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Margaret [11]
3 years ago
14

Calculate the force needed to move a 2kg mass with an acceleration of 5ms-2

Physics
1 answer:
kap26 [50]3 years ago
7 0

Answer:

\Huge \boxed{\mathrm{10 \ N}}

Explanation:

\sf Force \ (N)=mass \ (kg) \cdot acceleration  \ (ms^{-2})

F=ma

The mass is 2 kg.

The acceleration is 5 ms⁻².

F = 2 \cdot 5

F=10

The force is 10 N.

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Consider a plywood square mounted on an axis that is perpendicular to the plane of the square and passes through the center of t
belka [17]

Answer:

T= 8.061N*m

Explanation:

The first thing to do is assume that the force is tangential to the square, so the torque is calculated as:

T = Fr

where F is the force, r the radius.

if we need the maximum torque we need the maximum radius, it means tha the radius is going to be the edge of the square.

Then, r is the distance between the edge and the center, so using the pythagorean theorem, r i equal to:

r = \sqrt{(0.38m)^2+(0.38m)^2}

r = 0.5374m

Finally, replacing the value of r and F, we get that the maximun torque is:

T = 15N(0.5374m)

T= 8.061N*m

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4 years ago
As gravity continues to crush the inner core, the fusion of carbon and oxygen begins. What temperature is needed for the fusion
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Answer:

600,000,000 degree C

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The most heavier atom that can be formed out of this fusion is the iron. The moment all the atoms becomes of iron, no further fusion is possible hence that body emits radiation of high intensity and collapse causing a big supernova.

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A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s
Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g=4.67\times 10^{-3} kg

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1,m_1=1177g=1.177kg

Mass of block 2,m_2=1626 g=1.626 kg

Velocity of block 1,v_1=0.681m/s

(a)

Let velocity of the second block  after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

mv=m_1v_1+(m+m_2)v_2

4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2

1.66719=0.801537+1.63067v_2

1.66719-0.801537=1.63067v_2

0.865653=1.63067v_2

v_2=\frac{0.865653}{1.63067}

v_2=0.531m/s

Hence, the  velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

K_i=\frac{1}{2}mv^2

k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)

k_i=297.59 J

Final kinetic energy after collision

K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2

K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2

K_f=0.5028 J

Now, he ratio of the total kinetic energy after the collision to that before the collision

=\frac{k_f}{k_i}=\frac{0.5028}{297.59}

=0.00169

5 0
3 years ago
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