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oksian1 [2.3K]
3 years ago
10

An object starts from rest, and accelerates at 2m/s2 for 10s. How far has it gone in that time

Physics
1 answer:
blagie [28]3 years ago
7 0

Answer:

100m

Explanation:

s = ut +  \frac{1}{2} a {t}^{2}

u=0;t=10sec;a=2m/s²

s = 0(10) +  \frac{1}{2}(2 \times  {10}^{2} )

s=10²;100m

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1 (3 points)
trapecia [35]

Answer:

300

Explanation:

15x20

5 0
3 years ago
a force vector f has a magnitude of 12.0 n. it is oriented 60° to the left of the y ax what are its x and y components?
Arte-miy333 [17]

The  force vector that has a magnitude of 12.0 N. and is oriented 60° to the left of the (y) has the followings components:

  • v(x) =6 N
  • v(y) = 10.39 N

To solve this exercise the formulas and procedures we will use are:

  • v(x) = v * cosine (angle)
  • v(y) = v * sine (angle).

Where:

  • v= magnitude of the vector
  • v(x) = component of the vector on the (x) axis
  • v(y) = component of the vector on the (y) axis
  • angle = angle

Information about the problem:

  • angle = 60º
  • v = 12.0 N
  • v(x)= ?
  • v(y)= ?

Applying the formula of the component of the vector in the (x) axis we have:

v(x) = v * cosine (angle).

v(x) = 12.0 N * cosine (60º)

v(x) =6 N

Applying the formula of the component of the vector in the (y) axis we have:

v(y) = v * sine (angle)

v(y) = 12.0 N * sine (60º)

v(y) = 10.39 N

<h3>What is a vector?</h3>

It can be said to be a straight line described by a point (a) and (b) that has direction and sense.

Learn more about vector at: brainly.com/question/2094736

#SPJ4

5 0
1 year ago
:) What is practical machine? what is the reda<br>tion between MA and VR in a practical<br>machine?​
denis-greek [22]
Answer: For ideal machine efficiency = 1. Hence M.A = V. R. The V. R of an ideal machine and the practical machine is a constant or is the same for both
3 0
3 years ago
7. A beam of light converges at a point P. Now a lens is placed in the path of the convergent
Rudiy27

Answer:

Lens at a distance = 7.5 cm

Lens at a distance = 6.86 cm  (Approx)

Explanation:

Given:

Object distance u = 12 cm

a) Focal length = 20 cm

b) Focal length = 16 cm

Computation:

a. 1/v = 1/u + 1/f

1/v = 1/20 + 1/12

v = 7.5 cm

Lens at a distance = 7.5 cm

b.  1/v = 1/u + 1/f

1/v = 1/16 + 1/12

v = 6.86 cm  (Approx)

Lens at a distance = 6.86 cm  (Approx)

5 0
3 years ago
A cannonball is launched from the ground at an angle of 30 degrees above the horizontal and a speed of 30 m/s. Ideally (no air r
marusya05 [52]

As we know that here no air resistance while ball is moving in air

So here we will say that

initial total energy = final total energy

KE_i + U_i = KE_f + U_f

here we know that

Ui = U_f = 0 (as it will be on ground at initial and final position)

so we will say

KE_i = KE_f

since mass is always conserved

so we will say that final speed of the ball must be equal to the initial speed of the ball

so we have

v_f = v_i = 30 m/s

6 0
3 years ago
Read 2 more answers
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