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Liono4ka [1.6K]
2 years ago
10

Triton is a moon of Neptune. It has a

Physics
1 answer:
Alik [6]2 years ago
6 0

Answer:

840.96 mi

Explanation:

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after 2 seconds its velocity is -20 m/s. after 3 seconds its velocity is -30 m/s. after 10 seconds its velocity is -100 m/s.

Explanation:

This is my answer.

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2 years ago
What is the acceleration of a 10 kg mass pushed by a 5 N force
hichkok12 [17]

Answer:

The formula is a = F m so in this case a = 5 10 = 0.5 m s 2

Explanation:

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2 years ago
Quando aquecemos água em nossas casas, ao nível do mar, utilizando um recipiente aberto, sua temperatura nunca ultrapassa os 100
iren2701 [21]

Answer:

i dont speak spanish

Explanation:

4 0
2 years ago
In a game of basketball, a forward makes a bounce pass to the center. The ball is thrown with an initial speed of 4.8 m/s at an
tensa zangetsu [6.8K]
The equation to be used here is the trajectory of a projectile as written below:

y = xtanθ +/- gx²/2v²(cosθ)²
where
y is the vertical distance
x is the horizontal distance
θ is the angle of trajectory or launch angle
g is 9.81 m/s²
v is the initial velcity

Since the angle is below horizontal, let's use the minus equation. Substituting the values:

- 0.8 m = xtan15° - (9.81 m/s²)x²/2(4.8 m/s)²(cos15°)²
Solving for x,
x = 2.549 m

However, we only take half of this distance because it was specified that the distance asked before bouncing. Hence, the horizontal distance is equal to 1.27 m.
5 0
3 years ago
A residential subdivision encompasses 1100 acres with a housing density of four houses per acre. Assume that a high-value reside
aleksandrvk [35]

Answer:

(a). The average daily demand of this subdivision is 2444.44 gallon/min.

(b). The design-demand used to design the distribution system is 2444.44 gallon/min.

Explanation:

Given that,

Area = 1100 acres

Number of house in 1 acres = 4

\text{Number of house in 1100 acres} = 4\times1100

\text{Number of house in 1100 acres} = 4400

Per house water demand = 800 g/day/house

(a). We need to calculate the average daily demand of this subdivision

Using formula for average daily demand

\text{average daily demand}=house\times\text{Per house water demand}

\text{average daily demand}=4400\times800\ gallon/day

\text{average daily demand}=3520000\ gallon/day

\text{average daily demand}=\dfrac{3520000}{24\times60}\ gallon/min

\text{average daily demand}=2444.44\ gallon/min

The average daily demand of this subdivision is 2444.44 gallon/min.

(b). We need to calculate the design-demand used to design the distribution system

Using formula for the design-demand

\text{design demand}=(Q_{max})daily\times\text{fire flow}

\text{design demand}=1.64\times2444.4\times1000

\text{design demand}=4008816\ gallon/m

Hence, (a). The average daily demand of this subdivision is 2444.44 gallon/min.

(b). The design-demand used to design the distribution system is 2444.44 gallon/min.

8 0
3 years ago
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