb/ft²) + [(0.075 lb/ft³)*(150 ft/s)²]/[2*(32 ft/s²)
     = 314.37 lb/ft²
     = 314.37/144 = 2.18 lb/in²

Answer: 2.2 psi

5 0
3 years ago
1. The 10-min unithydrograph for a 2.25 km2 urban catchment is given by:(a) estimate the run off hydrograph for a 10-min rainfal
vivado [14]

Answer:

A ) Find the direct runoff depth; d =V A 22,698 _ 2.25x106 =0.01m =1cm The above result shows that the given hydrograph qualifies as a unit hydrograph. .

Explanation:

Brainlyist

5 0
3 years ago
How does kinetic energy affect the stopping distance of a small vehicle compared to a large vehicle?
Zarrin [17]
<span>The kinetic energy is the work done by the object due to its motion. It is represented by the formula of the half the velocity squared multiply by the mass of the object. In this problem, you have two  vehicles, the other one is large and the other one is small. Let us assume that they travel with the same velocity. Note that the kinetic energy is proportional to the mass of the object. So when you increase the mass of the other, it also increases the kinetic energy of that object. The same holds true for the two vehicles. The larger the vehicle, its kinetic energy is also large and therefore its stopping distance will be longer than that of the smaller vehicle.</span>
8 0
3 years ago
A 525 kg satellite is in a circular orbit at an altitude of 575 km above the Earth's surface. Because of air friction, the satel
Dafna1 [17]

Answer:

1.69\cdot 10^{10}J

Explanation:

The total energy of the satellite when it is still in orbit is given by the formula

E=-G\frac{mM}{2r}

where

G is the gravitational constant

m = 525 kg is the mass of the satellite

M=5.98\cdot 10^{24}kg is the Earth's mass

r is the distance of the satellite from the Earth's center, so it is the sum of the Earth's radius and the altitude of the satellite:

r=R+h=6370 km +575 km=6945 km=6.95\cdot 10^6 m

So the initial total energy is

E_i=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24} kg)}{2(6.95\cdot 10^6 m)}=-1.51\cdot 10^{10}J

When the satellite hits the ground, it is now on Earth's surface, so

r=R=6370 km=6.37\cdot 10^6 m

so its gravitational potential energy is

U = -G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24}kg)}{6.37\cdot 10^6 m}=-3.29\cdot 10^{10} J

And since it hits the ground with speed

v=1.90 km/s = 1900 m/s

it also has kinetic energy:

K=\frac{1}{2}mv^2=\frac{1}{2}(525 kg)(1900 m/s)^2=9.48\cdot 10^8 J

So the total energy when the satellite hits the ground is

E_f = U+K=-3.29\cdot 10^{10}J+9.48\cdot 10^8 J=-3.20\cdot 10^{10} J

So the energy transformed into internal energy due to air friction is the difference between the total initial energy and the total final energy of the satellite:

\Delta E=E_i-E_f=-1.51\cdot 10^{10} J-(-3.20\cdot 10^{10} J)=1.69\cdot 10^{10}J

8 0
3 years ago
The school bag of four students A,B,C,D measures 9kg, 2800gm, 2kg and 8000gm respectively. Whose bag is the lightest
inna [77]

Answer:

Student C

Explanation:

order from heaviest to lightest is...

9 kg (A) , 8000g (8 kg) (D) , 2800g (2.8kg) (B), 2 kg (C)

5 0
3 years ago
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