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nydimaria [60]
3 years ago
10

QUESTION 2 DOK 3 A Thompson's gazelle has a maximum acceleration of 4.5 m/s2 At this acceleration, how much time is required for

it to reach a speed of 40 km/h?​
Physics
1 answer:
Dominik [7]3 years ago
5 0

Answer:

2.47 s

Explanation:

Convert the final velocity to m/s.

  • 40 km/h → 11.1111 m/s

We have the acceleration of the gazelle, 4.5 m/s².

We can assume the gazelle starts at an initial velocity of 0 m/s in order to determine how much time it requires to reach a final velocity of 11.1111 m/s.

We want to find the time t.

Find the constant acceleration equation that contains all four of these variables.

  • v = v₀ + at

Substitute the known values into the equation.

  • 11.1111 = 0 + (4.5)t
  • 11.1111 = 4.5t
  • t = 2.469133333

The Thompson's gazelle requires a time of 2.47 s to reach a speed of 40 km/h (11.1111 m/s).

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6 0
2 years ago
A boy drags a suitcase along the ground with a force of 100 N. If the frictional force opposing the motion of the suitcase is 50
stira [4]
Fortunately, 'force' is a vector.  So if you know the strength and direction
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When we talk in vectors, one newton forward is the negative of
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the complete solution of the problem.


            (100 N forward) plus (50 N backward)

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That's it.
Is there any part of the solution that's not clear ?

4 0
3 years ago
your friend sit in a sked in the snow. if you apply a force of 120 N to them, they have an acceleration of 1.3 m/s2. what is the
levacccp [35]
Need to know the equation for force
F=MA
F is force
M is mass- we need to know the mass
A is acceleration
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7 0
3 years ago
A block of mass 0.510 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x. T
maria [59]

Answer:

x=0.46m, speed=7.9m/s

Explanation:

Using the concept of conservation of energy:

1. kinetic energy of mass m and velocity v: E_k=\frac{1}{2}mv^2

2. gravitational potential energy of mass m, grav. acc. g and height h: E_g=mgh

3. potential energy in a spring with spring constant k and displacement from equilibrium x: E_s=\frac{1}{2}kx^2

Calculating x:

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x=\sqrt{\frac{m}{k}}v_a

Calculating the speed:

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h_a=0, h_b=2R,W_{friction}=F_{friction}\times distance=7\pi R

\frac{1}{2}mv_a^2=\frac{1}{2}mv_b^2+2mgR+7\pi R

Solving for v_b:

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7 0
3 years ago
Which of the following would accurately label the X axis?
kogti [31]

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I hope this helps :)

6 0
3 years ago
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