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iogann1982 [59]
4 years ago
15

A ball A is thrown vertically upward from the top of a 30-m-high building with an initial velocity of 5 m/s. At the same instant

another ball B is thrown upward from the ground with an initial velocity of 20 m/s. Determine the height from the ground and the time at which they pass.

Physics
1 answer:
Serga [27]4 years ago
6 0

Explanation:

Below is an attachment containing the solution.

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Kryger [21]
The answer for this question is letter "B.Fission releases energy, and its products have greater stability."

Fission and Fusion are both nuclear reactions that when they release energy, they make the nuclei more stable. So among the choices, option B is the most fitting for the definition. 
4 0
4 years ago
Donna drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took 8 hours. When Donna drove
Dennis_Churaev [7]

Answer:

d=360 miles

Donna lives 360 miles from the mountains.

Explanation:

Conceptual analysis

We apply the formula to calculate uniform moving distance[

d=v*t   Formula (1)

d: distance in miles

t: time in hours

v: speed in miles/hour

Development of problem

The distance Donna traveled to the mountains is equal to the distance back home, equal to d,then,we pose the kinematic equations for d, applying formula 1:

travel data to the mountains: t₁= 8 hours ,  v=v₁

d= v₁*t₁=8*v₁ Equation (1)

data back home : t₂=4hours ,  v=v₂=v₁+45

d=v₂*t₂=(v₁+45)*4=4v₁+180 Equation (2)

Equation (1)=Equation (2)

8*v₁=4v₁+180

8*v₁-4v₁=180

4v₁=180

v₁=180÷4=45 miles/hour

we replace v₁=45 miles/hour in equation (1)

d=8hour*45miles/hour

d=360 miles

8 0
3 years ago
Help me pls its for my science class i need to show my work
Lostsunrise [7]

Answer:

P = 5880  J

Explanation:

Given that,

The mass of the block, m = 30 kg

The block is sitting at a height of 20 m.

The block will have gravitational potential energy. The formula for gravitational potential energy is given by :

P=mgh\\\\=30\times 9.8\times 20\\\\=5880\ J

So, the required potential energy is equal to 5880  J.

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3 years ago
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sveticcg [70]
Earths tilt making the sun go haywire lol XD
6 0
3 years ago
An aircraft is in level flight at 225 km/hr through air at standard conditions. The lift coefficient at this speed is 0.45 and t
mojhsa [17]

Answer:

- the effective lift area for the aircraft is 8.30 m²

- the required engine thrust is 1275 N

- required power is 79.7 kW

Explanation:

Given the data in the question;

Speed V = 225 km/hr = 62.5 m/s

The lift coefficient CL = 0.45

drag coefficient CD = 0.065

mass = 900 kg

g = 9.81 m/s²

a)  the effective lift area for the aircraft

we know that for a steady level flight, weight = lift and thrust = drag

Using the equation for the lift force

F_L = C_L\frac{1}{2}ρV²A = W

we substitute

0.45 × \frac{1}{2} × 1.21 × ( 62.5 )² × A = ( 900 × 9.81 )

1081.05 × A = 8829

A = 8829 / 1081.05

A = 8.30 m²

Therefore, the effective lift area for the aircraft is 8.30 m²

b) the required engine thrust and power to maintain level flight.

we use the expression for drag force

F_D = T = C_D\frac{1}{2}ρV²A

we substitute

= 0.065 × \frac{1}{2} × 1.21 × ( 62.5 )² × 8.30

T = 1275 N

Since drag and thrust force are the same,

Therefore, the required engine thrust is 1275 N

Power required;

P = TV

p = 1275 × 62.5

p = 79687.5 W

p = ( 79687.5 / 1000 )kW

p = 79.7 kW

Therefore, required power is 79.7 kW

8 0
3 years ago
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