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Phantasy [73]
3 years ago
8

A 5kg wheel rolls off a flat roof of a 50 m tall building at 12m/s.

Physics
1 answer:
Olegator [25]3 years ago
6 0

Explanation:

a) Given in the y direction (taking down to be positive):

Δy = 50 m

v₀ = 0 m/s

a = 10 m/s²

Find: t

Δy = v₀ t + ½ at²

50 m = (0 m/s) t + ½ (10 m/s²) t²

t = 3.2 s

b) Given in the x direction:

v₀ = 12 m/s

a = 0 m/s²

t = 3.2 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (12 m/s) (3.2 s) + ½ (0 m/s²) (3.2 s)²

Δx = 38 m

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If a car is speeding down a road at 40 miles/hour (mph), how long is the stopping distance d40 compared to the stopping distance
tia_tia [17]
Assume that the deceleration due to braking is a ft/s².

Note that
40 mph = (40/60)*88 = 58.667 ft/s
25 mph = (25/60)*88 = 36.667 ft/s

The final velocity is zero when the car stops, therefore
v² - 2ad = 0, or d = v²/(2a)
where
v = initial speed
a = deceleration
d = stopping distance.

The stopping distance, d₄₀, at 40 mph is
d₄₀ = 58.667²/(2a)
The stopping distance, d₂₅, at 25 mph is
d₂₅ = 36.667²/(2a)

Therefore
d₄₀/d₂₅ = 58.667²/(2a) ÷ 36.667²/(2a)
           = (58.667/36.667)²
           = 2.56

Answer:
The stopping distance at 40 mph is 2.56 times the stopping distance at 25 mph.
8 0
3 years ago
Read 2 more answers
The metallic elements which all react in water are found in the:
MAXImum [283]
The metallic elements which all react to water are found in the same group, the groups name is the Alkali metals, or also known as group one. 
6 0
4 years ago
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If the mass of the book is 50 sliding with acceleration 1.2 m/s ^ 2 then the friction force is
Leokris [45]
73 Newton is the correct answer
6 0
3 years ago
What is a force?
Fudgin [204]

Answer:

2. A push or pull

Explanation:

I'm 100% sure

5 0
3 years ago
The lighting needs of a storage room are being met by six fluorescent light fixtures, each fixture containing four lamps rated a
Amanda [17]

Answer:

amount of energy  = 4730.4 kWh/yr

amount of money = 520.34 per year

payback period = 0.188 year

Explanation:

given data

light fixtures = 6

lamp = 4

power = 60 W

average use = 3 h a day

price of electricity = $0.11/kWh

to find out

the amount of energy and money that will be saved and simple payback period if the purchase price of the sensor is $32 and it takes 1 h to install it at a cost of $66

solution

we find energy saving by difference in time the light were

ΔE = no of fixture × number of lamp × power of each lamp × Δt

ΔE is amount of energy save and Δt is time difference

so

ΔE = 6 × 4 × 365 ( 12 - 9 )

ΔE = 4730.4 kWh/yr

and

money saving find out by energy saving and unit cost that i s

ΔM = ΔE × Munit

ΔM = 4730.4 × 0.11

ΔM = 520.34 per year

and

payback period is calculate as

payback period = \frac{excess initial cost}{\Delta M}

payback period = \frac{32 + 66}{520.34}

payback period = 0.188 year

8 0
3 years ago
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