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Bingel [31]
3 years ago
13

Does an object travel farther on a smooth or slippery surface or on a rough surface? Why?

Physics
1 answer:
Vitek1552 [10]3 years ago
8 0

Answer:

There is much more friction on the rough surface than there is on the smooth surface.

Explanation:

You might be interested in
Electromagnetic radiation behaves both as particles (called photons) and as waves. Wavelength (λ) and frequency (ν) are related
inessss [21]
<h2>Answer: 136.363 m</h2>

Explanation:

We can find the wavelength of the radiation produced by the microwave oven by using the following given equation:

c=\lambda.\nu   (1)

Clearing \lambda :

\lambda=\frac{c}{\nu}   (2)

Knowing \nu=2.20 GHz=2.20(10)^{6}Hz=2.20(10)^{6}s^{-1}

\lambda=\frac{3(10)^{8}m/s}{2.20(10)^{6}s^{-1}}   (3)

\lambda=136.363m This is the wavelength of the radiation produced by the microwave

8 0
3 years ago
In 1999, Robbie Knievel was the first to jump the Grand Canyon on a motorcycle. At a narrow part of the canyon (65 m wide) and t
vfiekz [6]

Answer:

His launching angle was 14.72°

Explanation:

Please, see the figure for a graphic representation of the problem.

In a parabolic movement, the velocity and displacement vectors are two-component vectors because the object moves along the horizontal and vertical axis.

The horizontal component of the velocity is constant, while the vertical component has a negative acceleration due to gravity. Then, the velocity can be written as follows:

v = (vx, vy)

where vx is the component of v in the horizontal and vy is the component of v in the vertical.

In terms of the launch angle, each component of the initial velocity can be written using the trigonometric rules of a right triangle (see attached figure):

sin angle = opposite / hypotenuse

cos angle = adjacent / hypotenuse

In our case, the side opposite the angle is the module of v0y and the side adjacent to the angle is the module of vx. The hypotenuse is the module of the initial velocity (v0). Then:

sin angle = v0y / v0  then: v0y = v0 * sin angle

In the same way for vx:

vx = v0 * cos angle

Using the equation for velocity in the x-axis we can find the equation for the horizontal position:

dx / dt = v0 * cos angle

dx = (v0 * cos angle) dt (integrating from initial position, x0, to position at time t and from t = 0 and t = t)

x - x0 = v0 t cos angle

x = x0 + v0 t cos angle

For the displacement in the y-axis, the velocity is not constant because the acceleration of the gravity:

dvy / dt = g ( separating variables and integrating from v0y and vy and from t = 0 and t)

vy -v0y = g t

vy = v0y + g t

vy = v0 * sin angle + g t

The position will be:

dy/dt = v0 * sin angle + g t

dy = v0 sin angle dt + g t dt (integrating from y = y0 and y and from t = 0 and t)

y = y0 + v0 t sin angle + 1/2 g t²

The displacement vector at a time "t" will be:

r = (x0 + v0 t cos angle, y0 + v0 t sin angle + 1/2 g t²)

If the launching and landing positions are at the same height, then the displacement vector, when the object lands, will be (see figure)

r = (x0 + v0 t cos angle, 0)

The module of this vector will be the the total displacement (65 m)

module of r = \sqrt{(x0 + v0* t* cos angle)^{2} }  

65 m = x0 + v0 t cos angle ( x0 = 0)

65 m / v0 cos angle = t

Then, using the equation for the position in the y-axis:

y = y0 + v0 t sin angle + 1/2 g t²

0 =  y0 + v0 t sin angle + 1/2 g t²

replacing t =  65 m / v0 cos angle and y0 = 0

0 = 65m (v0 sin angle / v0 cos angle) + 1/2 g (65m / v0 cos angle)²  

cancelating v0:

0 = 65m (sin angle / cos angle) + 1/2 g * (65m)² / (v0² cos² angle)

-65m (sin angle / cos angle) = 1/2 g * (65m)² / (v0² cos² angle)  

using g = -9.8 m/s²

-(sin angle / cos angle) * (cos² angle) = -318.5 m²/ s² / v0²

sin angle * cos angle = 318.5 m²/ s² / (36 m/s)²

(using trigonometric identity: sin x cos x = sin (2x) / 2

sin (2* angle) /2 = 0.25

sin (2* angle) = 0.49

2 * angle = 29.44

<u>angle = 14.72°</u>

3 0
3 years ago
A car initially traveling at 24 m/s slams on the brakes and moves forward 196 m before coming to a complete halt. What was the m
Marrrta [24]

Answer:

-1.47 m/s^2

Explanation:

We can use the following SUVAT equation to solve the problem:

v^2 - u^2 = 2ad

where

v = 0 is the final velocity of the car

u = 24 m/s is the initial velocity

a is the acceleration

d = 196 m is the displacement of the car before coming to a stop

Solving the equation for a, we find the acceleration:

a=\frac{v^2-u^2}{2d}=\frac{0-(24)^2}{2(196)}=-1.47 m/s^2

4 0
3 years ago
‏A 50 - N x m torque acts on a wheel with a moment of inertia 150 kg x m² . If the wheel starts from rest , how long will it tak
denis-greek [22]

Answer:

t = 6.17 s

Explanation:

For a 1 revolution movement, \triangle \theta = 2\pi

Torque, \tau = 50 Nm

Moment of Inertia, I = 150 kg m^2

If the wheel starts from rest, w_{0} = 0 rad/s

The angular displacement of the wheel can be given by the formula:

\triangle \theta = \omega_0 t + 0.5 \alpha t^2................(1)

Where \alpha is the angular acceleration

\tau = I \alpha\\\alpha = \frac{\tau}{I} \\\alpha = 50/150\\\alpha = 0.33 rad/s^2

To get t, put all necessary parameters into equation (1)

2\pi = 0(t) + 0.5(0.33)t^2\\2\pi =0.5(0.33)t^2\\t^2 = \frac{4 \pi}{0.33} \\t^2 = 38.08\\t = 6.17 s

3 0
3 years ago
Please help me with this question ASAP.
victus00 [196]

Answer:

i. 6.923 V

ii. The e.m.f. = 22.5 V

Explanation:

i. The given parameters are;

Length of potentiometer = 1 m

The resistance of the potentiometer = 10 Ω

The e. m. f. of the attached cell = 9 V

The current, I flowing in the circuit = e. m. f/(Total resistance)

The current, I flowing in the circuit = 9 V/(10 + 3) = 9/13 A

The potential difference, p.d. across the 1 m potentiometer wire = I × Resistance of the potentiometer wire

The p.d. across the potentiometer wire = 9/13×10 = 90/13 = 6.923 V

ii) Given that the 1 m potentiometer wire has a resistance of 10 Ω, 75 cm which is 0.75 m will have an e.m.f. given by the following relation;

\dfrac{E}{R_{balance}} = \dfrac{V}{R_{cell}}

Where:

E = e.m.f. of the balance point cell

R_{balance} = Resistance of 75 cm of potentiometer wire  = 0.75×10 = 7.5 Ω

R_{cell} = Resistance of the cell in the circuit = 3 Ω

V = e.m.f. attached cell = 9 V

\dfrac{E}{7.5} = \dfrac{9}{3}

E = 7.5*3 = 22.5 V

The e.m.f. = 22.5 V

7 0
3 years ago
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