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stira [4]
3 years ago
6

A small space probe of mass 170 kg is launched from a spacecraft near Mars. It travels toward the surface of Mars, where it will

eventually land. At a time 22.9 seconds after it is launched, the probe is at location <5600, 7200, 0> m, and at this same instant its momentum is <51000, -7000, 0> kg·m/s. At this instant, the net force on the probe due to the gravitational pull of Mars plus the air resistance acting on the probe is <-4000, -780, 0> N.
Assuming that the net force on the probe is approximately constant during this time interval, what is the change of the momentum of the probe in the time interval from 22.6 seconds after the probe is launched to 22.9 seconds after the launch?
Physics
1 answer:
alukav5142 [94]3 years ago
6 0

Answer:

The change  in momentum is  \Delta p =   kg \cdot m/s      

Explanation:

From the question we are told that  

       The mass of the probe is  m = 170 kg

       The location of the prob at time t = 22.9 s is  A  =

       The  momentum at time  t = 22.9 s is  p = < 51000, -7000, 0> kg m/s

        The net force on the probe is  F =  N

Generally the change in momentum is mathematically represented as

              \Delta p = F * \Delta t

The initial time is   22.6 s

 The final time  is  22.9 s

             Substituting values  

           \Delta p =  * (22.9 - 22.6)

            \Delta p =  * (0.3)  

              \Delta p =   kg \cdot m/s        

 

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F = 50 N<br> m = 72 kg<br> m/s2
lyudmila [28]

Answer:

Explanation:

F=ma

a=F/m

a=50/72=

a=0.694

3 0
3 years ago
What affects the amount of potential energy stored in the magnetic field when a magnet is moved against a magnetic force?
LekaFEV [45]

Answer: Strength of magnet and distance from magnetic material

Explanation:

The potential energy of a magnet is determined by the strength of the magnet and the distance between a magnet and another magnet or a magnetic material. Magnetic materials are materials that would be attracted when brought close to a magnet, example of magnetic materials are most metals.

8 0
3 years ago
A playground merry-go-round has a mass of 115 kg and a radius of 2.50 m and it is rotating with an angular velocity of 0.520 rev
tatuchka [14]

Answer:

W_f = 2.319 rad/s

Explanation:

For answer this we will use the law of the conservation of the angular momentum.

L_i = L_f

so:

I_mW_m = I_sW_f

where I_m is the moment of inertia of the merry-go-round, W_m is the initial angular velocity of the merry-go-round, I_s is the moment of inertia of the merry-go-round and the child together and W_f is the final angular velocity.

First, we will find the moment of inertia of the merry-go-round using:

I = \frac{1}{2}M_mR^2

I = \frac{1}{2}(115 kg)(2.5m)^2

I = 359.375 kg*m^2

Where M_m is the mass and R is the radio of the merry-go-round

Second, we will change the initial angular velocity to rad/s as:

W = 0.520*2\pi rad/s

W = 3.2672 rad/s

Third, we will find the moment of inertia of both after the collision:

I_s = \frac{1}{2}M_mR^2+mR^2

I_s = \frac{1}{2}(115kg)(2.5m)^2+(23.5kg)(2.5m)^2

I_s = 506.25kg*m^2

Finally we replace all the data:

(359.375)(3.2672) = (506.25)W_f

Solving for W_f:

W_f = 2.319 rad/s

7 0
3 years ago
A bird lands on a bare copper wire carrying a current of 51
nirvana33 [79]
The resistance of the piece of wire is
R= \frac{\rho L}{A}
where
\rho = 1.68 \cdot 10^{-8}\Omega m is the resistivity of the copper
L=5.1 cm=0.051 m is the length of the piece of wire
A=0.13 cm^2 = 0.13 \cdot 10^{-4} m^2 is the cross sectional area of the wire
By substituting these values, we find the value of R:
R= \frac{\rho L}{A}=6.6 \cdot 10^{-5} \Omega

Then, by using Ohm's law, we find the potential difference between the two points of the wire:
V=IR=(51 A)(6.6 \cdot 10^{-5} \Omega )=3.4 \cdot 10^{-3} V
7 0
3 years ago
Definition of distance travelled?​
Mekhanik [1.2K]

Answer:

distance traveled is a total length of the path traveled between two positions.

6 0
3 years ago
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