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jeka94
3 years ago
15

8

Physics
1 answer:
Doss [256]3 years ago
4 0

Answer:

The resultant velocity is <u>169.71 km/h at angle of 45° measured clockwise with the x-axis</u> or the east-west line.

Explanation:

Considering west direction along negative x-axis and north direction along  positive y-axis

Given:

The car travels at a speed of 120 km/h in the west direction.

The car then travels at the same speed in the north direction.

Now, considering the given directions, the velocities are given as:

Velocity in west direction is, \overrightarrow{v_1}=-120\ \vec{i}

Velocity in north direction is, \overrightarrow{v_2}=120\ \vec{j}

Now, since v_1\ and\ v_2 are perpendicular to each other, their resultant magnitude is given as:

|\overrightarrow{v_{res}}|=\sqrt{|\overrightarrow{v_1}|^2+|\overrightarrow{v_2}|^2}

Plug in the given values and solve for the magnitude of the resultant.This gives,

|\overrightarrow{v_{res}}|=\sqrt{(120)^2+(120)^2}\\\\|\overrightarrow{v_{res}}|=120\sqrt{2} = 169.71\ km/h

Let the angle made by the resultant be 'x' degree with the east-west line or the x-axis.

So, the direction is given as:

x=\tan^{-1}(\frac{|v_2|}{|v_1|})\\\\x=\tan^{-1}(\frac{120}{-120})=\tan^{-1}(-1)=-45\ deg(clockwise\ angle\ with\ the\ x-axis)

Therefore, the resultant velocity is 169.71 km/h at angle of 45° measured clockwise with the x-axis or the east-west line.

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Can someone help me with this Physics question please?
strojnjashka [21]

Answer:

Explanation:

The formula for this, the easy one, is

N=N_0(\frac{1}{2})^{\frac{t}{H} where No is the initial amount of the element, t is the time in years, and H is the half life. Filling in:

N=48.0(\frac{1}{2})^{\frac{49.2}{12.3} and simplifying a bit:

N=48.0(.5)^4 and

N = 48.0(.0625) so

N = 3 mg left after 12.3 years

6 0
2 years ago
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A hollow spherical shell has mass 8.20 kg and radius 0.220 m. It is initially at rest and then rotates about a stationary axis t
Likurg_2 [28]

Answer:

8.91 J

Explanation:

mass, m = 8.20 kg

radius, r = 0.22 m

Moment of inertia of the shell, I = 2/3 mr^2

                                                    = 2/3 x 8.2 x 0.22 x 0.22 = 0.265 kgm^2

n = 6 revolutions

Angular displacement, θ = 6 x 2 x π = 37.68 rad

angular acceleration, α = 0.890 rad/s^2

initial angular velocity, ωo = 0 rad/s

Let the final angular velocity is ω.

Use third equation of motion

ω² = ωo² + 2αθ

ω² = 0 + 2 x 0.890 x 37.68

ω = 8.2 rad/s

Kinetic energy,

K = \frac{1}{2}I\omega ^{2}

K = 0.5 x 0.265 x 8.2 x 8.2

K = 8.91 J

6 0
3 years ago
Savanna regions developed during the Triassic period. true or false
Inga [223]
Savanna regions developed during the Triassic period. is true
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3 years ago
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An alert driver can apply the brakes fully in about 0.5 seconds. How far would the car travel if it
dybincka [34]

Answer:

The car would travel after applying brakes is, d = 14.53 m

Explanation:

Given that,

The time taken to apply brakes fully is, t = 0.5 s

The velocity of the car, v = 29.06 m/s

The distance traveled by the car in 0.5 s, d = ?

The relation between the velocity, displacement, and time is given by the formula                

                                d = v x t    m

Substituting the values in the above equation,

                                  d = 29.06 m/s x 0.5 s

                                     = 14.53 m

Therefore, the car would travel after applying brakes is, d = 14.53 m

8 0
3 years ago
An alien spaceship traveling at 0.600 c toward the Earth launches a landing craft. The landing craft travels in the same directi
Arturiano [62]

The kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

To find the answer, we have to know about the Lorentz transformation.

<h3>What is its kinetic energy as measured in the Earth reference frame?</h3>

It is given that, an alien spaceship traveling at 0.600 c toward the Earth, in the same direction the landing craft travels with a speed of 0.800 c relative to the mother ship. We have to find the kinetic energy as measured in the Earth reference frame, if the landing craft has a mass of 4.00 × 10⁵ kg.

                  V_x'=0.8c\\V=0.6c\\m=4*10^5kg

  • Let us consider the earth as S frame and space craft as S' frame, then the expression for KE will be,

                  KE=m_0c^2=\frac{mc^2}{1-(\frac{v_x^2}{c^2} )}

  • So, to V_x=(0.8+0.6)c-[\frac{0.6c*(0.8c)^2}{c^2}]=1.016find the KE, we have to find the value of speed of the approaching landing craft with respect to the earth frame.
  • We have an expression from Lorents transformation for relativistic law of addition of velocities as,

                      V_x'=\frac{V_x-V}{1-\frac{VV_x}{c^2} } \\thus,\\V_x=V_x'(1-\frac{VV_x}{c^2} )+V

  • Substituting values, we get,

          V_x=0.8c(1-\frac{0.8c*0.6c}{c^2} )+0.6c=(0.8c*0.52)+0.6c=1.016c

  • Thus, the KE will be,

              KE=\frac{4*10^5*(3*10^8)^2}{\sqrt{1-\frac{(1.016c)^2}{c^2} } } =\frac{1.2*10^{22}}{0.179}=6.704*10^{22}J

Thus, we can conclude that, the kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

Learn more about frame of reference here:

brainly.com/question/20897534

SPJ4

3 0
2 years ago
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