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matrenka [14]
3 years ago
12

What is the displacement of a jogger that went 30 m to the east and 40 m to the north?

Physics
1 answer:
navik [9.2K]3 years ago
7 0
The displacement in 50m NE
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A sample of an unknown material weighs 500.0 N in air and 400.0 N when submerged in an alcohol solution with a density of 0.70 ×
Papessa [141]

Answer:

500

Explanation:

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4 0
3 years ago
A crash test dummy with a mass of 55kg is placed into a car traveling at 40m/s. During the first trial the dummy does not wear a
s2008m [1.1K]

a. The force experienced by the dummy with a seatbelt is 11,000 Newton.

b. The force experienced by the dummy without a seatbelt is 4,400 Newton.

<u>Given the following data:</u>

  • Mass = 55 kg
  • Initial velocity = 0 m/s
  • Final velocity = 40 m/s
  • Time A = 0.2 seconds
  • Time B = 0.5 seconds

a. To calculate the force experienced by the dummy with a seatbelt:

<h3>Newton's Second Law of Motion</h3>

In order to solve for the force acting on the dummy, we would apply Newton's Second Law of Motion.

<u>Note:</u> The acceleration of an object is equal to the rate of change in velocity with respect to time.

Mathematically, Newton's Second Law of Motion is given by this formula;

F = \frac{M(v\;-\;u)}{t}

Substituting the given parameters into the formula, we have;

F = \frac{55(40\;-\;0)}{0.2}\\\\F = \frac{55\times 40}{0.2}\\\\F = \frac{2200}{0.2}

Force = 11,000 Newton.

b. To calculate the force experienced by the dummy without a seatbelt:

F = \frac{55(40\;-\;0)}{0.5}\\\\F = \frac{55\times 40}{0.5}\\\\F = \frac{2200}{0.5}

Force = 4,400 Newton.

Read more on force here: brainly.com/question/1121817

4 0
3 years ago
According to this graph, the acceleration
skelet666 [1.2K]

Answer:

1 m/s^2

Explanation:

a=(v-u)/t

where,

a=acceleration

v=final velocity

u=initial velocity

(4.5-0)/4.5

1 m/s^2

7 0
3 years ago
You're driving down the highway late one night at 20 m/s when a deer steps onto the road 39 m in front of you. Your reaction tim
miss Akunina [59]

Answer:

a) 10.8 m

b) 24.3 m/s

Explanation:

a)

  • In order to get the total distance traveled since you see the deer till the car comes to an stop, we need to take into account that this distance will be composed of two parts.
  • The first one, is the distance traveled at a constant speed, before stepping on the brakes, which lasted the same time that the reaction time, i.e., 0.5 sec.
  • We can find this distance simply applying the definition of average velocity, as follows:

       \Delta x_{1} = v_{1o} * t_{react} = 20 m/s * 0.5 s = 10 m (1)

  • The second part, is the distance traveled while decelerating at -11 m/s2, from 20 m/s to 0.
  • We can find this part using the following kinematic equation (assuming that the deceleration keeps the same all time):

       v_{1f} ^{2}  - v_{1o} ^{2} = 2* a* \Delta x  (2)

  • where v₁f = 0, v₁₀ = 20 m/s, a = -11 m/s².
  • Solving for Δx, we get:

       \Delta x_{2} = \frac{-(20m/s)^{2}}{2*(-11m/s)} = 18.2 m (3)

  • So, the total distance traveled was the sum of (1) and (3):
  • Δx = Δx₁ + Δx₂ = 10 m + 18.2 m = 28.2 m (4)
  • Since the initial distance between the car and the deer was 39 m, after travelling 28.2 m, the car was at 10.8 m from the deer when it came to a complete stop.

b)

  • We need to find the maximum speed, taking into account, that in the same way that in a) we will have some distance traveled at a constant speed, and another distance traveled while decelerating.
  • The difference, in this case, is that the total distance must be the same initial distance between the car and the deer, 39 m.
  • ⇒Δx = Δx₁ + Δx₂ = 39 m. (5)
  • Δx₁, is the distance traveled at a constant speed during the reaction time, so we can express it as follows:

       \Delta x_{1} = v_{omax} * t_{react} = 0.5* v_{omax} (6)

  • Δx₂, is the distance traveled while decelerating, and can be obtained  using (2):

        v_{omax} ^{2} = 2* a* \Delta x_{2} (7)

  • Solving for Δx₂, we get:

       \Delta x_{2} = \frac{-v_{omax} ^{2} x}{2*a}  = \frac{-v_{omax} ^{2}}{(-22m/s2)} (8)

  • Replacing (6) and (8) in (5), we get a quadratic equation with v₀max as the unknown.
  • Taking the positive root in the quadratic formula, we get the following value for vomax:
  • v₀max = 24.3 m/s.
6 0
3 years ago
Rusty Pickins, a gold miner, has just found a chunk of gold. Rusty would like to change the gold from a solid to a liquid, so th
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<span>b.) Add heat to increase the molecular motion of the atoms</span>
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4 years ago
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