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Igoryamba
2 years ago
5

PLEASE ANSWER ASAP!! YOU WILL GET BRAINLIEST IF RIGHT!

Physics
2 answers:
Elena-2011 [213]2 years ago
6 0
D just east of station b
yarga [219]2 years ago
3 0
D will be your answer because if I am right point a is the pressure and tge line is heading to the east side of point b. I hope this helps
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Please help ASAP. In what direction do you need to apply force to move an object vertically?
PilotLPTM [1.2K]
The answer is up . tylrhscjwizn
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3 years ago
- How much force is needed to accelerate a 26 kg skier at 4 m/sec??
kondor19780726 [428]

Answer:

<h2>104 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 26 × 4

We have the final answer as

<h3>104 N</h3>

Hope this helps you

8 0
2 years ago
Starting from rest, a particle that is confined to move along a straight line is accelerated at a rate of 5.0 m/s2. Which one of
Elanso [62]

Answer:

c) The slope is not constant and increases with increasing time.

Explanation:

The equation for the position of this particle (starting from rest is)

s = at^2/2 = 5t^2/2 = 2.5t^2

We can take derivative of this with respect to time t to get the equation of slope:

s' = (2.5t^2)' = 2*2.5t = 5t

As time t increase, the slope would increases with time as well.

6 0
3 years ago
At one instant, the center of mass of a system of two particles is located on the x-axis at 2.0 cm and has a velocity of (5.0 m/
Nata [24]

Answer:

Explanation:

Given that,

At one instant,

Center of mass is at 2m

Xcm = 2m

And velocity =5•i m/s

One of the particle is at the origin

M1=? X1 =0

The other has a mass M2=0.1kg

And it is at rest at position X2= 8m

a. Center of mass is given as

Xcm = (M1•X1 + M2•X2) / (M1+M2)

2 = (M1×0 + 0.1×8) /(M1 + 0.1)

2 = (0+ 0.8) /(M1 + 0.1)

Cross multiply

2(M1+0.1) = 0.8

2M1 + 0.2 =0.8

2M1 = 0.8-0.2

2M1 = 0.6

M1 = 0.6/2

M1 = 0.3kg

b. Total momentum, this is an inelastic collision and it momentum after collision is given as

P= (M1+M2)V

P = (0.3+0.1)×5•i

P = 0.4 × 5•i

P = 2 •i kgm/s

c. Velocity of particle at origin

Using conversation of momentum

Momentum before collision is equal to momentum after collision

P(before) = M1 • V1 + M2 • V2

We are told that M2 is initially at rest, then, V2=0

So, P(before) = 0.3V1

We already got P(after) = 2 •i kgm/s in part b of the question

Then,

P(before) = P(after)

0.3V1 = 2 •i

V1 = 2/0.3 •i

V1 = 6 ⅔ •i m/s

V1 = 6.667 •i m/s

4 0
3 years ago
Question 4 of 10
Alexus [3.1K]

Answer:

22.3 kg•m/s

Explanation:

Apex;)

7 0
3 years ago
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