When a star is moving away the light waves appear to be....

What is the frequency of a wave having a period equal to 18 seconds

Orlov [11]

Frequency is 1 over the period. Hence, the frequency is 1/18.

6 0

3 years ago

A ball of mass m is thrown into the air in a 45° direction of the horizon, after 3 seconds the ball is seen in a direction 30° f

Rzqust [24]

Answer:

Velocity (magnitude) is 98.37 m/s

Explanation:

We use the vertical component of the initial velocity, which is:

$v_{0y}=v_0*sin(45)=\frac{\sqrt{2} }{2}v_0$

Using kinematics expression of vertical velocity (in y direction) for an accelerated motion (constant acceleration, which is gravity):

$v_{y}=v_{0y}+a*t=\frac{\sqrt{2} }{2}v_0-9.8t$

Now we need to find $v_y$ as a function of $v_0$ . We use the horizontal velocity, which is always the same as follow:

$v_x=v_0cos(45\º)=\frac{\sqrt{2} }{2}v_0=v_{t=3}*cos(30\º) \\$

We know the angle at 3 seconds:

$v_y(t=3)=v_{t=3}*sin(30\º)\\v_{t=3}=\frac{v_y}{sin(30\º)}$

Substitute $v_{t=3}$ in $v_x$ and then solve for $v_y$

$\frac{\sqrt{2} }{2}v_0=\frac{v_y*cos(30\º) }{sin(30\º)} \\v_y=\frac{\sqrt{6} }{6}v_0$

With this expression we go back to the kinematic equation and solve it for initial speed

$\frac{\sqrt{6} }{6} v_0 =\frac{\sqrt{2} }{2}v_0-29.4\\v_0(\frac{\sqrt{6}-3\sqrt{2}}{6} )=-29.4\\v_0=98.37 m/s$

3 0

3 years ago

A stuntman with a mass of 80.5 kg swings across a moat from a rope that is 11.5 m. At the bottom of the swing the stuntman's spe

goldenfox [79]

Answer:

No
5.49 m/s

Explanation:

The net force required to accelerate the stuntman in a circular arc of radius 11.5 m will be ...

F = mv²/r . . . . where this m is the mass being accelerated, v is the tangential velocity, and r is the radius.

Here, the net force needs to be ...

F = (80.5 kg)(8.45 m/s)²/(11.5 m) . . . . . where this m is meters

≈ 499.8175 kg·m/s² = 499.8 N

Gravity exerts a force on the stuntman of ...

F = mg = (80.5 kg)(9.8 m/s²) = 788.9 kg·m/s² = 788.9 N

Then the tension required in the rope/vine is ...

499.8 N+788.9 N= 1288.7 N

This is more than the capacity of the rope, so we do not expect the stuntman to make it across the moat.

_____

The allowed net force for centripetal acceleration is ...

1000 N -788.9 N = 211.1 N

Then the allowed velocity is ...

211.1 = 80.5v²/11.5

30.16 = v² . . . . multiply by 11.5/80.5

5.49 = v . . . . . . take the square root

The maximum speed the stuntman can have is 5.49 m/s.

_____

<em>Comment on crossing the moat</em>

The kinetic energy at the bottom of the swing translates to potential energy at the end of the swing. At the lower speed, the stuntman cannot rise as high, so will traverse a shorter arc. At 8.45 m/s, the moat could be about 16.8 m wide; at 5.49 m/s, it can only be about 11.5 m wide.

5 0

2 years ago

Give 10 examples of units you might use or see in any given day

Ierofanga [76]

Cups
teaspoon
tablespoon
liters
milliliters
gallons
pints
tons
inches

3 0

3 years ago

Why do atoms do group 1 element lose electrons to form cations

Mrac [35]

Answer:

when a element of 1 group take part in reaction, its atom looses outer electron and form positively charged ions called Cation.

Explanation:

8 0

2 years ago