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Oxana [17]
3 years ago
14

Question 1 (1 point)

Physics
1 answer:
puteri [66]3 years ago
3 0

Answer:

250 N

Explanation:

a =  \frac{vf - vi}{t}

Vf=final velocity

Vi =initial velocity

=70-20/2=25m/s^2

F=ma

=10kg * 25m/s^2

=250N

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A 1.0-kg ball has a velocity of 12 m/s downward just before it strikes the ground and bounces up with a velocity of 12 m/s upwar
Nezavi [6.7K]

Answer:

The change in momentum of the ball is 24 kg-m/s  

Explanation:

It is given that,

Mass of the ball, m = 1 kg

Initial velocity of the ball, u = -12 m/s (in downwards)

Final velocity of the ball, v = +12 m/s (in upward)

We need to find the change in momentum of the ball.

Initial momentum of the ball, p_i=mu=1\ kg\times (-12\ m/s)=-12\ kg-m/s

Final momentum of the ball, p_f=mv=1\ kg\times (12\ m/s)=12\ kg-m/s

Change in momentum of the ball, \Delta p=p_f-p_i

\Delta p=12-(-12)=24\ kg-m/s

So, the change in momentum of the ball is 24 kg-m/s. Hence, this is the required solution.

3 0
3 years ago
How can you use the graph of velocity versus time to estimate the acceleration of the ball?
kenny6666 [7]

Answer: acceleration = slope  graph velocity vs time

Explanation: if you have the graph of velocity vs time , the slope of that graph equals the acceleration of our object assuming constant acceleration...but remenber por a real object is really hard to keep constant acceleration

7 0
3 years ago
Read 2 more answers
A student applies a force of 245 N to move a bicycle 12.0 meters. How much work is done on the bicycle
atroni [7]

Answer:

<h2>2940 J</h2>

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question we have

workdone = 245 × 12

We have the final answer as

<h3>2940 J</h3>

Hope this helps you

8 0
3 years ago
A carnival game consists of a two masses on a curved frictionless track, as pictured below. The player pushes the larger object
Harman [31]

Answer:

v₁₀ = 1.90 m / s

Explanation:

In this exercise we are given the maximum height data, with energy we can know how fast the body came out

Final mechanical energy, maximum height

    Em_{f} = U = m g h

Initial mechanical energy, in the lower part of the track

    Em₀ = K = ½ m v²

    Em=   Em_{f}

    ½ m v² = m g h

    v = √ 2gh

Now we can use the moment to find the speed with which objects collide

The large object has a mass M = 5.41 kg a velocity starts v₁₀, the small object has a mass m = 1.68 kg an initial velocity of zero v₂₀ = 0 and  final velocity v

Initial before the crash

    p₀ = M v₁₀ + 0

Final after the crash

      p_{f} = M v1f + m v

   p₀ =   p_{f}

   M v₁₀ = M v_{1f}+ m v

As the shock is elastic the kinetic energy is conserved

     K₀ = K_{f}

    ½ M v₁₀² = ½ M v_{1f}² + ½ m v²

Let's write the system of equations

    M v₁₀ = M  v_{1f} + m v

    M v1₁₀² = M v_{1f}² + m v²

We cleared v1f in the first we replaced in the second

   v_{1f} = (M v₁₀ - mv) / M

    M v₁₀² = M (M v₁₀ - mv)² / M² + m v²

    M v₁₀² = 1 / M (M² v₁₀² - 2mM v v₁₀ + m² v²) +m v²

     v₁₀² (M - M) + 2 m v v₁₀ - v² (m2 + m) / M = 0

     2 m v₁₀ - v (m + 1) m/ M = 0

     v₁₀ = v (m +1) / (2M)

Let's substitute the value of v

     v1₁₀= √ (2gh) (m +1) / (2M)

Let's calculate

    v₁₀ = √ (2 9.8 3) (1+ 1.68) / (2  5.41)

    V₁₀ = 7.668 (2.68) / 10.82

   v₁₀ = 1.90 m / s

5 0
3 years ago
Class II levers like ankles and wheelbarrows are useful because they provide mechanical advantage, by amplifying the input force
marusya05 [52]

Answer:

The solution and the explanation are in the Explanation section.

Explanation:

According to the diagram that is in the attached image, the EFFORT force at point A and the load is at O point. The torque due to weight is:

TA = W * (a * cosθ)

The torque due to effort at C point is:

TC = E * (b * cosθ)

The net torque is equal to 0, we have:

Tnet = 0

W * (a * cosθ) - E * (b * cosθ) = 0

E=W\frac{a}{b}

From the figure, you can observe that a/b < 1, thus E < W

8 0
3 years ago
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