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Readme [11.4K]
3 years ago
8

Please help, if u can. i linked the picture down below :)

Physics
1 answer:
natali 33 [55]3 years ago
5 0

Answer:

For Parent #1 missing box it's t and for Parent #2 missing box is t

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A distant large asteroid is detected that might pose a threat to Earth. If it were to continue moving in a straight line at cons
Vlada [557]

Answer:

The minimum speed required is 5.7395km/s.

Explanation:

To escape earth, the kinetic energy of the asteroid must be greater or equal to its gravitational potential energy:

K.E\geq P.E

or

\dfrac{1}{2}mv^2 \geq  G\dfrac{Mm}{R}

where m is the mass of the asteroid, R= 24,000,000\:m is its distance form earth's center, M = 5.9*10^{24}kg is the mass of the earth, and G = 6.7*10^{-11}m^3/kg\: s^2 is the gravitational constant.

Solving for v we get:

v \geq \sqrt{\dfrac{2GM}{R} }

putting in numerical values gives

v \geq \sqrt{\dfrac{2(6.7*10^{-11})(5.9*10^{24})}{(24,000,000)} }

\boxed{v\geq 5739.5m/s}

in kilometers this is

v\geq5.7395m/s.

Hence, the minimum speed required is 5.7395km/s.

5 0
3 years ago
What is the weight of an astronaut who has a mass of 90 kg on the moon? (Note: acceleration due to gravity of the moon is 1.62 N
motikmotik

Answer:

W = 145.8 [N]

Explanation:

To solve this problem we must remember that weight is defined as the product of mass by gravity, in this case lunar gravity.

W = m*g

where:

m = mass = 90 [kg]

g = gravity acceleration = 1.62 [kg/m²]

W = 90*1.62

W = 145.8 [N]

8 0
3 years ago
Without steering, a car (mass 1501 kg) can travel with a certain speed around a banked frictionless corner angled at 25.0° to th
natima [27]

Answer:

Its tricky but am still working on it

6 0
3 years ago
Light of wavelength 480 nm illuminates a pair of slits separated by 0.27 mm. If a screen is place 1.7 m from the slits, determin
Dahasolnce [82]

Answer:

  Δy = 6.05 mm

Explanation:

The double slit phenomenon is described by the expression

      d sin θ = m λ                constructive interference

      d sin θ = (m + ½) λ       destructive interference

      m = 0,±1, ±2, ...

As they tell us that they measure the dark stripes, we are in a case of destructive interference, let's use trigonometry to find the sins tea

      tan θ = y / x

      y = x tan θ

In the interference experiments the measured angle is very small so we can approximate the tangent

      tan θ = sin θ / cos θ

     cos θ = 1

     tan θ = sin θ

     y = x sin θ

We substitute in the destructive interference equation

     d (y / x) = (m + ½) λ

    y = (m + ½) λ x / d

The first dark strip occurs for m = 0 and the third dark strip for m = 2. Let's find the distance for these and subtract it

 m = 0

      y₀ = (0+ ½) 480 10⁻⁹ 1.7 / 0.27 10⁻³

      y₀ = 1.511 10⁻³ m

 m = 2

     y₂ = (2 + ½) 480 10⁻⁹ 1.7 / 0.27 10⁻³

     y₂ = 7.556 10⁻³ m

The separation between these strips is Δy

     Δy = y₂-y₀

     Δy = (7.556 - 1.511) 10⁻³

     Δy = 6.045 10⁻³ m

     Δy = 6.05 mm

5 0
3 years ago
In a machine shop, a hydraulic lift is used to raise heavy equipment for repairs. The system has a small piston with a cross-sec
Mrac [35]

Answer:

F_s=1075.9493\ N

Explanation:

Given:

  • area of piston on the smaller side of hydraulic lift, a_s=0.075\ m^2
  • area of piston on the larger side of hydraulic lift, a_l=0.237\ m^2
  • Weight of the engine on the larger side, W_l=3400\ N

Now, using Pascal's law which state that the pressure change in at any point in a confined continuum of an incompressible fluid is transmitted throughout the fluid at its each point.

P_s=P_l

\frac{F_s}{a_s}=\frac{W_l}{a_l}

\frac{F_s}{0.075} =\frac{3400}{0.237}

F_s=1075.9493\ N is the required effort force.

5 0
3 years ago
Read 2 more answers
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