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erastova [34]
3 years ago
9

How long would it take the wind in a martian dust storm, moving at a speed of 150km/h, to encircle the planets equator?

Physics
1 answer:
goldfiish [28.3K]3 years ago
7 0
<span>21343 km / 150 km/hr = 142.29 hrs</span>
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A current of 1.41 A in a long, straight wire produces a magnetic field of 5.61 uT at a certain distance from the wire. Find
pshichka [43]

Answer:

0.050 m

Explanation:

The strength of the magnetic field produced by a current-carrying wire is given by

B=\frac{\mu_0 I}{2\pi r}

where

\mu_0=4\pi \cdot 10^{-7} H/m is the vacuum permeability

I is the current in the wire

r is the distance from the wire

And the magnetic field around the wire forms concentric circles, and it is tangential to the circles.

In this problem, we have:

I=1.41 A (current in the wire)

B=5.61\mu T=5.61\cdot 10^{-6} T (strength of magnetic field)

Solving  for r, we find the distance  from the wire:

r=\frac{\mu_0 I}{2\pi B}=\frac{(4\pi \cdot 10^{-7})(1.41)}{2\pi (5.61\cdot 10^{-6})}=0.050 m

4 0
3 years ago
A fullback preparing to carry the football starts from rest and accelerates straight ahead. He is handed the ball just before he
RideAnS [48]

Answer:

x=4.06m

Explanation:

A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t  (1)\\\\

{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\

X=Xo+ VoT+0.5at^{2}    (3)\\

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve

for this problem

Vf=7.6m/s

t=1.07

Vo=0

we can use the ecuation number one to find the acceleration

a=(Vf-Vo)/t

a=(7.6-0)/1.07=7.1m/s^2

then we can use the ecuation number 2 to find the distance

{Vf^{2}-Vo^2}/{2.a} =X

(7.6^2-0^2)/(2x7.1)=4.06m

4 0
3 years ago
If a fisherman applies a horizontal force with magnitude 47.0 NN to the box and produces an acceleration of magnitude 3.20 m/s2m
musickatia [10]

Answer:

The correct solution is "14.6875 kg".

Explanation:

Given values:

Force,

F = 47.0 N

Acceleration,

a = 3.20 m/s²

Now,

⇒ Force=Mass\times Acceleration

or,

⇒       F=ma

⇒    47.0=m\times 3.20

⇒       m=\frac{47.0}{3.20}

⇒           =14.6875 \ kg

4 0
2 years ago
MA of the first class lever may be equal to, greater than 1.Why​
Anika [276]

Answer:

hjhjhjhjhjhj

Explanation:

jmjk

4 0
3 years ago
A ball is projected upward at time t = 0.0 s, from a point on a roof 60 m above the ground. The ball rises, then falls until it
musickatia [10]
<span> y=y0 + vt +1/2gt^2
(solve for t here) cause you know y,y0,v,g
you will do quad formula here

then:
v=v0 +at solve for v
(remember the direction of the ball too (signs))

The main thing to remember here is that when the ball passes exactly (height) where it was launched it will travel the speed at which it was launched. *its almost like the ball was thrown in the downward direction. </span>
7 0
3 years ago
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