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Nutka1998 [239]
3 years ago
9

El carrito a control remoto de Juanito corre en una pista recta de 40 m de un extremo a otro en 90 segundos; luego de regreso so

bre la pista recorre 15 metros y se ubica en medio en 25 segundos, cual es la rapidez?
y cual es la velocidad?
Physics
1 answer:
PSYCHO15rus [73]3 years ago
8 0

Explanation:

hir ceh3co8d ueirfkgrewkfgv ruk gr

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What type of image can be larger or smaller than the object?
Stolb23 [73]
It’s D. An enlargement (hope this helps!)
4 0
3 years ago
Submarines need to be extremely strong to withstand the extremely high pressure of water pushing down on them. An experimental r
Brrunno [24]
1500 I think so but I not sure
8 0
3 years ago
The magnitude of a force vector is 89.6 newtons (N). The x component of this vector is directed along the +x axis and has a magn
insens350 [35]

Answer:

(a) θ = 33.86°

(b) Ay = 49.92 N

Explanation:

You have that the magnitude of a vector is A = 89.6 N

The x component of such a vector is Ax = 74.4 N

(a) To find the angle between the vector and the x axis you use the following formula for the calculation of the x component of a vector:

A_x=Acos\theta       (1)

Ax: x component of vector A

A: magnitude of vector A

θ: angle between vector A and the x axis

You solve the equation (1) for θ, by using the inverse of cosine function:

\theta=cos^{-1}(\frac{A_x}{A})=cos{-1}(\frac{74.4N}{89.6N})\\\\\theta=33.86\°

the angle between the A vector and the x axis is 33.86°

(b) The y component of the vector is given by:

A_y=Asin\theta\\\\A_y=(89.6N)sin(33.86\°)=49.92N

the y comonent of the vecor is Ay = 49.92 N

3 0
3 years ago
On a snowy day, max (mass = 15 kg) pulls his little sister maya in a sled (combined mass = 20 kg) through the slippery snow. max
sesenic [268]

Work done by a given force is given by

W = F.d

here on sled two forces will do work

1. Applied force by Max

2. Frictional force due to ground

Now by force diagram of sled we can see the angle of force and displacement

work done by Max = W_1 = Fdcos\theta

W_1 = 12*5cos15

W_1 = 57.96 J

Now similarly work done by frictional force

W_2 = Fdcos\theta

W_2 = 4*5cos180

W_2= -20 J

Now total work done on sled

W_{net}= W_1 + W2

W_{net} = 57.96 - 20 = 37.96 J

7 0
3 years ago
Three small masses are positioned as follows: 2.0 kg at (0.0 m, 0.0 m), 2.0 kg at (2.0 m, 0.0 m), and 4.0 kg at (2.0 m, 1.0 m).
melamori03 [73]
Refer to the diagram shown below.

The given data is

mass, kg   Coordinates. m
-------------   -----------------
   2               (0, 0)
   2               (2, 0)
   4               (2, 1)

Total mass, M = 2+2+4 = 8kg
Let (x,y) be the coordinates of M.

Then, taking moments about the origin, we obtain
8x = 2*0 + 2*2 + 4*2 = 12
x = 1.5 m

8y = 2*0 + 2*0 + 4*1 = 4
y = 0.5 m

Answer:  (1.5, 0.5) m




6 0
3 years ago
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