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Darina [25.2K]
3 years ago
12

A 5,000 kg truck moving at 8 m/s has the same momentum as a 2,500 kg car. What is the velocity of

Physics
1 answer:
a_sh-v [17]3 years ago
8 0

Answer:

16m/s

Explanation:

Mv=mv

5000x8=2500x v

V=5000x8/2500

V=40000/2500

= 16m/s

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6)An electric field of 6 N/C points in the positive X direction. What is the electric flux through a surface that is 4 m2, if it
Salsk061 [2.6K]

Answer:

Flux is 21 Nm^2/C.

Explanation:

Electric field, E = 6 N/C along X axis

Electric filed vector, E = 6 i N/C

Area, A = 4 square meter

Area vector

\overrightarrow{A} = 4 (cos30 \widehat{i} + sin 30 \widehat{j})\\\\\overrightarrow{A} = 3.5 \widehat{i} + 2 \widehat{j}\\

The flux is given by

\phi= \overrightarrow{E}.\overrightarrow{A}\\\\\phi = 6 \widehat{i} .  \left (3.5 \widehat{i} + 2 \widehat{j}  \right )\\\\\phi = 21 Nm^2/C

3 0
3 years ago
Is there life on other planets ?
RUDIKE [14]
Supposedly there is life on Mars but it’s early to say for sure
7 0
3 years ago
Read 2 more answers
A force of 80 N is exerted on an object on a frictionless surface for a distance of 4 meters. If the object has a mass of 10 kg,
solmaris [256]
There are two ways to find energy. Energy=F*d=mv^2. We can use this relationship to find v:
Fd=0.5mv^{2} \\ 80*4=0.5*10v^{2} \\ 64=v^{2} \\ v=8 m/s
6 0
3 years ago
A dart is thrown at a dartboard 3.66 m away. When the dart is released at the same height as the center of the dartboard, it hit
lapo4ka [179]

Answer:

The  angle is  \theta  =  15.48^o

Explanation:

From the question we are told that  

     The distance of the dartboard from the dart is  d  =  3.66  \ m

     The time taken is  t =  0.455 \ s

   

The  horizontal component of the speed of the dart is mathematically represented as

      u_x =  ucos \theta

where u is the the velocity at dart is lunched

  so

      distance =  velocity \ in \ the\  x-direction  *  time

substituting values

      3.66 =   ucos  \theta *  (0.455)

 =>   ucos \theta =  8.04  \ m/s

From projectile kinematics the time taken by the dart can be mathematically represented as

         t  =  \frac{2usin \theta }{g}

=>    usin \theta =  \frac{g  * t}{2 }

       usin \theta =  \frac{9.8  * 0.455}{2 }

      usin \theta = 2.23

=>   tan \theta =  \frac{usin\theta }{ucos \theta }  =  \frac{2.23}{8.04}

       \theta  =  tan^{-1} [0.277]

      \theta  =  15.48^o

     

4 0
3 years ago
If a plane is flying level at 910 km/h and the banking angle is not to exceed 50 ∘, what's the minimum curvature radius for the
hoa [83]

Answer:

5.5 km

Explanation:

First, we convert the distance from km/h to m/s

910 * 1000/3600

= 252.78 m/s

Now, we use the formula v²/r = gtanθ to get our needed radius

making r the subject of the formula, we have

r = v²/gtanθ, where

r = radius of curvature needed

g = acceleration due to gravity

θ = angle of banking

r = 252.78² / (9.8 * tan 50)

r = 63897.73 / (9.8 * 1.19)

r = 63897.73 / 11.662

r = 5479 m or 5.5 km

Thus, we conclude that the minimum curvature radius needed for the turn is 5.5 km

4 0
3 years ago
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