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2 years ago

Name 3 different types of graphs that can be used to plot data.

Physics

2 answers:

siniylev [52]2 years ago

7 0

Answer:

Bar graph

Pie graph

Line Graph

Explanation:

mark as brainliest and drop some thanks!!!

Natali5045456 [20]2 years ago

3 0

Line graphs, bar graphs and histograms, pie charts, and Cartesian graphs.

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A curve in a stretch of highway has radius R. The road is not banked in any way. The coefficient of static friction between the

adelina 88 [10]

Answer:

maximum possible velocity = $\sqrt{ugR}$

Explanation:

centripetal acceleration when the car is going in the circle must be less than the maximum friction for the car to not slip.

centripetal acceleration $\frac{mv^{2}}{r}$

where v is the velocity of car and r is the radius of circle

maximum friction = umg

where u is the coefficient of static friction.

therefore $umg\geq \frac{mv^{2}}{R}$

therefore maximum possible velocity = $\sqrt{ugR}$

6 0

3 years ago

A diffraction pattern forms when light passes through a single slit. The wavelength of the light is 691 nm. Determine the angle

expeople1 [14]

Explanation:

Given that,

Wavelength of the light, $\lambda=691\ nm=691\times 10^{-9}\ m$

(a) Slit width, $a=3.8\times 10^{-4}\ m$

The angle that locates the first dark fringe is given by :

$sin\theta=\dfrac{\lambda}{a}$

$sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-4}}$

$\theta=0.104^{\circ}$

(b) Slit width, $a=3.8\times 10^{-6}\ m$

The angle that locates the first dark fringe is given by :

$sin\theta=\dfrac{\lambda}{a}$

$sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-6}}$

$\theta=10.47^{\circ}$

Hence, this is the required solution.

7 0

3 years ago

In 8.4 s a fisherman winds 2.9 m of fishing line onto a reel whose radius is 3.0 cm (assumed to be constant as an approximation)

alukav5142 [94]

Answer:

The angular speed of the reel is 11.33 rad/s

Explanation:

Given

The fisherman takes t = 8.4 s to wind distance x = 2.9 m into a circle radius of r = 3 cm = 0.03 m

Than the tangencial speed equals the change in the distance to the time

v = $\frac{x}{t}$ = $\frac{2.9 m}{8.4 s} = 0.34 \frac{m}{s}$

Knowing the tangencial velocity is proportional to the radius r and the angular velocity

v = r*w

w = $\frac{0.34 m/s}{0.03 m} = 11.33 \frac{rad}{s}$

3 0

3 years ago

Energy cannot be created nor destroyed. A. Ella said "a lot of energy in the hot water has disappeared." Explain why her stateme

harkovskaia [24]

Answer:

See Explanation

Explanation:

The principle of conservation of energy states that; energy can neither be created nor destroyed but is converted from one form to another.

In view of this principle, Ella can not be correct when she says that a lot of energy has disappeared. The use of the term "disappeared" connotes the idea that the energy no longer exists which does not happen.

Hence, energy can not "disappear" from hot water rather the energy in the water may be transferred to the surroundings.

6 0

2 years ago

A 56 kg diver runs and dives from the edge of a cliff into the water which is located 4.0 m below. If she is moving at 8.0 m/s t

Reil [10]

Answer:

1) 2197.44 J

2) 0 J

3) 2197.44 J = Constant

4) 2197.44 J

5) Approximately 8.86 m/s

Explanation:

The given parameters are;

The mass of the diver, m = 56 kg

The height of the cliff, h = 4.0 m

The speed with which the diver is moving, vₓ = 8.0 m/s

The gravitational potential energy = Mass, m × Height of the cliff, h × Acceleration due to gravity, g

1) Her gravitational potential energy = 56 × 4.0 × 9.81 = 2197.44 J

2) The kinetic energy = 1/2·m·u²

Where;

u = Her initial velocity = 0 when she just leaves the cliff

Therefore;

Her kinetic energy when she just leaves the cliff = 1/2 × 56 × 0² = 0 J

3) The total mechanical energy = Kinetic energy + Potential energy

The total mechanical energy is constant

Her total mechanical energy relative to the water surface when she leaves the cliff = Her gravitational potential energy = 2197.44 J = Constant

4) Her total mechanical energy relative to the water surface just before she enters the water = 2197.44 J

5) The speed with which she enters the water, v, is given from, v² = u² + 2·g·h

Where;

u = The initial velocity at the top of the cliff before she jumps= 0 m/s

∴ v² = 0² + 2 × 9.81 × 4 = 78.48

v = √78.48 ≈ 8.86 m/s

The speed with which she enters the water, v ≈ 8.86 m/s

7 0

2 years ago