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sertanlavr [38]
3 years ago
11

If you go to space at 15000 mph. how long would it take you at the same speed to reach mars

Physics
1 answer:
madreJ [45]3 years ago
6 0

Answer:

if your going 15000 mph all you can do is accelerate  or go fast  unless there is a opposite

Explanation:

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A motorboat heads due east at 16 m/s across a river that flows due south at 9.0 m/s. a.) What is the resultant velocity of the b
alukav5142 [94]

Answer:

a)V=18.35 m/s (South -East)

b) t =7.41 m/s

c)D= 66.70 m

Explanation:

Given that

Velocity of boat in east direction = 16 m/s

Velocity of river = 9 m/s

a)The resultant velocity V

V=\sqrt{16^2+9^2}\ m/s

V=18.35 m/s (South -East)

b)

We know that

Distance = Velocity x time

Lets t time takes to cross the river

136 = 18.35 x t

t =7.41 m/s

c)

   The distance covered downstream  

We know that

Distance = Velocity x time

t= 7.41 s

D= 7.41 x 9 m

D= 66.70 m

3 0
3 years ago
Can you guys please help me on this one?​
LenKa [72]

If the object is not at rest how?

4 0
2 years ago
Read 2 more answers
A circular rod has a radius of curvature R = 9.09 cm and a uniformly distributed positive charge Q = 6.49 pC and subtends an ang
Digiron [165]

Answer:

E = 1.19 N/C

Explanation:

Let's first determine the length of the arc which can be given as:

L= Rθ

where:

L = length of the arc

R = radius of curvature

θ = angle in radius

L = (9.09×10⁻²m)(2.59)

L = (0.0909)(2.59)

L = 0.235431 m

Then, the magnitude of electric field that Q produces at the center of curvature can be calculated by using the formula:

E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}-sin(-\frac{\theta}{2})]

E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}+sin(\frac{\theta}{2})]

E= \frac{2\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}]

Since \lambda = \frac{Q}{L}

where;

L = length

Q = charge

λ =  density of the charge;

then substituting \frac{Q}{L} for λ, we have :

E= \frac{2(\frac{Q}{L})}{4 \pi E_oR}[sin\frac{\theta}{2}]

E= \frac{2Q[sin\frac{\theta}{2}]}{4 \pi E_oLR}

substituting our given parameter; we have:

E= \frac{2(6.26*10^{-12}C)[sin\frac{2.59rad}{2}]}{4 \pi (8.85*10^{-12}C^2/N.m^2)(0.235431)(0.0909)}

E = 1.1889 N/C

E = 1.19 N/C

∴ the magnitude of the electric field that Q produces at the center of curvature = 1.19 N/C

4 0
3 years ago
Explain why the direction of the south equatorial current changes
son4ous [18]
Some scientist believe that the ozone layers of the earth had been weakening and the waters or current changes direct ever 5-7 month.
7 0
3 years ago
When the forces acting on a particle are resolved into cylindrical components, friction forces always act in the
Aleksandr [31]

Answer:

Tangential

Explanation: This is a kind of force which act on a moving body in such a way that it is curved in the direction of the path of the body. This implies that when the velocity of the object is positive, the acceleration will be negative.

7 0
3 years ago
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