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Reil [10]
3 years ago
12

two boxes sit on a frictionless surface and are in contact with one another. the first box has a mass of 7 kg and the second box

has a mass of 8 kg. A 25 N force is applied to the 7 kg box. Find the acceleration of the boxes. find the force between the boxes
Physics
1 answer:
egoroff_w [7]3 years ago
7 0
The acceleration of the boxes depends on the mass and weight. 

we have a mass of 7 and 8 kilograms

if it took 25 N force to move box A, then you would take 25 and multiply by 8 then divide by 2. 

It will leave you with 100 N. 

finally take the sq rt of 100 to get 10
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A 85 kg lineman tackles a 90 kg receiver. The receiver is running 5.8 m/s, and the lineman is moving 4.1 m/s, at a right angle t
weeeeeb [17]

Answer:

3.59 m/s

Explanation:

We are given that

Mass of lineman,m=85 kg

Mass of receiver,m'=90 kg

Speed of receiver,v'=5.8 m/s

Speed of lineman,v=4.1 m/s

\theta=90^{\circ}

We have to find the their velocity immediately after the tackle.

Initial momentum,P_i=\sqrt{p^2_1+p^2_2}=\sqrt{(85\times 4.1)^2+(90\times 5.8)^2}=627.6 kgm/s

According to law of conservation of momentum

Initial momentum=Final momentum=(m+m')V

627.6=(85+90)V=175V

V=\frac{627.6}{175}=3.59 m/s

3 0
3 years ago
The density of table sugar is 1.59 g/cm3. What is the volume of 7.85 g of sugar?
Likurg_2 [28]
\frac{7.85g }{1.59 \frac{g}{ cm^{3} } } = 4.937 cm^{3}
4 0
3 years ago
You kick a ball with a speed of 14 m/s at an angle of 51°. How far away does the ball land?
In-s [12.5K]
-- The vertical component of the ball's velocity is 14 sin(<span>51°) = 10.88 m/s

-- The acceleration of gravity is 9.8 m/s².

-- The ball rises for 10.88/9.8 seconds, then stops rising, and drops for the
same amount of time before it hits the ground.

-- Altogether, the ball is in the air for (2 x 10.88)/(9.8) = 2.22 seconds
==================================

-- The horizontal component of the ball's velocity is  14 cos(</span><span>51°) = 8.81 m/s

-- At this speed, it covers a horizontal distance of (8.81) x (2.22) = <em><u>19.56 meters</u></em>
before it hits the ground.


As usual when we're discussing this stuff, we completely ignore air resistance.
</span>
4 0
3 years ago
Read 2 more answers
What will happen if a car experiences a 300 N force to the right from the engine and a separate 150 N force due to friction and
gayaneshka [121]

Force applied on the car due to engine is given as

F_1 = 300 N towards right

Also there is a force on the car towards left due to air drag

F_2 = 150 N towards left

now the net force on the car will be given as

\vec F_{net} = \vec F_1 + \vec F_2

now we can say that since the two forces are here opposite in direction so here the vector sum of two forces will be the algebraic difference of the two forces.

So we can say

F_{net} = F_1 - F_2

F_{net} = 300 - 150

F_[net} = 150 N

So here net force on the car will be 150 N towards right and hence it will accelerate due to same force.

5 0
3 years ago
Action-reaction forces are not balanced forces because of what?
lara [203]

Oh but they are !

Newton's 3rd law of motion says that for every action, the <em><u>re</u></em>action is
equal and opposite.  That's as balanced as you can get.


4 0
3 years ago
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