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Natasha_Volkova [10]
2 years ago
11

Ms. Perry travels north 5 km, 10 Km west, and then turns and travels 5 Km 20 points

Physics
1 answer:
Lyrx [107]2 years ago
4 0

Answer:

A

Explanation:

traveling north no displacement ,west displacement is 10,

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Complete the following​
Furkat [3]

answer is down hera

3 0
2 years ago
Water (2510 g ) is heated until it just begins to boil. if the water absorbs 5.01×105 j of heat in the process, what was the ini
natka813 [3]
E=energy=5.09x10^5J = 509KJ 
<span>M=mass=2250g=2.25Kg </span>
<span>C=specific heat capacity of water= 4.18KJ/Kg </span>
<span>ΔT= change in temp= ? </span>
<span>E=mcΔT </span>
<span>509=(2.25)x(4.18)xΔT </span>
<span>509=9.405ΔT </span>
<span>ΔT=509/9.405=54.1degrees </span>
<span>Initial temp = 100-54 = 46 degrees </span>
<span>Hope this helps :)</span>
5 0
3 years ago
Two forces, F⃗ 1F→1F_1_vec and F⃗ 2F→2F_2_vec, act at a point,F⃗ 1F→1F_1_vec has a magnitude of 8.80 NN and is directed at an an
castortr0y [4]

Answer:

  • Fx = -9.15 N
  • Fy = 1.72 N
  • F∠γ ≈ 9.31∠-10.6°

Explanation:

You apparently want the sum of forces ...

  F = 8.80∠-56° +7.00∠52.8°

Your angle reference is a bit unconventional, so we'll compute the components of the forces as ...

  f∠α = (-f·cos(α), -f·sin(α))

This way, the 2nd quadrant angle that has a negative angle measure will have a positive y component.

  = -8.80(cos(-56°), sin(-56°)) -7.00(cos(52.8°), sin(52.8°))

  ≈ (-4.92090, 7.29553) +(-4.23219, -5.57571)

  ≈ (-9.15309, 1.71982)

The resultant component forces are ...

  • Fx = -9.15 N
  • Fy = 1.72 N

Then the magnitude and direction of the resultant are

  F∠γ = (√(9.15309² +1.71982²))∠arctan(-1.71982/9.15309)

  F∠γ ≈ 9.31∠-10.6°

4 0
3 years ago
a container of water is knocked off a 10.0 meter high ledge with a horizontal velocity of 1.00 meters/second. calculate the time
Evgen [1.6K]

Answer:

1.43 s

Explanation:

The time it takes for the container to reach the ground is determined only by the vertical motion of the container, which is a free-fall motion, so a uniformly accelerated motion with a constant acceleration of g=9.8 m/s^2 towards the ground.

The vertical distance covered by an object in free fall is given by

S=ut + \frac{1}{2}at^2

where

u = 0 is the initial vertical speed

t is the time

a= g = 9.8 m/s^2 is the acceleration

since u=0, it can be rewritten as

S=\frac{1}{2}gt^2

And substituting S=10.0 m, we can solve for t, to find the duration of the fall:

t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(10.0 m)}{9.8 m/s^2}}=1.43 s

3 0
3 years ago
Example 4.6 provides a nice example of the overlap between kinematics and dynamics. It is known that the plane accelerates from
kodGreya [7K]

Answer:

ax = 2.60m/s^{2}, t = 26.92s

Explanation:

The acceleration of the plane can be determined by means of the kinematic equation that correspond to a Uniformly Accelerated Rectilinear Motion.

(vx)f^{2} = (vx)i^{2} + 2ax \Lambda x (1)

Where (vx)f^{2} is the final velocity, (vx)i^{2} is the initial velocity, ax is the acceleration and  \Lambda x is the distance traveled.

Equation (1) can be rewritten in terms of ax:

(vx)f^{2} - (vx)i^{2} = 2ax \Lambda x

2ax \Lambda x = (vx)f^{2} - (vx)i^{2}

ax = \frac{(vx)f^{2} - (vx)i^{2}}{2 \Lambda x}  (2)

Since the plane starts from rest, its initial velocity will be zero ((vx) = 0):

Replacing the values given in equation 2, it is gotten:

ax = \frac{(70m/s)^{2} - (0m/s)^{2}}{2(940m)}

ax = \frac{4900m/s}{2(940m)}

ax = \frac{4900m/s}{1880m}

ax = 2.60m/s^{2}

So, The acceleration of the plane is 2.60m/s^{2}    

Now that the acceleration is known, the next equation can be used to find out the time:

(vx)f = (vx)i + axt (3)

Rewritten equation (3) in terms of t:

t = \frac{(vx)f - (vx)i}{ax}

t = \frac{70m/s - 0m/s}{2.60m/s^{2}}

t = 26.92s

<u>Hence, the plane takes 26.92 seconds to reach its take-off speed.</u>

5 0
3 years ago
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