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iren2701 [21]
3 years ago
6

A car is traveling at 60 mph. If the rate of speed increases 4 mph each hour,

Physics
2 answers:
Sauron [17]3 years ago
6 0

Answer:

5 hours ................

Vikentia [17]3 years ago
6 0

Answer:

B=5 hrs

explanation:

becoz if you add 4 mph/hr

it will graually be 80 in 5 hrs

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One litre of crude oil weighs 9.6N. Calculate its specific weight, density and specific gravity.​
Zepler [3.9K]

Answer:

The answer is "\bold{9600 \frac{N}{m^3}, 978.59 \frac{kg}{m^3}, and \ 0.978}"

Explanation:

Given:

\to v=1\ liter= 10^{-3} \ m^3\\\\\to  w= 9.6 \ N\\

calculation:

Specific \ weight =\frac{w}{v}=\frac{9.6}{10^{-3}}=9600 \frac{N}{m^3} \\\\w=mg\\\\m= \frac{w}{g}=\frac{9.6}{9.81}=0.9785\ kg\\\\\rho\ (density)=\frac{m}{v}=\frac{0.9785}{10^{-3}}=978.59 \frac{kg}{m^3}\\\\specific \ gravity = \frac{\prho \ obj}{\rho w}=\frac{978.54}{1000}=0.978

4 0
2 years ago
Which of the following is a correct equation for total energy?
Liula [17]

Answer:

kinetic energy + potential energy

6 0
3 years ago
Read 2 more answers
What is the displacement of the armadillo between 0s and 24s ?
Ann [662]

Answer:

Displacement: 6 meters

Distance: 24 meters

Explanation:

4 0
3 years ago
Vectors A and B lie in the xy-plane. Vector A has a magnitude of 19.1 and is at an angle of 125.5º counterclockwise from the +x-
Nady [450]

Answer:

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Explanation:

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8 0
3 years ago
A spotlight on a boat is y = 2.2 m above the water, and the light strikes the water at a point that is x = 8.5 m horizontally di
slava [35]

Answer:

The answer to the question is

The distance d, which locates the point where the light strikes the bottom is   29.345 m from the spotlight.

Explanation:

To solve the question we note that Snell's law states that

The product of the incident index and the sine of the angle of incident is equal to the product of the refractive index and the sine of the angle of refraction

n₁sinθ₁ = n₂sinθ₂

y = 2.2 m and strikes at x = 8.5 m, therefore tanθ₁ = 2.2/8.5 = 0.259 and

θ₁ =  14.511 °

n₁ = 1.0003 = refractive index of air

n₂ = 1.33 = refractive index of water

Therefore sinθ₂ =  \frac{n_1sin\theta_1}{n_2}  = \frac{1.003*0.251}{1.33} = 0.1885 and θ₂ = 10.86 °

Since the water depth is 4.0 m we have tanθ₂ = \frac{4}{x_2} or x₂ = \frac{4}{tan\theta_2 } =\frac{4}{tan(10.86)} = 20.845 m

d = x₂ + 8.5 = 20.845 m + 8.5 m = 29.345 m.

5 0
3 years ago
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