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2 years ago

In Florida, the standard defined speed limit in a residential area is ____ mph.

1 answer:

7 0

The correct answer among the choices given is the last option. In Florida, the standard defined speed limit in a residential area is 30 mph. In the United States, the speed limit in urban areas is 30 mph and in unpaved roads is 35 mph. In interstate roads the limit is at 70 mph.

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An iron railroad rail is 800 ft long when the temperature is 31°C. What is its length (in ft) when the temperature is −17°C?

natta225 [31]

Answer:

799.54 ft

Explanation:

Linear thermal expansion is:

ΔL = α L₀ ΔT

where ΔL is the change in length,

α is the linear thermal expansion coefficient,

L₀ is the original length,

and ΔT is the change in temperature.

Given:

α = 1.2×10⁻⁵ / °C

L₀ = 800 ft

ΔT = -17°C − 31°C = -48°C

Find: ΔL

ΔL = (1.2×10⁻⁵ / °C) (800 ft) (-48°C)

ΔL = -0.4608

Rounded to two significant figures, the change in length is -0.46 ft.

Therefore, the final length is approximately 800 ft − 0.46 ft = 799.54 ft.

7 0

2 years ago

The radius of curvature of a spherical mirror is 20cm.What is its focal length?

hichkok12 [17]

Answer:

$\boxed{\sf Focal \ length = 10 \ cm}$

Given:

Radius of curvature (R) of a spherical mirror = 20

To Find:

Focal length (f)

Explanation:

Formula:

$\boxed{ \bold{\sf Focal \ length \ (f) = \frac{Radius \ of \ curvature \ (R)}{2}}}$

Substituting value of R in the equation:

$\sf \implies f = \frac{20}{2}$

$\sf \implies f = \frac{ \cancel{2} \times 10}{ \cancel{2}}$

$\sf \implies f = 10 \: cm$

5 0

3 years ago

Read 2 more answers

A tensile test specimen has a gage length = 50 mm and its cross-sectional area = 100 mm2. The specimen yields at 48,000 N, and t

Shalnov [3]

Answer:

a) yield strength

$\sigma_y = \dfrac{F_y}{A} = =\dfrac{48000}{100} = 480 MPa$

b) modulus of elasticity

strain calculation

$\varepsilon_0=\dfrac{L-L_0}{L_0}=\dfrac{50.23-50}{50} = 0.0046$

strain for offset yield point

$\varepsilon_{new} = \varepsilon_0 -0.002$

=0.0046-0.002 = 0.0026

now, modulus of elasticity

$E = \dfrac{\sigma_y}{\varepsilon_{new}}=\dfrac{480}{0.0026}$

= 184615.28 MPa = 184.615 GPa

c) tensile strength

$\sigma_u =\dfrac{F_{max}}{A}=\dfrac{87000}{100}=870MPa$

d) percentage elongation

$\% Elongation = \dfrac{L-L_0}{L_0}\times 100 = \dfrac{67.3-50}{50}\times 100 = 34.6\%$

e) percentage of area reduction

$\% Area\ reduction = \dfrac{A-A_f}{A}\times 100=\dfrac{100-53}{100}= 47 \%$

7 0

3 years ago

Protons and neutrons grouped in a specific pattern

alexgriva [62]

Answer b protons and electrons

5 0

3 years ago

A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the

BartSMP [9]

Answer:

$r=1.14m$

Explanation:

$\theta$ is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

$F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB$

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

$F_m=F_c\\qvB=F_c(1)$

$F_c$ is the centripetal force and is defined as:

$F_c=m\frac{v^2}{r}$

Here $v$ is the proton's speed and $r$ is the radius of the circular motion. Replacing this in (1) and solving for r:

$qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}$

Recall that 1 J is equal to $6.242*10^{12}MeV$ , so:

$4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J$

We can calculate $v$ from the kinetic energy of the proton:

$K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}$

Finally, we calculate the radius of the proton path:

$r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m$

8 0

3 years ago