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3 years ago

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A loaded truck is going at a speed of 36 km/hr and a car starts at a point 3 km behind the truck with an acceleration 2 m/s^2 .

When will the car overtake the truck?What is distance covered by car before it overtakes the truck

1 answer:

kozerog [31]3 years ago

6 0

It's 3600 m and 60 sec

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Suppose you have a Frisbee with a mass of 0.10 kg. You first throw it with a force of 5N and then change the force you throw it

natali 33 [55]

Answer:

The acceleration increases.

Explanation:

From Newton's 2nd Law, we have $\Sigma F=ma$ . We can see that force is directly proportional to mass and acceleration. Therefore, as force increases, either mass or acceleration must increase as well, and vice versa. Since mass is maintained here, if you increase the force applied to the Frisbee, the acceleration will increase as well.

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3 years ago

the system shown above is released from rest. if friction is negligible, the acceleration of the 4.0 kg block sliding on the tab

JulsSmile [24]

The acceleration of the first block (4 kg) is -9.8 m/s².

The given parameters:

<em>Mass of the first block, m₁ = 4.0 kg</em>
<em>Mass of the second block, m₂ = 2.0 kg</em>

The net force on the system of the two blocks is calculated as follows;

$m_2 g - T = m_1 a$

where;

<em>T </em><em>is the tension in the connecting string due weight of the first block</em>

$m_2 g - m_1 g = m_1 a\\\\a = \frac{m_2 g - m_1g}{m_1} \\\\a = \frac{g(m_2 - m_1)}{m_1} \\\\a = \frac{9.8(2-4)}{2} \\\\a = -9.8 \ m/s^2$

Thus, the acceleration of the first block (4 kg) is -9.8 m/s².

Learn more about net force on two connected blocks here: brainly.com/question/13539944

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3 years ago

An object in space is initially stationary relative to the Earth. Then, a force begins acting on the object, starting with a for

Thepotemich [5.8K]

Answer:

21741 s i believe

Explanation:

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Earth, has a bigger mass than the Moon .

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klasskru [66]

Answer with Explanation:

We are given that

Force acts on a pebble= $2\hat{i}-3\hat{k}$ N

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Torque= $r\times F$

Substitute the values then we get

$Torque= (0.5j-2k)\times (2i-3k)$

Torque= $-k-1.5i-4j$ N-m

By using $i\times j=k,j\times k=i,k\times i=j,j\times i=-k,k\times j=-i,i\times k=-j,i\times i=j\times j=k\times k=0$

b. $r=2i-3k$

$r-r_1=(0.5j-2k)-(2i-3k)=-2i+0.5j+k$

Torque about point (2,0,-3)

$\tau=(-2i+0.5j+k)\times (2i-3k)$

$\tau=-6j-k-1.5i+2j=-1.5i-4j-k$ N-m

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3 years ago