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3 years ago

2.08

Physics

1 answer:

natka813 [3]3 years ago

5 0

Calculate its average speed in meters per second

Answer:

5.77 m/s

Explanation:

Speed= Distance/Time

Distance= 40+ half of 40= 40+20= 60 m

Time= 8.8+1.6=10.4 s

Average speed= 60/10.4=5.769230769 m/s

Approximately, the average speed is 5.77 m/s

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The speed of light is about 3.00 × 10 meters per second. What is the frequency of green light that has a wavelength about 500 na

RideAnS [48]

Answer:

The frequency of the green light is $6x10^{14}Hz$

Explanation:

The visible region is part of the electromagnetic spectrum, any radiation of that electromagnetic spectrum has a speed of $3.00x10^{8}m/s$ in the vacuum.

Green light is part of the visible region. Therefore, the frequency can be determined by the following equation:

$c = \lambda \cdot \nu$ (1)

Where c is the speed of light, $\lambda$ is the wavelength and $\nu$ is the frequency.

Notice that since it is electromagnetic radiation, equation 1 can be used. Remember that light propagates in the form of an electromagnetic wave (that is a magnetic field perpendicular to an electric field).

Then, $\nu$ can be isolated from equation 1

$\nu = \frac{c}{\lambda}$ (2)

Notice that it is necessary to express the wavelength in units of meters.

$\lambda = 500nm . \frac{1m}{1x10^{9}nm}$ ⇒ $5x10^{-7}m$

$\nu = \frac{3.00x10^{8}m/s}{5x10^{-7}m}$

$\nu = 6x10^{14}s^{-1}$

$\nu = 6x10^{14}Hz$

Hence, the frequency of the green light is $6x10^{14}Hz$

4 0

4 years ago

How high would a projectile go if it was launched from ground level with an initial speed of 26 m/s at an angle of 30 degrees ab

tigry1 [53]

Answer:

Vy = 26 m/s sin 30 = 13 m/s vertical speed

t = Vy / a = 13 m/s / 9.80 m/s^2 = 1.33 sec time to reach Vy = 0

H = Vy t + 1/2 g t^2

H = 13 m/s * 1.33 sec - 1.33^2 * 9.8 / 2 m = 8.62 m

4 0

1 year ago

A car on a level road turns a quarter circle ccw. You learned in Chapter 8 that static friction causes the centripetal accelerat

pav-90 [236]

Answer:

Zero

Explanation:

As we know that the force and the motion direction should always be perpendicular to each other due to which the work is done by static friction be zero

Therefore

F.dcos(theta) = F.d cos(90) = 0

Hence, the work done by static friction is zero

Therefore the same is to be considered

3 0

3 years ago

If the range of a projectile's trajectory is six times larger than the height of the trajectory, then what was the angle of laun

zvonat [6]

Answer:

H = 1/2 g t^2 where t is time to fall a height H

H = 1/8 g T^2 where T is total time in air (2 t = T)

R = V T cos θ horizontal range

3/4 g T^2 = V T cos θ 6 H = R given in problem

cos θ = 3 g T / (4 V) (I)

Now t = V sin θ / g time for projectile to fall from max height

T = 2 V sin θ / g

T / V = 2 sin θ / g

cos θ = 3 g / 4 (T / V) from (I)

cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ

tan θ = 2/3

θ = 33.7 deg

As a check- let V = 100 m/s

Vx = 100 cos 33.7 = 83,2

Vy = 100 sin 33,7 = 55.5

T = 2 * 55.5 / 9.8 = 11.3 sec

H = 1/2 * 9.8 * (11.3 / 2)^2 = 156

R = 83.2 * 11.3 = 932

R / H = 932 / 156 = 5.97 6 within rounding

3 0

3 years ago

A spaceship negotiates a circular turn of radius 2925 km at a speed of 29960 km/h. (a) What is the magnitude of the angular spee

emmainna [20.7K]

a) 0.0028 rad/s

b) $23.68 m/s^2$

c) $0 m/s^2$

Explanation:

When an object is in circular motion, the angular speed of the object is the rate of change of its angular position. In formula, it is given by

$\omega = \frac{\theta}{t}$

where

$\theta$ is the angular displacement

t is the time interval

The angular speed of an object in circular motion can also be written as

$\omega = \frac{v}{r}$ (1)

where

v is the linear speed of the object

r is the radius of the orbit

For the spaceship in this problem we have:

$v=29,960 km/h$ is the linear speed, converted into m/s,

$v=8322 m/s$

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Subsituting into eq(1), we find the angular speed of the spaceship:

$\omega=\frac{8322}{2.925\cdot 10^6}=0.0028 rad/s$

When an object is in circular motion, its direction is constantly changing, therefore the object is accelerating; in particular, there is a component of the acceleration acting towards the centre of the orbit: this is called centripetal acceleration, or radial acceleration.

The magnitude of the radial acceleration is given by

$a_r=\omega^2 r$

where

$\omega$ is the angular speed

$r$ is the radius of the orbit

For the spaceship in the problem, we have

$\omega=0.0028 rad/s$ is the angular speed

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Substittuing into the equation above, we find the radial acceleration:

$a_r=(0.0028)^2(2.925\cdot 10^6)=23.68 m/s^2$

When an object is in circular motion, it can also have a component of the acceleration in the direction tangential to its motion: this component is called tangential acceleration.

The tangential acceleration is given by

$a_t=\frac{\Delta v}{\Delta t}$