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SCORPION-xisa [38]
3 years ago
15

A curve of radius 53.1 m is banked so that a car of mass 2.9 Mg traveling with uniform speed 67 km/hr can round the curve withou

t relying on friction to keep it from slipping on the surface.
Physics
1 answer:
prisoha [69]3 years ago
6 0

Answer:

33.65°

Explanation:

radius, r = 53.1 m

m = 2.9 Mg = 2.9 x 10^6 g = 2900 kg

v = 67 km/h

convert km/h into m/s

v = 18.61 m/s

Let the angle of banking of road is θ, without friction

tan\theta =\frac{v^{2}}{rg}

tan\theta =\frac{18.61^{2}}{53.1\times 9.8}

tan  θ = 0.6655

θ = 33.65°

Thus, the angle of banking of road is 33.65°.

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A block of lead has dimensions of 4.50cm by 5.20cm by 6.00cm. The block weighs 1587g. From this information, calculate the densi
postnew [5]
Density = Mass / Volume
Volume = Area x Length = (4.50 x 5.20) x 6.00 = 140.4
Density = 1587 x 140.4 = 222,814.8 g/cm3 (cubed)
222,814.8 / 1000 = 222.8148 kg/m3 (cubed)
4 0
3 years ago
You are given aqueous solutions of six different substances and asked to determine whether they are strong, weak, or nonelectrol
kogti [31]

Answer:

Answer is explained below;

Explanation:

Electrolytes are any substances that dissociate into charged particles called ions when dissolved in water. The positively charged ions called cations and the negatively charged ions called anions move toward the negative and positive terminals (cathode and anode) of an electric circuit.

When a substance dissolved in water completely dissociates into ions, it is called a strong electrolyte. The aqueous solutions containing strong electrolytes conduct electricity very well and the examples include strong acids and soluble ionic compounds such as barium chloride, sodium hydroxide, etc.  

When a substance dissolved in water does not completely dissociate into ions, it is called a weak electrolyte. Since the aqueous solutions containing weak electrolytes have relatively few ions, their electrical conductivity is very low compared to the solutions containing strong electrolytes. Examples of weak electrolytes include weak acids and bases like acetic acid, ammonia, etc.

When a substance does not dissociate into ions when dissolved in water, it is called a nonelectrolyte. Since the aqueous solutions containing nonelectrolytes do not contain any ions, such solutions do not conduct electricity. Examples of nonelectrolytes are ethanol, aldehydes, glucose, ketones, etc.

If a solution contains dissolved ions, it conducts electricity and as the ion concentration increases, the conductivity also increases. To determine whether the aqueous solutions of six different substances are strong, weak, or nonelectrolytes, we can test them by applying a voltage to electrodes immersed in the solutions and a light bulb. By observing the brightness of the light bulb or by measuring the flow of electrical current, we can find out which solution contains a strong electrolyte or weak electrolyte, or nonelectrolyte.

If the solution contains a nonelectrolyte, the current flow is nil and the light bulb does not glow. If the solution contains a strong electrolyte, the current flow is very strong and so the brightness of the light bulb is very high. If the solution contains a weak electrolyte, the current flow is much low compared to the strong electrolyte and the light bulb glows, but the brightness is very low.

3 0
3 years ago
Two blocks of clay, one of mass 1.00 kg and one of mass 7.00 kg, undergo a completely inelastic collision. Before the collision
RideAnS [48]

Answer:8 m/s

Explanation:

Given

m_1=1 kg

m_2=7 kg

kinetic Energy of m_1=32 J

initially m_2 is at rest and let say m_1 is moving with velocity u

kinetic Energy of m_1 is =\frac{m_1u^2}{2}=32

u^2=64

u=8 m/s

In Completely inelastic collision both mass stick together and move with common velocity

Suppose v is the common velocity

m_1u+0=(m_1+m_2)v

v=\frac{1\times 8}{1+7}=1 m/s

therefore Final velocity with which both blocks moves is 1 m/s

7 0
3 years ago
A camera lens focuses on an object 75.0 cm from the lensThe image forms 3.50 cm behind the lens. What is the magnification of th
N76 [4]

Answer:

7/150

Explanation:

The following data were obtained from the question:

Object distance (u) = 75cm

Image distance (v) = 3.5cm

Magnification (M) =..?

Magnification is simply defined as:

Magnification (M) = Image distance (v)/ object distance (u)

M = v /u

With the above formula, we can obtain the magnification of the image as follow:

M = v/u

M = 3.5/75

M = 7/150

Therefore, the magnification of the image is 7/150.

6 0
3 years ago
A race car traveling at 10 meters per second accelerates at 1.5 meters per second squared while moving a distance of 600 meters.
Mekhanik [1.2K]

Answer:

Explanation:

Givens

vi = 10 m/s

a = 1.5 m/s^2

d = 600 m

vf = ?

Formula

vf^2 = vi^2 + 2*a*d

Solution

vf^2 = 10^2 + 2*1.5 * 600

vf^2 = 100 + 1800

vf^2 = 1900

sqrt(vf^2) = sqrt(1900)

vf = 43.59 m/s

7 0
3 years ago
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