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saul85 [17]
3 years ago
6

What is the force needed to accelerate a 1,357 kg car forward at 8.0 m/s squared?

Physics
1 answer:
USPshnik [31]3 years ago
8 0
Formula for Force is:
F=mass x Acceleration
Force= 1,357x8.0m/s
Force= 10,856N
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Answer:

an example of an exthermic process is combustion

Explanation:

combustion is like lighting a candle

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Find the Y component of 35m/s at 57 degrees from the X-axis.
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35m/s[57o].

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3 years ago
Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial spe
Lisa [10]

Answer:

1) v₀x = 13.76 m/s

2) v₀y = 19.66 m/s

3) ymax = 21.199 m

4) X = 55.1746 m

5) and 6) y = 18.4 m

Explanation:

1) v₀x = v₀*Cos α = 24 m/s* Cos 55° = 13.76 m/s

2) v₀y = v₀*Sin α = 24 m/s* Sin 55° = 19.66 m/s

3) ymax = y₀ + (v₀y²/(2g)) = 1.5 m + ((19.66 m/s)²/(2*9.81 m/s²)) = 21.199 m

4) We can use this equation

y = y₀ + (tan α)*x – (g / (2* v₀x²))*x²

where y = y₀ = 1.5 m

then

1.5 = 1.5 + tan (55°)*x - (9.81 / (2* (13.76)²))*x²

⇒   0.02588 x² - 1.42815 x = 0

Solving this equation we get

x₁ = 0     and    x₂ = 55.1746 m

The distance between the two girls is 55.1746 m

5) and 6) If   v₀x = 15 m/s = vx   and   ymax = 24 m

y = ?   when x = (xmax/2)

ymax = y₀ + (v₀y²/(2g)) ⇒ v₀y = √(2g*(ymax - y₀))

⇒ v₀y = √(2(9.81 m/s²)(24 m - 1.5 m)) = 21.01 m/s

then we get α' as follows

α' = tan⁻¹(v₀y/v₀x) = tan⁻¹(21.01 m/s/15 m/s) = 54.47°

v₀ = √(v₀x² + v₀y²) = √((15 m/s)² + (21.01 m/s)²) = 25.81 m/s

Now we can apply the equation of the path

y = ymax - ((gx²)/(2v₀²))

⇒  y = 24m - ((9.81)(55.1746/2)²/(2*25.81²))

⇒  y = 18.4 m

7 0
3 years ago
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What is true about solar energy hitting earth?
polet [3.4K]
B. It’s the same roughly at all latitudes
6 0
2 years ago
Given the indices of refraction n_1 and n_2 of material 1 and material 2, respectively, rank these scenarios on the basis of the
aniked [119]

Answer:

a)  order of refraction is a, e, de, c , c) a f g

Explanation:

a) when lightning is refracted it must comply with the law of refraction

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          sin θ / sin δδa) when lightning is refracted it must comply with the law of refraction

          .n1 sin θ₁ = n2 sin θ ₂

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1 we see that the ray deviation is promotional to the ratio of the refractive indices

The order of refraction is a, e, de, c

.b) When a ray passes from a medium with less indicated refraction to a higher index the part of the reflected ray has a phase change of 180º

a) no phase change

b) reflected ray has a phase change of 180º

c) no phase change

d) no phase change

e) there is a phase change

f) these phase changeo  

1 we see that the ray deviation is promotional to the ratio of the refractive indices

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.b) When a ray passes from a medium with less indicated refraction to a higher index the part of the reflected ray has a phase change of 180º

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d) no phase change

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f) ay phase vabio

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