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Anni [7]
3 years ago
5

vector a having mignitude 3.2 makes 50degre with x-axis and vector b with magnitude 5.2 makes 110degre with x-axis what is the m

agnitude of their dot and cross product?
Physics
1 answer:
Svetllana [295]3 years ago
8 0
Yep ask google good idea
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Giving brainliest!! if you answer correctly :) (30pts)
AnnZ [28]

Answer:

32000joule.

Explanation:

given

mass. (m)=160kg

speed (v)=20m/s

now

kinetic energy =1/2 (mv²)=1/2 ×{160×20²}=32000joule.

8 0
2 years ago
Read 2 more answers
You punt a ball straight up at 20 m/s. What is the balls hangtime
Natalija [7]

The hang time of the ball is 4.08 s

Explanation:

The ball is in free fall motion: this means that it is acted upon gravity only, so its acceleration is the acceleration of gravity,

a=g=-9.8 m/s^2

downward (the negative sign refers to the downward direction).

Since this is a uniformly accelerated motion, we can solve the problem by using the following suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

First we calculate the time it takes for the ball to reach the maximum height, where the velocity is zero:

v = 0

Substituting:

u = +20 m/s

a=-9.8 m/s^2

we find t

t=\frac{v-u}{a}=\frac{0-20}{-9.8}=2.04 s

The motion of the ball is symmetrical, so the total time of flight is just twice the time needed to reach the maximum height, therefore:

T=2t=2(2.04)=4.08 s

Learn more about free fall:

brainly.com/question/1748290

brainly.com/question/11042118

brainly.com/question/2455974

brainly.com/question/2607086

#LearnwithBrainly

4 0
2 years ago
Help with these please?
Anuta_ua [19.1K]

E=kq/r^2

q=(E*r^2)/k

q=(.086N/C)(1.7m^2)/(8.99*10^9N*m^2/C^2)

q=2.76*10^-11 C

q=2.8*10^-11 C

5 0
3 years ago
Read 2 more answers
Amoebas have projections called pseudopods. What are they used for
Dvinal [7]

Answer:

Amoebas have projections called pseudopods

Explanation:

Pseudopodia is the locomatary organ of amoeba. It helps them in movement and transportation.

8 0
2 years ago
amping equipment weighing 6.0 kN is pulled across a frozen lake by means of ahorizontal rope. The coecient of kinetic friction i
Harman [31]

Answer:

Work done = 422.45 kJ

Explanation:

given,                                  

weight of equipment = 6 kN      

coefficient of kinetic friction = 0.05          

distance up to which it is pulled = 1000 m

constant acceleration = 0.2 m/s²                    

Work done by the camper = ?                

actual acceleration acting a'      

m a = m a' - μ mg            

a' = a + μ g                      

a' = 0.2 + 0.05 x 9.8                

a' = 0.69 m/s²                              

Work done = Force x distance

F = m a'                                                            

F = \dfrac{6000}{9.8} \times 0.69

F = 422.44897 N                            

Work done = F x d                          

Work done = 422.44897 x 1000

Work done = 422449 J                  

Work done = 422.45 kJ            

4 0
2 years ago
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