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3 years ago

Need help on this thank you

Physics

1 answer:

Semmy [17]3 years ago

7 0

Answer:

TRUE - In any collision between two objects, the colliding objects exert equal and opposite force upon each other. This is simply Newton's law of action-reaction.

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Explain your answer. What parts of your hypothesis were strong correct? What parts were weak?

Anastaziya [24]

This question is from quizlet.

So better check this question!

5 0

2 years ago

A rowboat passenger uses an oar to push the boat off the dock by exerting a force of 40N for 3.0s. What impulse acts on the boat

jarptica [38.1K]

Answer:

120 Ns

Explanation:

The impulse exerted on an object is given by:

$I=Ft$

where

F is the force applied

t is the time taken

In this problem, we have:

F = 40 N

t = 3.0 s

So, the impulse acting on the boat is

$I=Ft=(40 N)(3.0 s)=120 Ns$

8 0

3 years ago

As an intern at an engineering firm, you are asked to measure the moment of inertia of a large wheel for rotation about an axis

klio [65]

Hi there!

We can begin by finding the acceleration of the block.

Use the kinematic equation:

$d = v_0t + \frac{1}{2}at^2$

The block starts from rest, so:

$d = \frac{1}{2}at^2\\\\12 = \frac{1}{2}a(4^2)\\\\\frac{24}{16} = a = 1.5 m/s^2$

Now, we can do a summation of forces of the block using Newton's Second Law:

$F = ma = m_bg - T$

mb = mass of the block

T = tension of string

Solve for tension:

$T = m_bg - ma = 8.2(9.8) - 8.2(1.5) = 68.06 N$

Now, we can do a summation of torques for the wheel:

$\Sigma \tau = rF\\\\\Sigma\tau = rT$

Rewrite:

$I\alpha = rT$

We solved that the linear acceleration is 1.5 m/s², so we can solve for the angular acceleration using the following:

$\alpha = a/r\\\\\alpha = 1.5/.42= 3.57 rad/sec^2$

Now, plug in the values into the equation:

$I(3.57) = (0.42)(68.06)\\\\I = (0.42)(68.06)/(3.57) = \boxed{8.00 kgm^2}$

8 0

2 years ago

77. There is a 250-m-high cliff at Half Dome in Yosemite National Park in California. Suppose a boulder breaks loose from the to

lisabon 2012 [21]

Answer:70 m/s or 6.84 s

Explanation:

8 0

3 years ago

An elevator car weighs 5500 N. If the car accelerates upwards at a rate of 4.0 m/s2, what is the tension in the support cable li

nadya68 [22]

Answer:

Explanation:

The equation for this, since we are talking about weight on an elevator, is Newton's 2nd Law adjusted to fit our needs:

$F_n=ma+w$ where the Normal Force needed to lift that elevator car is the tension. So the equation then becomes

T = ma + w where T is the tension in the cable to lift the elevator, m is the mass of the elevator (which we have to solve for), a is the acceleration of the elevator (positive since it's going up), and w is the weight of the elevator (which we have as 5500 N). Solving first for mass:

w = mg and

5500 =- m(10) so

m = 550 kg. Now we have what we need to solve for the tension:

T = 550(4.0) + 5500 and

T = 2200 + 5500 so

T = 7700 N

4 0

3 years ago