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NeTakaya
3 years ago
14

An interplanetary spacecraft is moving at

Physics
2 answers:
jenyasd209 [6]3 years ago
8 0

Answer:

1,728,000 Km

Explanation:

Relationship of Speed, Distance and Time is

Distance = Speed × time

Given: Speed= 20000 m/s = 20 Km/s

Time =one day = 24 hrs × 60 mins × 60 sec = 86,400 seconds

so Distance = 20 Km/sec × 86,400 sec = 1,728,000 Km

mixer [17]3 years ago
7 0
The awnser is. 1728000 kilometers
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Which of the following waves travel at the greatest speed
NeTakaya
HEY mate here your answer

p waves travel at the greatest speed and these waves can travel at solid and liquid also. p waves means primary waves.
4 0
3 years ago
Which property of the wave makes it-(C)
AlekseyPX

Answer:

low amplitude hope it will help you

4 0
3 years ago
Read 2 more answers
A ball is thrown directly downward with an initial speed of 7.70 m/s, from a height of 30.2 m. After what time interval does it
KatRina [158]

Answer:

t = 1.82

Explanation:

Given

u = 7.70m/s -- initial velocity

s = 30.2m --- height

Required

Determine the time to hit the ground

This will be solved using the following motion equation.

s = ut + \frac{1}{2}gt^2

Where

g = 9,8m/s^2

So, we have:

30.2 = 7.70t + \frac{1}{2} * 9.8 * t^2

30.2 = 7.70t + 4.9 * t^2

Subtract 30.2 from both sides

30.2 -30.2  = 7.70t + 4.9 * t^2 - 30.2

0  = 7.70t + 4.9 * t^2 - 30.2

0  = 7.70t + 4.9t^2 - 30.2

7.70t + 4.9t^2 - 30.2  = 0

4.9t^2 + 7.70t - 30.2  = 0

Solve using quadratic formula:

t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}

Where

a = 4.9;\ b = 7.70;\ c = -30.2

t = \frac{-7.70\±\sqrt{7.70^2 - 4*4.9*-30.2}}{2*4.9}

t = \frac{-7.70\±\sqrt{651.21}}{9.8}

t = \frac{-7.70\±25.52}{9.8}

Split the expression

t = \frac{-7.70+25.52}{9.8} or t = \frac{-7.70-25.52}{9.8}

t = \frac{17.82}{9.8} or t = -\frac{33.22}{9.8}

Time can't be negative;  So, we have:

t = \frac{17.82}{9.8}

t = 1.82

Hence, the time to hit the ground is 1.82 seconds

7 0
3 years ago
a driver travels 135 km, east in 1.5 h, stops for 45 minutes for lunch, and then resumes driving for the next 2.0 h through a di
daser333 [38]

Answer:

v = 98.75 km/h

Explanation:

Given,

The distance driver travels towards the east, d₁ = 135 km

The time period of the travel, t₁ = 1.5 h

The halting time, tₓ = 46 minutes

The distance driver travels towards the east, d₂ = 215 km

The time period of the travel, t₁ = 2 h

The average speed of the vehicle before stopping

                                    v₁ = d₁/t₁

                                        = 135/1.5

                                       = 90 km/h

The average speed of vehicle after stopping

                                    v₂ = d₂/t₂

                                         = 215/2

                                        = 107.5 km/h

The total average velocity of the driver

                                      v = (v₁ +v₂) /2

                                         = (90 + 107.5)/2

                                         = 98.75 km/h

Hence, the average velocity of the driver, v = 98.75 km/h

7 0
3 years ago
A football is kicked from the ground with a velocity of 38m/s at an angle of 40 degrees and eventually lands at the same height.
Anastasy [175]

Initially, the velocity vector is \langle 38cos(40^{\circ}),38sin(40^{\circ}) \rangle=\langle 29.110, 24.426 \rangle. At the same height, the x-value of the vector will be the same, and the y-value will be opposite (assuming no air resistance). Assuming perfect reflection off the ground, the velocity vector is the same. After 0.2 seconds at 9.8 seconds, the y-value has decreased by 4.9(0.2)^2, so the velocity is \langle 29.110, 24.426-0.196 \rangle = \langle 29.110, 24.23 \rangle.

Converting back to direction and magnitude, we get \langle r,\theta \rangle=\langle \sqrt{29.11^2+24.23^2},tan^{-1}(\frac{29.11}{24.23}) \rangle = \langle 37.87,50.2^{\circ}\rangle

4 0
3 years ago
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